Chapter 14 And 15- Genetics 61gh

Chapter 14 and 15 Genetics Due: 11:59pm on Wednesday, March 18, 2015 You will receive no credit for items you complete after the assignment is due. Grading Policy

Activity: Dihybrid Cross

Click here to complete this activity. Then answer the questions.

Part A What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.)

ANSWER: bbGg BBGG bbGG bbgg BbGg

Correct This result of the cross indicates that both orange eyes and white skin are recessive.

Part B Black eyes are dominant to orange eyes, and green skin is dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. Carole is a MendAlien with orange eyes and white skin. If Sam and Carole were to mate, the predicted phenotypic ratio of their offspring would be _____. ANSWER: There is insufficient information to determine Sam's genotype. 1 black eyes, green skin : 1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin 9 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 1 orange eyes, white skin 3 black eyes, green skin : 3 black eyes, white skin : 9 orange eyes, green skin : 1 orange eyes, white skin 1 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 9 orange eyes, white skin

Correct Sam's genotype is BbGg, and Carole's genotype is bbgg.

Part C In order to determine the genotype of a MendAlien with black eyes and green skin, you would cross this individual with a(n) _____ individual. ANSWER: BBGG BBgg BbGg bbgg bbGG

Correct This is an extension of the single character testcross you worked with in the activity on monohybrid crosses.

Part D A cross between an individual with orange eyes and green skin and an individual with black eyes and white skin is an example of a _____ cross. ANSWER: monohybrid test trihybrid difficult dihybrid

Correct The cross is examining two characters.

Part E A phenotypic ratio of 9:3:3:1 in the offspring of a cross indicates that _____. ANSWER: both parents are homozygous dominant both parents are heterozygous for both genes one parent is heterozygous and one parent is homozygous recessive one parent is homozygous dominant and one parent is heterozygous one parent is homozygous dominant and one parent is homozygous recessive

Correct Such a result indicates that the genes assort independently and that, for each gene, the alleles exhibit a dominant/recessive relationship.

Part F The observed distribution of alleles into gametes is an illustration of _____.

ANSWER:

Mendel's law of independent assortment only Mendel's law of segregation only Thienemann's rule Mendel's laws of segregation and independent assortment Eimer's principle

Correct The events seen here illustrate both the law of segregation and the law of independent assortment.

Part G An individual heterozygous for eye color, skin color, and number of eyes mates with an individual who is homozygous recessive for all three characters; what would be the expected phenotypic ratio of their offspring? [Hint: B = black eyes, b = orange eyes; G = green skin, g = white skin; C = two eyes, c = one eye] ANSWER: 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye : 1 orange eyes, green skin, two eyes : 1 orange eyes, green skin, one eye : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye 9 black eyes, green skin, two eyes : 3 black eyes, green skin, one eye : 3 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye 1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 orange eyes, green skin, two eyes : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye 9 black eyes, green skin, two eyes : 3 black eyes, green skin, one eye : 3 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye : 9 orange eyes, green skin, two eyes : 3 orange eyes, green skin, one eye : 3 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye 9 orange eyes, green skin, two eyes : 9 orange eyes, green skin, one eye : 9 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye

Correct This is a tough problem; you had to expand your Punnett square to accommodate another character. However, 1:1:1:1:1:1:1:1 ratio is the expected outcome of a BbGgCc x bbggcc cross.

Part H A BbGg x bbgg cross yields a phenotypic ratio of approximately 5 black eyes, green skin : 5 orange eyes, white skin : 1 black eyes, white skin : 1 orange eyes, green skin. Which of the following best explains these results? ANSWER: Mendel's law of independent assortment is being violated. The heterozygous individual is male, and the homozygous individual is female. Mendel's laws of segregation and independent assortment are being violated. The genes for eye color and skin color are codominant. Mendel's law of segregation is being violated.

Correct If the genes for eye color and skin color assorted independently, then the outcome of this cross would have been a 1:1:1:1 ratio.

Part I In the following cross the genotype of the female parent is BbGg. What is the genotype of the male parent? [Hint: B = black eyes, b = orange eyes, G = green skin, g = white skin]

ANSWER: BBGg BbGg BbGG bbGG BBGG

Correct All of the offspring have black eyes, and there is a 3:1 ratio of skin color.

Part J In a situation in which genes assort independently, what is the ratio of the gametes produced by an AaBB individual? ANSWER: 1 A : 1 B 3 AA : 1 BB 3 AB : 1 ab 1 AB : 1 aB 3 A : 1 B

Correct This is the correct ratio of gametes.

Activity: The Principle of Independent Assortment

Click here to view this animation. Then answer the questions.

Part A What process is responsible for the independent assortment of alleles?

Hint 1. How are traits inherited under the theory of blending inheritance? ANSWER: Mitosis Meiosis. Cytokinesis. DNA replication.

Correct The role of meiosis is to separate homologous chromosomes and their respective alleles, which are assorted in an independent manner.

Part B How do cells acquire homologous chromosome pairs that carry the alleles that are independently assorted? ANSWER: Fusion of gametes Meiosis DNA replication Mitosis

Correct During fertilization, gametes bring together homologous chromosomes to generate a diploid individual.

Part C Which of the following statements most accurately describes the process of independent assortment? ANSWER: Genes located on the same chromosome assort as an independent unit. Dominant alleles segregate with recessive alleles. Alleles found on separate chromosomes segregate based upon their origin. Alleles of different genes segregate from one another in a random manner.

Correct The random distribution of alleles is required for independent assortment.

Part D True or false? The principle of independent assortment is best illustrated by events that take place during metaphase II, during which sister chromatids segregate independently of each other.

Hint 1. Consider which meiotic division involves the segregation of alleles of different genes. ANSWER:

True False

Correct The principle of independent assortment is best illustrated by events that take place during metaphase I, during which nonhomologous chromosomes segregate independently of each other.

Part E How many genetically unique types of gametes could be produced by an individual with the genotype RrYY?

Hint 1. Assume that the alleles assort independently of each other during gamete formation. ANSWER: Two One Eight Three

Correct The gametes would be either RY or rY.

Part F Which of the following parental genotypes would yield a 1:1:1:1 phenotypic ratio in the offspring?

Hint 1. Use a Punnett square to diagram the genotypic and phenotypic ratios that would result from the crosses. ANSWER: aaBb, aabb AaBb, AaBb AaBb, aabb aaBb, AaBb

Correct The offspring of these parents would have a phenotypic ratio of 1:1:1:1.

Chapter 14 PreTest Question 1 Part A In his breeding experiments, Mendel first crossed truebreeding plants to produce a second generation, which were then allowed to selfpollinate to generate the offspring. How do we name these three generations?

Hint 1. P stands for parental, and F stands for filial.

ANSWER: P ... F1 ... F2 P1 ... P2 ... P3 F ... P1 ... P2 P1 ... P2 ... F F1 ... F2 ... F3

Correct Mendel started with truebreeding P generation plants, and in a typical experiment generated two subsequent generations, called F1 and F2.

Chapter 14 PreTest Question 2 Part A Which of the following is true about a plant with the genotype AABbcc?

Hint 1. Picture the homologous pairs of chromosomes on which these genes are located. ANSWER: It has recessive alleles at three loci. It is triploid. It is homozygous at two loci. It will not express the recessive c allele. It is heterozygous at two loci.

Correct An organism having a pair of identical alleles at a given locus is said to be homozygous for that locus. Here, the plant is homozygous at loci A and C.

Chapter 14 PreTest Question 3 Part A What is an allele?

Hint 1. How are genes related to alleles? What do loci have to do with it? ANSWER: the recessive form of a gene a variety of pea plant used by Mendel the dominant form of a gene an alternative version of a gene a type of chromosome

Correct A diploid organism carries two alleles for each autosomal gene. The two alleles are found at comparable locations (loci) on homologous chromosomes. The alleles may be identical or slightly different, but they affect the same genetic character.

Chapter 14 Question 2

Part A What was the most significant conclusion that Gregor Mendel drew from his experiments with pea plants? ANSWER: Recessive genes occur more frequently in the F1 generation than do dominant ones. Genes are composed of DNA. There is considerable genetic variation in garden peas. Traits are inherited in discrete units, and are not the results of "blending." An organism that is homozygous for many recessive traits is at a disadvantage.

Correct


Part A The individual with genotype AaBbCCDdEE can make many kinds of gametes. Which of the following is the major reason? ANSWER: recurrent mutations forming new alleles different possible assortment of chromosomes into gametes crossing over during prophase I segregation of maternal and paternal alleles the tendency for dominant alleles to segregate together

Correct


Part A Why did Mendel continue some of his experiments to the F2 or F3 generation? ANSWER: to observe whether or not the dominant trait would reappear to obtain a larger number of offspring on which to base statistics to be able to describe the frequency of recombination to observe whether or not a recessive trait would reappear to distinguish which alleles were segregating

Correct


Part A Which of the following differentiates between independent assortment and segregation? ANSWER: The law of segregation requires describing two or more genes relative to one another. The law of segregation requires having two or more generations to describe. The law of independent assortment is ed for by observations of prophase I. The law of independent assortment requires describing two or more genes relative to one another. The law of segregation is ed for by anaphase of mitosis.

Correct


Part A A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which of the following genotypes is possible in a gamete from this organism? ANSWER: Hh HhTt HT T tt

Correct


Part A Mendel ed for the observation that traits which had disappeared in the F1 generation reappeared in the F2 generation by proposing that ANSWER: the mechanism controlling the appearance of traits was different between the F1 and the F2 plants. new mutations were frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1. traits can be dominant or recessive, and the recessive traits were obscured by the dominant ones in the F1. of the F1 generation had only one allele for each trait, but  of the F2 had two alleles for each trait. the traits were lost in the F1 due to dominance of the parental traits.

Correct


Part A Use the figure and the following description to answer the question below. In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A truebreeding darkleaved plant is crossed with a lightleaved one, and the F1 offspring is allowed to self pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.

Which of the boxes marked 14 correspond to plants with dark leaves? ANSWER: 2 and 3 1 only 1 and 2 4 only 1, 2, and 3

Correct



Which of the boxes correspond to plants with a heterozygous genotype? ANSWER:

1 and 2 1 1, 2, and 3 2 and 3 2, 3, and 4

Correct



Which of the plants will be truebreeding? ANSWER: 1, 2, 3, and 4 1 and 4 only 2 and 3 only 1 and 2 only 1 only

Correct

Mendel's Law of Segregation This tutorial introduces Mendel’s law of segregation of alleles using monohybrid crosses, the addition and multiplication rules, conditional probability, and binomial probability. In this tutorial you will investigate how alleles segregate during meiosis. You will also explore how this segregation, in conjunction with random gamete fusion, allows geneticists to predict the outcome of genetic crosses using simple rules of probability.

Part A Allele segregation and gamete formation One character in peas that Mendel studied was yellow versus green seeds. A cross between a homozygous yellow line (YY) and a homozygous green line (yy) will result in F1 plants that are heterozygous (Yy) for this trait and produce yellow seeds.

When an F1 plant undergoes meiosis, what gamete types will it produce, and in what proportions? Use your understanding of Mendel’s law (or principle) of segregation to label each gamete possibility as “correct” or “incorrect.”

Hint 1. How do alleles segregate during gamete formation? According to Mendel’s law of segregation, a gamete ANSWER: always receives both of the parent’s alleles for each gene. always receives only one of the parent’s alleles for each gene. may receive either one or two of the parent’s alleles for each gene.

Hint 2. Does allele segregation determine gamete frequencies (i.e., proportions)? According to Mendel’s law of segregation, ANSWER: dominant alleles segregate into gametes more frequently than recessive alleles. allele frequencies (i.e., proportions) in gametes cannot be predicted in advance. alleles segregate into gametes with equal frequency.

ANSWER:

Correct Mendel’s law of segregation states that allele pairs segregate equally into gametes during meiosis. This means that a gamete will have only one allele of any given gene, and that the probability of a gamete having one allele or the other is equal (and therefore ½, or 50%, for either allele).

Part B Punnett square Punnett squares are convenient ways to represent the types and frequencies of gametes and progeny in experimental crosses. This Punnett square shows the results of a Yy x Yy cross to form F2 progeny. Use your understanding of Mendel’s law of segregation and the rules of probability to complete the Punnett square for this cross. First identify the gametes. Use pink labels to identify the male and female gamete types and white labels to identify the gamete frequencies. Then identify the F2 progeny. Use pink labels to identify the progeny genotypes and white labels to identify the progeny frequencies.

Hint 1. Setting up a Punnett square To create a Punnett square, list the possible gametes from one parent down one side and the possible gametes from the other parent across the top. Each gamete is assigned a frequency according to Mendel’s law of segregation (as explained in Part A). The boxes inside the Punnett square represent the genotypes of the progeny that result from random gamete fusion. Progeny frequencies are then calculated using the frequencies of the gametes that formed them. (See Hint 2.)

Hint 2. How do I calculate the frequencies of the F2 progeny? Which statement correctly explains how to determine the frequencies of the F2 progeny? ANSWER: The progeny frequencies are determined by the addition rule (adding the gamete frequencies together). The progeny frequencies are determined by the multiplication rule (multiplying the gamete frequencies together). The progeny frequencies cannot be predicted from gamete frequencies.

ANSWER:

Correct A Punnett square is a convenient method for representing Mendel’s law of segregation in a visual form. Using a Punnett square allows one to easily see gamete types and frequencies, as well as the genotypes and frequencies of progeny formed by random gamete fusion. The genotype frequencies inside a Punnett square are calculated using the multiplication rule: The probability of two independent events occurring simultaneously is the product of their individual probabilities. In this example, the genotype frequencies inside the square (¼) are the product of the gamete frequencies that led to their formation (½ x ½ = ¼).

Part C Using the Punnett square to make predictions Use the completed Punnett square in Part B to answer the questions below about the F2 generation. Answer each question in the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

Hint 1. How to choose between the addition rule and the multiplication rule The addition (or sum) rule is used when there is more than one way to fulfill the requirements of the event of interest. Often the word “or” acts as a hint that the addition rule is needed. For example, you might say, “Yy individuals in the F2 can be formed by either a Y from the female parent and a y from the male parent OR a y from the female parent and a Y from the male parent." This indicates that there are two ways to fulfill the requirement of “an F2 progeny with genotype Yy” in this case, and tells you that the addition rule should be used. The multiplication (or product) rule is used to determine the probability of simultaneous events with individual probabilities. Often the word “and” acts as a hint for the multiplication rule. For example, you might say “YY individuals in the F2 require a Y from the female parent AND a Y from the male parent." In this case, there is only one way to achieve the desired event. Therefore, the addition rule is not needed, and the probability of the YY genotype in the F2 is determined with the multiplication rule alone.

Hint 2. For question 2, how do you  for conditional probabilities? Conditional probabilities apply when the pool of possible outcomes is reduced in some way by additional criteria. For example, determining probabilities for genotypes within only the yellow seeds in the F2 restricts the total number of possibilities because it excludes green seeds by definition. Use your knowledge of conditional probabilities and the Punnett square in Part B to answer this question. What is the probability that an F2 seed chosen at random from among only the yellow seeds will have a genotype of Yy? ANSWER:

1/2 2/3 3/4

Hint 3. How to approach questions 3 and 4 For questions 3 and 4, you need to consider all possible groupings and orders of three F2 seeds and their probabilities (determined using the multiplication rule). The table below shows all the possible groupings of three F2 seeds. Two of the probabilities have been filled in for you. Complete the table by calculating all of the other probabilities.

Once you’ve filled in the table, you will be able to calculate the probability of any defined outcome, employing the multiplication and addition rules as needed. (See Hint 1.) For example, to determine the probability of getting two green seeds and one yellow seed in a threeseed grouping, there are three groupings that qualify: Yellow, Green, Green Green, Yellow, Green Green, Green, Yellow After you've calculated the individual probabilities of these three groupings (using the multiplication rule) you can then employ the addition rule to calculate their combined probability (because any of these three outcomes meets the stated criterion of having one yellow seed and two green seeds). ANSWER:

Correct Question 1 simply asks about the probability of forming a yellow seed in the F2, which is 3/4. Question 2 introduces the concept of conditional probability. Although there are two genotypes in the F2 that will breed true (YY and yy), only one of them is yellow. Thus, the question adds a new condition: Among yellow seeds only, 1/3 are truebreeding. Questions 3 and 4 require a full evaluation of all eight possible groupings of three randomly selected F2 seeds. Each grouping has its own probability, shown in the table below. For question 3, there are seven possible groupings that have at least one yellow seed (shown in blue in the table); the sum of their individual probabilities yields the final answer (63/64). Note that you could have also calculated this answer by subtracting the probability of the one category that doesn’t fit the criterion (Green, Green, Green) from the total probability (1 – 1/64 = 63/64). For question 4, there are three possible groupings that have one green and two yellow seeds (shown in bold in the table); the sum of their individual probabilities yields the final answer (27/64).

Activity: Gregor's Garden


Part A Mendel worked _____. ANSWER: in a monastery on gene mapping with fruit flies on the Human Genome Project on problems involving characters that were inherited together more often than expected by chance

Correct Mendel was a monk who worked in his monastery garden.

Chapter 14 PreTest Question 6 Part A You cross a truebreeding redflowered snapdragon with a truebreeding whiteflowered one. All of the F1 are pink. What does this say about the parental traits?

Hint 1. Consider where pink falls relative to red and white. ANSWER: Red shows incomplete dominance over white. Both red and white are pleiotropic. Pink is dominant, and red and white are recessive. Red is completely dominant.

Correct Red shows incomplete dominance over white, and the F1 is therefore pink.

Chapter 14 PreTest Question 7 Part A Height in humans generally shows a normal (bellshaped) distribution. What type of inheritance most likely determines height?

Hint 1. Is height a strictly Mendelian trait? ANSWER: a combination of complete dominance and environmental factors a combination of epistasis and environmental factors a combination of polygenic inheritance and environmental factors incomplete dominance a combination of multiple alleles and codominance

Correct Several genes (polygenic inheritance) control height in humans, giving an overall normal distribution. Environmental factors such as nutrition smooth out the curve.


Part A In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white? ANSWER:

red × white red × roan roan × roan white × roan The answer cannot be determined from the information provided.

Correct


Part A Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following  best describes this? ANSWER: incomplete dominance epistasis multiple alleles codominance pleiotropy

Correct


Part A Which of the following is an example of polygenic inheritance? ANSWER: Huntington's disease in humans pink flowers in snapdragons skin pigmentation in humans the ABO blood group in humans white and purple flower color in peas

Correct


Part A Use the following information to answer the question below. Radish flowers may be red, purple, or white. A cross between a redflowered plant and a whiteflowered plant yields allpurple offspring. The part of the radish we eat may be oval or long, with long being the dominant trait.

The flower color trait in radishes is an example of which of the following? ANSWER: sex linkage codominance a multiple allelic system epistasis incomplete dominance

Correct


Part A Use the following information to answer the question below. Skin color in a certain species of fish is inherited via a single gene with four different alleles.

How many different types of gametes would be possible in this system? ANSWER: 4 2 1 16 8

Correct


Part A Use the following information to answer the question below. Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is incompletely dominant to white (r). The heterozygous condition results in pink (Rr) flower color.

If snapdragons are heterozygous for height as well as for flower color, a mating between them will result in what ratio? ANSWER: 9:3:3:1 9:4:3 6:3:3:2:1:1 1:2:1 27:9:9:9:3:3:3:1

Correct


Part A Which of the following provides an example of epistasis? ANSWER: Recessive genotypes for each of two genes (aabb) results in an albino corn snake. In rabbits and many other mammals, one genotype (ee) prevents any fur color from developing. In cacti, there are several genes for the type of spines. In Drosophila (fruit flies), white eyes can be due to an Xlinked gene or to a combination of other genes. The allele b17 produces a dominant phenotype, although b1 through b16 do not.

Correct


Part A Refer to the following information to answer the question below. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.

A cross between a truebreeding sharpspined cactus and a spineless cactus would produce ANSWER: 25% sharpspined, 50% dullspined, 25% spineless progeny. all sharpspined progeny. all spineless progeny. 50% sharpspined, 50% dullspined progeny. It is impossible to determine the phenotypes of the progeny.

Correct


Part A Use the following information to answer the question below. A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative and is designated by the alleles R and r, respectively. A third gene for the MN blood group has codominant alleles M and N.

Which of the following is a possible partial genotype for the son? ANSWER:

IAIA ii IBIA IBi IBIB

Correct



Which of the following is a possible genotype for the mother? ANSWER: IAi ii IBIB IAIB IAIA

Correct



Which of the following is a possible phenotype for the father? ANSWER: AB negative O negative A negative B positive impossible to determine

Correct

Incomplete Dominance and Codominance You are studying leaf development in a member of the mustard family. You identify several mutants of interest in this plant and make pure (truebreeding) lines of each mutant for further study.

Part A Determining relationships between alleles You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure wildtype line. The numbers in the F2 indicate the number of progeny in each phenotypic class.

From these results, determine the relationship between the mutant allele and its corresponding wildtype allele in each line. Label each mutant line with the best statement from the list below. Labels may be used once, more than once, or not at all.

Hint 1. How can I tell if an allele is dominant or recessive? You have obtained several color variants of a berry that has a black wildtype phenotype. You have also determined that all the variants are due to different alleles of the same gene. Crosses between pure lines of the variants and wild type produce the following results:

Based on these results, select the four true statements from the list below. Check all that apply. ANSWER:

The blue allele is recessive to the wildtype allele. The blue allele is dominant. The green allele is dominant to the wildtype allele. The blue allele is dominant to the green allele. The green allele is dominant to the blue allele. The green allele is recessive to the blue allele. The green allele is recessive to the wildtype allele. The green allele is recessive. The blue allele is recessive. The blue allele is dominant to the wildtype allele. The green allele is dominant. The blue allele is recessive to the green allele.

Hint 2. How to determine allele relationships in cases with intermediate F1 phenotypes Some heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes. This situation is the result of incomplete dominance: Neither allele is completely dominant to the other. ANSWER:

Correct An allele is never intrinsically "dominant" or "recessive." Instead, these  describe a relationship between two alleles. This relationship is evaluated by examining a heterozygote: The allele that determines the phenotype of the heterozygote is dominant to the other (recessive) allele. Some heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes. This situation is the result of incomplete dominance: Neither allele is completely dominant to the other.

Part B Crossing the forked and pale mutants You continue your genetic analysis by crossing the forked and pale mutant lines with each other. The leaves of the F1 are light green (intermediate between pale and wildtype leaves) and forked. The F2 has six phenotypic classes, as shown below. You designate the forked mutant allele as F (wild type = f+ ) and the pale mutant allele as p (wild type = P). 1. Consider the alleles for leaf color first. Drag the white labels to the white targets to identify the genotype of each F2 class. that p (the pale mutant allele) and P (the wildtype allele) are incompletely dominant to each other. 2. Consider the alleles for leaf shape next. Drag the blue labels to the blue targets to identify the genotype of each F2 class.

that F (the forked mutant allele) is dominant to f + (the wildtype allele). Labels may be used once, more than once, or not at all. For help getting started, see the hints.

Hint 1. Asg genotypes for alleles that exhibit complete dominance In cases where an allele is completely dominant to another allele, only two phenotypic categories are present in a monohybrid cross: the dominant phenotype and the recessive one. Individuals with the dominant phenotype may have either of two genotypes: homozygous dominant or heterozygous. Because progeny in this category cannot be assigned a genotype on the basis of their phenotype alone, their genotypes are often summarized by writing the dominant allele followed by an underscore ( _ ), which indicates that either the dominant or the recessive allele may be present at the second position.

Hint 2. How do I assign genotypes for alleles that exhibit incomplete dominance? Select the two phrases that correctly complete this sentence: In a monohybrid cross, incomplete dominance results in ANSWER: an F1 that matches the phenotype of one of the parental lines. an F1 that has an intermediate phenotype unlike that of either parental line. three phenotypic categories in the F2: homozygous dominant, heterozygous, and homozygous recessive progeny all have distinct phenotypes. two phenotypic categories in the F2: homozygous dominant and heterozygous progeny have the same phenotype, and homozygous recessive progeny have a different phenotype.

ANSWER:

Correct Alleles P and p are incompletely dominant to each other. Therefore, each genotype has a distinct phenotype. That is why you are able to assign definite genotypes for leaf color to each F2 plant. Allele F is dominant to f +. Therefore, you cannot be certain whether forked leaves are homozygous dominant or heterozygous. That is why you must assign the genotype F_ to the F2 plants in the top row. For the unforked leaves in the bottom row, you know they must be homozygous for the wild type allele.

Part C Crossing the forked and twist mutants You continue your analysis by crossing the forked and twist lines. Your results are as follows:

Which of the following statements best explains the outcome of this cross?

Hint 1. Distinguishing between incomplete dominance, codominance, and recombination Incomplete dominance and codominance apply to alleles at one locus. Recall that incomplete dominance means that a heterozygote has an intermediate phenotype. Codominance occurs when both alleles at a single locus express their phenotype simultaneously. Recombination applies to alleles at different loci. It occurs in a dihybrid when an allele at one locus is placed into a gamete with an allele at a second locus in a combination not present in either parental line.

Hint 2. Why are there no wildtype progeny in the F2 of this cross? Why are there no wildtype progeny in the F2 of this cross? Check all that apply. ANSWER: because there are no wildtype forked alleles in either parental line because there are no wildtype forked alleles in the F1 plants because the wildtype forked alleles in the F2 are hidden by recombination

ANSWER: The forked mutation is incompletely dominant to the twist mutation. The forked mutation has recombined with the twist mutation. The forked mutation and the twist mutation are codominant alleles of the same locus. The twist mutation is incompletely dominant to the forked mutation.

Correct The absence of wildtype progeny in the F2 of a cross indicates a monohybrid cross: The two lines crossed are mutant at the same locus. Any other explanation that involves two loci (for example, recombination) fails because any cross with two loci will produce at least some F2 individuals with a wildtype genotype for both loci and therefore a wildtype phenotype. In this case, the F1 expresses both mutant phenotypes, indicating that the forked and twist alleles are codominant.

Part D Asg genotypes for codominant alleles You decide to designate the twist allele as FT to distinguish it from the forked allele F. Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels may be used once, more than once, or not at all.

Hint 1. Asg genotypes in cases involving codominant alleles Asg genotypes in cases of codominance is very straightforward because a codominant allele will express its phenotype even in heterozygotes. Individuals with both phenotypes are heterozygous; individuals with only one phenotype are homozygous for the corresponding allele. ANSWER:

Correct Like incomplete dominance, codominance produces a distinct phenotype for each of the three genotypes in a monohybrid cross. Unlike incomplete dominance, however, codominance results in both alleles fully expressing their phenotype in the heterozygote.

Chapter 14 PreTest Question 8 Part A Folk singer Woody Guthrie died of Huntington's disease, an autosomal dominant disorder. Which statement below must be true?

Hint 1. Draw a pedigree including Guthrie, his parents, and his children. ANSWER: All of his children will develop Huntington's disease. It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease. His daughters will die of Huntington's disease but not his sons. His sons will develop Huntington's disease but not his daughters. There is not enough information to answer the question.

Correct Unless the disease is caused by a new mutation, which is quite rare, individuals with a dominant condition must have inherited the dominant allele from one of their parents. As it happens, Guthrie's mother also died of Huntington's disease.

Chapter 14 PreTest Question 9 Part A Various procedures can be used to detect genetic disorders before birth. Among the tests discussed in this chapter, which is the least invasive (list first), and which two allow the chromosomes of the fetus to be examined?

Hint 1. Which techniques allow actual cells to be examined? ANSWER: amniocentesis ... ultrasound imaging ... chorionic villus sampling ultrasound imaging ... fetoscopy... amniocentesis amniocentesis ... fetoscopy ... chorionic villus sampling ultrasound imaging ... chorionic villus sampling ... amniocentesis chorionic villus sampling ... fetoscopy ... amniocentesis

Correct Both CVS and amniocentesis involve the preparation of a karyotype using fetal cells.

Chapter 14 PreTest Question 10 Part A Which of the following do you expect if an individual is heterozygous for the sicklecell trait?

Hint 1. Think about the term "heterozygote advantage." ANSWER: He or she will be more apt to acquire a serious case of malaria. He or she will have fullblown sicklecell disease because the allele is dominant. He or she will not develop sicklecell disease. He or she will show some symptoms of the disease. None of the above.

Correct A heterozygote will produce both normal and abnormal hemoglobin.


Part A An obstetrician knows that one of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is detectable biochemically in fetal cells. The obstetrician would most reasonably offer which of the following procedures to her patient? ANSWER: Xray ultrasound imaging amniocentesis or CVS karyotyping of the woman’s somatic cells blood transfusion

Correct


Part A The frequency of heterozygosity for the sicklecell anemia allele is unusually high, presumably because this reduces the frequency of malaria. Such a relationship is related to which of the following? ANSWER: Darwin's explanation of natural selection Darwin's observations of competition Mendel's law of independent assortment the malarial parasite changing the allele Mendel's law of segregation

Correct


Part A Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Therefore, the most efficient and effective treatment is which of the following? ANSWER: Regulate the diet of the affected persons to severely limit the uptake of the amino acid. Feed them the substrate that can be metabolized into this amino acid. Feed the patients the missing enzymes in a regular cycle, such as twice per week. Feed the patients an excess of the missing product. Transfuse the patients with blood from unaffected donors.

Correct


Part A One of two major forms of a human condition called neurofibromatosis (NF 1) is inherited as a dominant gene, although it may range from mildly to very severely expressed. If a young child is the first in her family to be diagnosed, which of the following is the best explanation? ANSWER: The child has a different allele of the gene than the parents. One of the parents has very mild expression of the gene. The mother carries the gene but does not express it at all. The condition skipped a generation in the family.

Correct


Part A The following question refers to the pedigree chart in the figure for a family, some of whose  exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

What is the genotype of individual II5? ANSWER: WW or ww WW ww ww or Ww Ww

Correct


Part A Use the following pedigree (the figure) for a family in which darkshaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.

From this pedigree, how does this trait seem to be inherited? ANSWER:

as an autosomal dominant as an autosomal recessive from mothers as a result of epistasis as an incomplete dominant

Correct


Part A Use the following pedigree (the figure) for a family in which darkshaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis.

In each generation of this family after generation I, the age at diagnosis is significantly lower than would be found in nonfamilial (sporadic) cases of this cancer (~ 63 years). What is the most likely reason? ANSWER: This is pure chance; it would not be expected if you were to look at a different family. This cancer requires mutations in more than this one gene. Affected  of this family are born with colon cancer, and it can be detected whenever they are first tested. of this family know to be checked for colon cancer early in life. Hereditary (or familial) cases of this cancer typically occur at earlier ages than do nonfamilial forms.

Correct

Pedigree Analysis: Galactosemia Galactosemia is a human genetic disorder caused by a deficiency of the enzyme galactose1phosphate uridylyltransferase (commonly called GALT). Children with galactosemia seem normal at birth, but begin to show symptoms as soon as they drink milk. This is because they cannot metabolize galactose, a simple sugar that results from the breakdown of lactose (the sugar found in milk). Early symptoms of galactosemia include vomiting and diarrhea. If untreated, affected babies will develop enlarged livers, cataracts, and mental retardation. Depending on the alleles the baby has inherited, early death may result.

Part A Construct a pedigree John and Jane Jones’ newborn baby girl, Leah, has just been diagnosed with galactosemia. Prior to their genetic counseling appointment, they submitted this family history: John and Jane’s older child, a son named Lee, does not have galactosemia. John is the only child of Hanna and Harry. Harry was an only child. Hanna has two older sisters, Hope (the oldest) and Holly. Both Harry and Hanna’s parents lived in good health into their 80s. Jane’s brother, Joe, is married to Jen. They have a son, Les, who is a healthy nineyearold with myopia (nearsightedness) but no sign of galactosemia.

Jane’s mother, Hilda, is alive and healthy; Jane’s father, Henry, ed away last year of a sudden heart attack. Hilda and Henry have no family history of galactosemia. Use the information in the family history to construct a pedigree for this family. Drag the correct symbols to their appropriate locations on the pedigree below. Not all symbols will be used. See the Hints for help getting started.

Hint 1. Can you identify the pedigree symbols? Drag the correct label under each pedigree symbol below. Labels can be used once or not at all. ANSWER:

Hint 2. How to approach this problem To construct a pedigree, you need to do the following: 1. Make sure to place each family member in the correct generation, and in the correct family relationship within the generation. 2. Use the correct symbols to accurately show each person’s sex and status as a couple, child, or sibling. 3. Use shaded symbols for people with the disorder and unshaded symbols for people who do not have the disorder. For this family, begin with John and Jane as a couple; then add their children, Lee and Leah. Next, add John’s parents, Harry and Hanna, and Hanna’s two sisters. Then add Jane’s brother Joe, his wife Jen, their son Les, and Jane’s parents, Hilda and Henry. ANSWER:

Correct A pedigree provides a pictorial representation of the genetic relationships in a family and shows each person’s sex and phenotype with respect to the trait in question. In this family, Leah is the only family member (on both sides) that has galactosemia.

Part B Identify the mode of inheritance Look over the pedigree you constructed in Part A. Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia?

Hint 1. Definitions of autosomal, sexlinked, dominant, and recessive autosomal gene: a gene that is not located on a sex chromosome (X or Y chromosome). sexlinked gene: a gene located on a sex chromosome (usually the X chromosome) for which there is no corresponding gene on the other sex chromosome (usually the Y chromosome). dominant: a phenotype that is fully expressed in a heterozygote. recessive: a phenotype that is not expressed in a heterozygote.

Hint 2. Can you identify inheritance patterns from pedigrees? Drag the appropriate label under each pedigree to correctly identify the most likely inheritance pattern it shows. ANSWER:

ANSWER: autosomal recessive sexlinked recessive sexlinked dominant autosomal dominant

Correct Because neither Jane nor John has the same condition as their daughter, and there is no evidence of sexlinkage, galactosemia must be an autosomal recessive trait.

Part C Predicting the probability of inheritance Jane and John are considering having another child. Given the pedigree you constructed and the mode of inheritance for galactosemia, what is the risk that their next child will have the disorder?

Hint 1. What are the genotypes of the parents? What are Jane and John’s genotypes? (Use G for the normal allele and g for the allele for galactosemia.) ANSWER: Jane is gg and John is Gg. Jane is GG and John is gg. Jane is Gg and John is gg. Both Jane and John are Gg.

Hint 2. Can you construct a Punnett square for this problem? Complete this Punnett square to determine the chance that Jane and John’s child will have galactosemia. (Use G for the normal allele and g for the allele for galactosemia.) ANSWER:

ANSWER: 1/16 1/4 0

(because they each have 1/4 chance of ing on the allele)

(because they are both heterozygotes)

(because they already have one child with the disorder)

3/4

(because they are both heterozygotes)

Correct A Punnett square for two heterozygous (Gg) parents reveals that each child would have a 1/4 chance of receiving the gg allele combination, resulting in galactosemia.

Offspring genotype probabilities: 1/4 GG; 1/2 Gg; 1/4 gg

Part D Genetic testing If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactose and galactosefree diet as soon as possible after birth. Even on this diet, affected individuals may still suffer from learning disabilities, ovarian failure (in young women), lateonset cataracts, and early death. Which of the following tests would be most useful for Jane and John to have?

Hint 1. When and on whom are genetic tests performed?

Drag the  on the left to the appropriate blanks on the right to complete the sentences. ANSWER:

ANSWER: no testing newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn’s blood) fetal chromosome analysis (examining fetal chromosomes for abnormalities as early as 11 weeks into the pregnancy) carrier screening of Jane and John (to identify whether either or both of them carry the allele for galactosemia)

Correct Newborn screening by either of these methods would be effective in identifying the condition in a newborn. It would provide definitive information early enough to begin dietary restrictions, if needed. Fetal testing for this genetic mutation does not make sense because it carries greater risk than newborn screening, and there is no treatment prior to birth.

ABC News Video: The Genetics of Short Sleepers Watch the ABC News video (1:53 minutes). Then answer the questions below.

Part A What defines a short sleeper? ANSWER: They are highly productive throughout the day. They are in excellent health. They can function well on 6 hours of sleep or less. They are able to bank longer sleep hours from the weekend as savings for their short sleep during the week.

Correct

Part B While it is estimated that less than 5% of the population are short sleepers, ANSWER: less than half of short sleepers carry the genetic mutation. an estimated 30% of the population sleep only 6 hours per night or less. very few short sleepers lead healthy lifestyles. the majority of short sleepers are obese.

Correct

Part C The risks of sleep deprivation include which of the following? ANSWER: obesity cardiovascular disease memory problems all of the above

Correct

Part D The discovery of a gene associated with short sleepers means that ANSWER: most people don't need 8 hours of sleep per night. there is potential to manipulate sleep pathways with medication. a person's physical environment has no effect on their sleep patterns. humans may eventually need less sleep.

Correct

Part E Since there is a genetic explanation for short sleepers, you would expect ANSWER: that short sleepers have higher metabolisms. a higher percentage of women to be short sleepers. that people can learn to be short sleepers. that this trait appears more frequently among family .

Correct

Misconception Question 66

Part A Quantitative characters vary in a population along a continuum. How do such characters differ from the characters investigated by Mendel in his experiments on peas? ANSWER: The nature of inheritance of quantitative characters is poorly understood, and Mendel understood the nature of inheritance for the characters he studied in his peas. Quantitative characters are due to polygenic inheritance, the additive effects of two or more genes on a single phenotypic character. A single gene affected all but one of the pea characters studied by Mendel. Environment and genes affect quantitative characters, whereas only genes determined the pea characters studied by Mendel.

Correct Read about Mendel’s experiments and about quantitative characters.


Part A

Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellowround seeds (YYRR) and a plant with greenwrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation. ANSWER: The chance that an individual taken at random from the F2 generation produces wrinkled seeds is 25% and the chance that the same individual produces yellow seeds is 75%. In a set of 16 individuals from the F2 generation, 3 will produce wrinkled yellow seeds. Wrinkled and yellow traits are linked in peas.

Correct Read about the Law of Independent Assortment and about how to solve complex genetics problems using the rules of probability.


Part A

Each chromosome in this homologous pair possesses a different allele for flower color. Which statement about this homologous pair of chromosomes is correct? ANSWER: These homologous chromosomes are formed by DNA replication. These homologous chromosomes represent a maternal and a paternal chromosome. Each of these homologous chromosomes consists of a single chromatid. Therefore, they must come from a haploid cell.

Correct Homologous chromosomes initially come together during fertilization with the contribution of a set of chromosomes from the sperm and the egg.


Part A When a dominant allele coexists with a recessive allele in a heterozygote individual, how do they interact with each other? ANSWER: The dominant allele prevents expression of the recessive allele by an unknown mechanism. The dominant allele prevents transcription of the recessive allele. They do not interact at all.

Correct An allele is called dominant because it appears in the phenotype of a heterozygote, not because it subdues a recessive allele. Read about dominance.

Chapter 15 PreTest Question 1 Part A What can we observe in order to visualize Mendel's Law of Segregation?

Hint 1. What are genes physically, and how do they behave during meiosis? ANSWER: homologous chromosomes separating during meiosis I the replication of DNA homologous chromosomes separating during meiosis II sister chromatids separating during mitosis the behavior of sexlinked genes

Correct Homologous chromosomes each carry an allele for every gene. Those alleles may be the same or different, but they are segregated during meiosis I and distributed to each gamete.

Chapter 15 PreTest Question 2 Part A What name is given to the most common phenotype in a natural population?

Hint 1. Where are you most likely to find a natural population? ANSWER:

genotype wild type mutant phenotype autosome locus

Correct The most common phenotype is the wild type.

Chapter 15 PreTest Question 3 Part A A whiteeyed female Drosophila is crossed with a redeyed male Drosophila. Which statement below correctly describes the results?

Hint 1. Red eyes is the wildtype trait. ANSWER: Fifty percent of the females will have red eyes. Twentyfive percent of the females will have white eyes. Twentyfive percent of the females will have red eyes. None of the females will have red eyes. None of the females will have white eyes.

Correct All of the females will have red eyes because they will inherit the dominant redeye allele from their male parent.


Part A Sturtevant provided genetic evidence for the existence of four pairs of chromosomes in Drosophila in which of these ways? ANSWER: The entire Drosophila genome has approximately 400 map units. There are four major functional classes of genes in Drosophila. Drosophila genes have, on average, four different alleles. Drosophila genes cluster into four distinct groups of linked genes. The overall number of genes in Drosophila is a multiple of four.

Correct


Part A Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early 20th century? ANSWER: Homologous chromosomes give rise to some genes and crossover chromosomes to other genes. No more than a single pair of chromosomes can be found in a healthy normal cell. Natural selection acts on certain chromosome arrays rather than on genes. Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis. Individuals inherit particular chromosomes attached to genes.

Correct


Part A A woman is found to have 47 chromosomes, including three X chromosomes. Which of the following describes her expected phenotype? ANSWER: masculine characteristics such as facial hair healthy female of slightly aboveaverage height sterile female excessive emotional instability enlarged genital structures

Correct

Chapter 15 PreTest Question 4 Part A In humans, what determines the sex of offspring and why?

Hint 1. Consider how sex is determined in humans and how these chromosomes are assorted during gamete formation. ANSWER: The chromosome contribution from both parents determines sex because the offspring uses all the parents' chromosomes. The female gamete determines sex because only the female gamete provides cytoplasm to the zygote. The male determines sex because the sperm can fertilize either a female egg or a male egg. The female gamete determines sex because only the female gametes can have one of two functional sex chromosomes. The male gamete determines sex because each male gamete can contribute either an X or a Y chromosome.

Correct In humans, the male produces both X and Ybearing sperm, whereas females only produce Xbearing eggs.

Chapter 15 PreTest Question 10

Part A Which of the following is true of an Xlinked gene, but not of a Ylinked gene?

Hint 1. Review the differences between X and Y chromosomes ANSWER: It does not segregate like other genes. It is expressed in half of the cells of either male or female. The gene is present in both males and females. Sister chromatids separate during mitosis. It is only expressed in female offspring.

Correct All cells have at least one functional X chromosome.


Part A SRY is best described in which of the following ways? ANSWER: an autosomal gene that is required for the expression of genes on the X chromosome a gene required for development, and males or females lacking the gene do not survive past early childhood an autosomal gene that is required for the expression of genes on the Y chromosome a gene region present on the Y chromosome that triggers male development a gene present on the X chromosome that triggers female development

Correct


Part A In cats, black fur color is caused by an Xlinked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male? ANSWER: orange females; black males tortoiseshell females; tortoiseshell males tortoiseshell females; black males black females; orange males orange females; orange males

Correct


Part A Redgreen color blindness is a sexlinked recessive trait in humans. Two people with normal color vision have a colorblind son. What are the genotypes of the parents? ANSWER: XNXn and XNY XNXN and XnY XnXn and XnY XnXn and XNY XNXN and XNY

Correct


Part A Normally, only female cats have the tortoiseshell phenotype because ANSWER: a male inherits only one allele of the Xlinked gene controlling hair color. the Y chromosome has a gene blocking orange coloration. only males can have Barr bodies. the males die during embryonic development. multiple crossovers on the Y chromosome prevent orange pigment production.

Correct


Part A In humans, clear gender differentiation occurs, not at fertilization, but after the second month of gestation. What is the first event of this differentiation? ANSWER: activation of SRY in male embryos and masculinization of the gonads anatomical differentiation of a penis in male embryos activation of SRY in females and feminization of the gonads formation of estrogens in female embryos formation of testosterone in male embryos

Correct


Part A Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition? ANSWER: Onehalf of the daughters of an affected man would have this condition. Onefourth of the daughters of an affected father and a carrier mother could have this condition. Only if a woman is XXX could she have this condition. Very rarely: it is rare that an affected male would mate with a carrier female. Women can never have this condition.

Correct


Part A Refer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a colorblind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and redgreen color blindness is Xlinked recessive.

What proportion of their sons would be colorblind and of normal height? ANSWER: half all three out of four none one out of four

Correct

Pedigrees and Sex Linkage In this tutorial, you will use pedigrees to explore the differences between Xlinked and autosomal inheritance patterns.

Part A Identifying Xlinked inheritance in pedigrees You are investigating the inheritance of two rare conditions (A and B) in an extended family of thoroughbred racehorses. You have constructed the following pedigree for these conditions.

Identify the inheritance mode of each condition by using the labels to complete the table. Labels can be used once, more than once, or not at all.

Hint 1. Pedigree symbols and numbering Pedigrees use standard symbols and labels to depict family trees. Circles represent females, squares represent males, and diamonds represent individuals of unspecified sex (often, unborn offspring). Filled circles and squares represent individuals who are affected by the condition being traced; open circles and squares represent unaffected individuals. (Diamonds are usually open, even though the individuals they represent may or may not be affected.) Two individuals connected by a horizontal line are parents. Two individuals connected by a double horizontal line are parents who are related to each other. (This is called a “consanguineous” pairing.) Individuals connected by lines from above are siblings. Generations are represented by Roman numerals; individuals within each generation are represented by Arabic numerals. (The Roman and Arabic numerals are not always shown on the pedigree, however.)

Hint 2. How to distinguish recessive conditions from dominant conditions Recessive conditions tend to “skip” generations, whereas dominant conditions cannot do so. If a condition affects an offspring whose parents are unaffected, then that condition must be recessive. For dominant conditions, every affected individual must have at least one affected parent. If both dominant and recessive inheritance modes are possible in a pedigree, the mode that requires fewer unrelated individuals to have the conditioncausing allele is more likely. This is especially true for rare conditions because the conditioncausing allele will be rare in the population.

Hint 3. How can I tell if a recessive condition is Xlinked or autosomal? The pedigrees below show the inheritance of two rare, recessive conditions (conditions C and D). The inheritance mode of both conditions could be explained as either autosomal recessive or Xlinked recessive, as shown below.

For both conditions, decide which of the following explanations is most likely by considering the genotype of the male indicated by the red arrow. Complete the table by selecting one pink label for condition C and one blue label for condition D. ANSWER:

ANSWER:

Correct Begin a pedigree analysis by determining the inheritance mode of the condition. Is the condition recessive or dominant? Dominant conditions require that an affected individual have at least one affected parent. An affected offspring of two unaffected parents indicates a recessive condition. (This is the case for both conditions A and B.) Is the condition autosomal or Xlinked? An affected female offspring of two unaffected parents indicates an autosomal recessive condition. (This is the case for condition A.) If the condition were Xlinked, the male parent would be affected. To distinguish between rare Xlinked and autosomal recessive conditions, choose the simpler mode for the observed pattern—that is, the mode that requires fewer unrelated individuals to carry a rare allele. In this example, the autosomal recessive mode would require two unrelated horses (II1 and II5) to carry a rare recessive allele for condition B, whereas the Xlinked recessive mode requires only one. Therefore, the Xlinked recessive mode is a better choice for condition B.

Part B Determining genotypes in pedigrees of Xlinked conditions The pedigree from Part A is shown below. Fill in the most likely genotypes of the indicated individuals in the pedigree. Note that a dominant allele followed by an underscore (_) indicates that either the dominant or the recessive allele may be present at the second position. Drag one pink label (for condition A, autosomal recessive) to each pink target. Drag one blue label (for condition B, Xlinked recessive) to each blue target. Labels can be used once, more than once, or not at all.

Hint 1. How can I deduce genotypes in autosomal recessive pedigrees? The simple pedigree below shows the inheritance of an autosomal recessive condition. Complete each statement by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all. ANSWER:

Hint 2. How can I deduce genotypes for Xlinked recessive conditions? Deducing male genotypes for Xlinked conditions is straightforward. Males have only one allele for genes on the X chromosome. That allele determines their phenotype: For Xlinked recessive conditions, males that have the recessive allele are affected, and males that have the dominant (wildtype) allele are not affected. Deducing female genotypes for Xlinked conditions is more challenging but relies on logic similar to that used for autosomal conditions. Consider the following simple pedigree for a rare Xlinked recessive condition. Fill in the most likely genotype for each of the three indicated females in the pedigree below. Labels can be used once, more than once, or not at all. ANSWER:

ANSWER:

Correct Use these rules when asg genotypes for Xlinked recessive conditions: Males have only one allele for every Xlinked gene. Affected males have the recessive allele, and nonaffected males have the dominant (wildtype) allele. Unaffected females with affected sons are heterozygous (carriers). Unaffected females with affected fathers are heterozygous. All sons of affected females will be affected. All daughters of affected females will be heterozygous. Use these rules when asg genotypes for autosomal recessive conditions: Unaffected parents with an affected offspring are both heterozygous. Unaffected offspring of one affected parent will be heterozygous. Unaffected offspring of two heterozygous parents may be homozygous dominant or heterozygous.

Part C Calculating probabilities in pedigrees of Xlinked conditions The pedigree from Part B is shown below.

Female III4 is pregnant via male III5. The owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (IV 3). Answer each question by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

Hint 1. For question 1, how to determine probabilities when certain genotypes can be eliminated To answer question 1, one of the things you need to calculate is the probability that individuals II5 and II7 have the genotype Aa. You can use the same logic described below for a different pedigree to help you calculate those probabilities. The simple pedigree below shows the inheritance of an autosomal recessive condition. In this pedigree, you know that the parents must be carriers because they have an affected offspring. The Punnett square shows all possible genotypes for their offspring.

Note that their unaffected offspring could have the genotype RR or Rr but not rr because they do not express the condition. Thus, you can eliminate rr as a possibility for these offspring, so you need to adjust the probabilities of the remaining genotypes. For example, the unadjusted probability that an offspring has the genotype RR is 1/4. However, the adjusted probability is 1/4 divided by the total of the remaining possibilities (3/4):

Hint 2. For question 2, how do I calculate probabilities for condition B? A good method for calculating probabilities in pedigrees is to consider the requirements for a certain outcome. What is required in order for IV3 to have condition B? Select the three correct requirements from the list below. ANSWER:

II4 es an X b chromosome to III4. II4 es an X B chromosome to III4. III4 es an X b chromosome to IV3. III4 es an X B chromosome to IV3. III5 es an X b chromosome to IV3. III5 es an X B chromosome to IV3. III5 es a Y chromosome to IV3.

Hint 3. How to approach question 3 Question 3 asks for the probability that an individual will have condition A AND condition B. Once you have calculated the probability of having each condition separately, think about whether you should use the product rule or the sum rule to find the probability of having both conditions.

Hint 4. How to approach question 4 Question 4 asks for the probability of having neither condition (not condition A AND not condition B). The probability of not condition A = 1 the probability of condition A, since these two situations are exclusive and must add up to 1. By the same reasoning, the probability of not condition B = 1 the probability of condition B.

Hint 5. How to approach question 5 Having at least one condition means that one of the following situations applies: (1) condition A AND condition B, OR (2) condition A AND not condition B, OR (3) condition B AND not condition A Therefore, one way to calculate the probability of having at least one condition is to apply the product and sum rules to situations 1, 2, and 3. Another way to calculate this probability is to realize that there is a fourth situation, which does not apply to individual IV3: (4) not condition A AND not condition B The probability of situations 1, 2, 3, and 4 must add up to 1. Therefore, the probability of having at least one condition = 1 the probability of (not condition A AND not condition B). ANSWER:

Correct A good method for calculating probabilities in pedigrees is to consider the requirements for a certain outcome. For question 1, the following are required for IV3 to have condition A (aa): II5 has the genotype Aa (probability = 2/3). If II5 has the genotype Aa (ed for by the above probability), then he es an a allele to III4 (probability = 1/2). If III4 has the genotype Aa (ed for by the above probability), then she es an a allele to IV3 (probability = 1/2). II7 has the genotype Aa (probability = 2/3). If II7 has the genotype Aa (ed for by the above probability), then he es an a allele to III5 (probability = 1/2). If III5 has the genotype Aa (ed for by the above probability), then he es an a allele to IV3 (probability = 1/2). All of these requirements are needed in sequence, so you apply the product rule (2/3 x 1/2 x 1/2 x 2/3 x 1/2 x 1/2 = 1/36). Similarly, for question 2, the following are required for IV3 to have condition B: II4 es an X b chromosome to III4 (probability = 1/2). If III4 has the genotype X B X b (ed for by the above probability), then she es an X b chromosome to IV3 (probability = 1/2). III5 es a Y chromosome to IV3 (probability = 1/2). All of these requirements are needed in sequence, so you apply the product rule here, too (1/2 x 1/2 x 1/2 = 1/8). Once the individual probabilities are known, the sum and/or product rules can be used for various combinations (both conditions, either condition, etc.).

Part D Adjusting probabilities based on new information Individual IV3 is born as a male affected with condition B but not condition A. His parents are bred again, and an ultrasound shows that the fetus is another male. The relevant portion of the pedigree is shown below. Use this new information to determine the parents’ genotypes (indicated by red arrows). Then calculate the probabilities that the second male offspring will have each condition. Drag one pink label to each pink target and one blue label to each blue target. Then use the white labels to answer questions 1 and 2. Labels can be used once, more than once, or not at all.

Hint 1. Revising genotypes based on new information When IV3 was born, his phenotype (affected with condition B) revealed that his mother (III4) must be a carrier (X B X b ). This information affects the probability calculations for the next pregnancy. See Hint 2 if you need help with these calculations.

Hint 2. How do I recalculate probabilities based on new information? There are two new pieces of information about the second pregnancy: 1. The ultrasound shows that the fetus (IV4) is male. 2. We now know that the mother (III4) is heterozygous for condition B (X B X b ) because she has an affected male offspring (IV3). We do not, however, have any new information about the genotypes of the breeding pair for condition A. Consider three of the requirements for IV3 that you determined in part C (before IV3 was born and the new information became available): II4 es an X b chromosome to III4 (probability = 1/2). III4 es an X b chromosome to IV3 (probability = 1/2). III5 es a Y chromosome to IV3 (probability = 1/2). For IV4, which of these requirements have altered probabilities based on the new information? Select all that apply. ANSWER: II4 es an X b chromosome to III4. III4 es an X b chromosome to IV4. III5 es a Y chromosome to IV4.

ANSWER:

Correct The new information allows you to be certain about two requirements that previously could only be assigned probabilities: 1. You now know that individual III4 has the genotype X B X b , meaning that she received an X b chromosome from II4. 2. You also now know that individual IV4 is a male, meaning that he received a Y chromosome from III5. This new information changes the probability calculations because you know that these two requirements have been met (thus, they have a probability of one certainty). The only requirement that is unchanged is that of III4 ing on an X b chromosome to IV4 the probability is still 1/2.

Video Tutor Session Quiz: SexLinked Pedigrees Click the play icon on the Video Tutor Session to your right. You can also  the video or view the text of the tutor session to read while you are watching. After you have watched the tutor session, answer the questions. Estimated time: 15 minutes

Part A

ANSWER: X H Y h X h Y 0 X H Y 0 X H Y H we cannot say for sure

Correct Since hemophilia is sexlinked, I must have only one copy of the gene, on my X chromosome. And since I don’t have the recessive disorder, I must have the dominant form of the gene.

Part B Click here to review the figure

ANSWER: X b X b X B X b X B X B X B Y it could be X B X B or X B X b

Correct We know that my mother ed on a dominant allele to me, so she must have an X B . But whether she has one or two copies of the X B , we cannot say for sure from the information shown here.

Part C

ANSWER: 0% 25% 50% 100% the answer cannot be determined from the information given

Correct If a woman is bald, she must have two copies of the recessive gene, so she must  one on to her children. Since her son would get his Y chromosome from his father and his X chromosome from his mother, he would inherit her baldness gene. That means that all of her sons will be bald.

Part D Click here to review the figure

ANSWER: X B X B X B X b X b X b it could be X B X B or X b X b we cannot say for sure

Correct Since the mother has an affected son who inherited his X from her, she must have an X b to  on to him. But she has another son who inherited an X from her but is not affected, so she must also have an X B to  on.

Part E Click here to review the figure

ANSWER: X B X B X B X b X b X b it could be X B X B or X B X b we cannot say for sure

Correct Since she is not affected, she must have at least one X B , which she inherited from her father. But her mother is heterozygous, so we cannot say for certain whether she is a carrier or not.

Chapter 15 PreTest Question 5 Part A

In general, the frequency with which crossing over occurs between two linked genes depends on what?

Hint 1. There are a limited number of crossover events per chromosome. ANSWER: whether the genes are on the X or some other chromosome how far apart they are on the chromosome the phase of meiosis in which the crossing over occurs whether the genes are dominant or recessive the characters the genes code for

Correct The farther apart two genes are, the greater the probability that a crossover will occur between them, and therefore the greater the recombination frequency.


Part A Which of the following statements is true of linkage? ANSWER: Crossing over occurs during prophase II of meiosis. Linked genes are found on different chromosomes. The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them. The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%. All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome.

Correct


Part A Three genes (A, B, and C) at three loci are being mapped in a particular species. Each gene has two alleles, one of which results in a phenotype that is markedly different from the wild type. The unusual allele of gene A is inherited with the unusual allele of gene B or C about 50% of the time. However, the unusual alleles of genes B and C are inherited together 14.4% of the time. Which of the following describes what is happening? ANSWER: Gene A is located 14.4 map units from genes B and C. Gene A is linked but genes B and C are not. Gene A is assorting independently of genes B and C, which are linked. The three genes are linked. The three genes are showing independent assortment.

Correct


Part A What is one map unit equivalent to? ANSWER: the distance between a pair of homologous chromosomes the physical distance between two linked genes the recombination frequency between two genes assorting independently 1 nanometer of distance between two genes 1% frequency of recombination between two genes

Correct


Part A Recombination between linked genes comes about for what reason? ANSWER: Crossovers between these genes result in chromosomal exchange. Nonrecombinant chromosomes break and then re with one another. Independent assortment sometimes fails because Mendel had not calculated appropriately. When genes are linked they always "travel" together at anaphase. Mutation on one homolog is different from that on the other homolog.

Correct


Part A The following is a map of four genes on a chromosome.

Between which two genes would you expect the highest frequency of recombination? ANSWER: A and W E and G A and E W and E A and G

Correct


Part A Use the following information to answer the question below. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows: A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): Phenotypes

Leaves

Stems

Roots

Number

1

a

+

+

14

2

a

+

c

0

3

a

b

+

32

4

a

b

c

440

5

+

b

+

0

6

+

b

c

16

7

+

+

c

28

8

+

+

+

470

Total

1,000

In which progeny phenotypes has there been recombination between genes A and B? ANSWER: in all 8 of them 1, 2, 5, and 6 2, 3, 5, and 7 1, 3, 6, and 7 2, 4, 5, and 8

Correct


Part A Use the following information to answer the question below. A plantlike organism on the planet Pandora can have three recessive genetic traits: bluish leaves, due to an allele (a) of gene A; a feathered stem, due to an allele (b) of gene B; and hollow roots due to an allele (c) of gene C. The three genes are linked and recombine as follows: A geneticist did a testcross with an organism that had been found to be heterozygous for the three recessive traits and she was able to identify progeny of the following phenotypic distribution (+ = wild type): Phenotypes

Leaves

Stems

Roots

Number

1

a

+

+

14

2

a

+

c

0

3

a

b

+

32

4

a

b

c

440

5

+

b

+

0

6

+

b

c

16

7

+

+

c

28

8

+

+

+

470

Total

1,000

Which of the following are the phenotypes of the parents in this cross? ANSWER: 2 and 5 4 and 8 3 and 7 1 and 2 1 and 6

Correct


Part A What is the reason that closely linked genes are typically inherited together? ANSWER: Genes align that way during metaphase I of meiosis. The number of genes in a cell is greater than the number of chromosomes. Alleles are paired together during meiosis. The likelihood of a crossover event between these two genes is low. Chromosomes are unbreakable.

Correct

Linked Genes and Linkage Mapping If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a crossover between them is very high, the genes are considered to be unlinked. Unlinked genes follow Mendel’s law of independent assortment. If, however, two genes tend to “travel together” because they are near one another on the same chromosome, they are said to be linked. Linked genes do not follow Mendel’s law of independent assortment. In this tutorial, you will compare the inheritance patterns of unlinked and linked genes.

Part A Independent assortment of three genes A wildtype tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP). Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp).

In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp). Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes:

solid, normal, smooth solid, normal, peach solid, dwarf, smooth solid, dwarf, peach

mottled, normal, smooth mottled, normal, peach mottled, dwarf, smooth mottled, dwarf, peach

Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.

Hint 1. How to approach this problem To calculate the phenotypic ratio of the offspring in the F2 generation, perform the following steps in this order: 1. Write out the genotypes of the parents (the two tomato plants in the testcross). 2. Complete three separate Punnett squares, one for each gene. 3. For each Punnett square, predict the probability of offspring with each of the two phenotypes. (Express the probabilities as fractions.) 4. Multiply together the individual fractions to determine the probability of offspring with each of the eight different combinations of phenotypes for the three traits. 5. To convert the fractions to ratios, first convert the fractions (if necessary) so they all have the same denominator. Then use the numerators of each fraction as the ratio.

Hint 2. Can you identify the genotypes of the offspring in the F2 generation? Match the genotypes to each of the eight possible phenotypes of the offspring in the F2 generation. Note that a blank ( _ ) is used if either a dominant allele or a recessive allele would result in the same phenotype. ANSWER:

Hint 3. Independent assortment and the multiplication rule When two or more genes are unlinked, their inheritance pattern follows Mendel’s law of independent assortment—the alleles for each gene segregate independently of the alleles for the other genes during gamete formation. When three events occur independently, you can use the multiplication rule to calculate the probability of the three events occurring together in a specific combination. This means that if you know the probability of each event occurring separately, you can find the probability of the events occurring together (in this case in the same plant) by multiplying the independent probabilities. Although you can achieve the same result using a single Punnett square that includes all the genes, this method becomes impractical when more than two genes are involved. Using the multiplication rule is preferable.

Hint 4. Can you calculate the probability of each phenotype? that when genes undergo independent assortment, you use the multiplication rule to find the probability of the independent events occurring together. For the cross MmDdPp x mmddpp, what is the probability of an offspring having the genotype mmDdPp (and the phenotype mottled, normal, smooth)? ANSWER: 1/4

x 1/4 x 1/2 = 1/32

1/4

x 1/2 x 1/2 = 1/16

1/2

x 1/2 x 1/2 = 1/8

1/4

x 1/4 x 1/4 = 1/64

ANSWER:

Correct The law of independent assortment states that each pair of alleles segregates independently of all other pairs of alleles during gamete formation. The three Punnett squares below show the predicted phenotypic ratios for each trait in the F2 offspring.

Thus, you would predict that all eight possible phenotypes from the cross MmDdPp x mmddpp are equally likely, resulting in a phenotypic ratio of 1:1:1:1:1:1:1:1.

Part B Gene linkage and phenotypic ratios Now, suppose that the three tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how? Which statement best predicts the results of the cross MmDdPp x mmddpp assuming that all three genes are linked?

Hint 1. Review of gene linkage Linked genes are genes that are located on the same chromosome and tend to be inherited together. The closer two genes are on the same chromosome, the less likely they are to be separated by a crossover event in meiosis I. The diagram shows what happens when crossing over occurs between two linked genes. Crossing over produces gametes with recombinant chromosomes, individual chromosomes that carry genes from two different parents.

Hint 2. What are the parental phenotypes in this cross? In a cross between the two parents, MmDdPp and mmddpp, which of the phenotypes listed below are the parental phenotypes? Select all that apply. ANSWER: solid, normal, smooth solid, normal, peach solid, dwarf, smooth solid, dwarf, peach mottled, normal, smooth mottled, normal, peach mottled, dwarf smooth mottled, dwarf, peach

ANSWER: All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes. Lack of independent assortment means that you cannot predict the frequencies of phenotypes in the offspring. Only the parental phenotypes could possibly occur in the offspring. All eight possible phenotypes would occur in equal numbers in the offspring (1:1:1:1:1:1:1:1).

Correct Because all three genes are linked, it is more likely that the parental allele combinations would stay together rather than be recombined through a crossover event. That is why a greater proportion of the offspring would have parental phenotypes. Nevertheless, some crossing over would likely occur, which is why a small proportion of the offspring would have recombinant phenotypes.

Part C Building a linkage map Suppose that you perform the cross discussed in Part B: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results:

Use the data to complete the linkage map below. Drag the labels onto the chromosome diagram to identify the locations of and distances between the genes. Use the blue labels and blue targets for the genes; use the white labels and white targets for the distances. Gene m has already been placed on the linkage map.

Hint 1. How to approach this problem To construct a linkage map, perform the following steps in this order: 1. Determine which offspring have parental genotypes and which have recombinant genotypes. 2. Determine which offspring are recombinants between genes m and d.  that identifying the original linkage in the trihybrid parent (MmDdPp) is key to determining this. (See Hints 2 and 3.) 3. Determine which offspring are recombinants between genes m and p. And finally, determine which offspring are recombinants between genes d and p. 4. Calculate the recombination frequency between each pair of genes. Do so by adding up the number of recombinants for a given pair and dividing that number by the total number of offspring (1000). Multiply by 100 to get the recombination frequency. Recombination f requency =

Number of recombinants Total number of of f spring

x 100

5. Use the recombination frequencies to construct your linkage map.  that a 1% recombination frequency is equal to one map unit, or one centimorgan (cM).

Hint 2. Which alleles are linked in the trihybrid parent of the F2 generation? For the parents in the testcross, it is clear that the homozygous recessive parent (mmddpp) has the allele combination mdp on each homologous chromosome. However, to know which alleles are linked on the homologous chromosomes in the trihybrid parent (MmDdPp), you must refer back to the homozygous parents in the P generation: MMDDPP x mmddpp. Which of the following correctly indicates the linkage in the trihybrid parent (MmDdPp)? ANSWER: mDP/Mdp MdP/mDp MDP/mdp MDp/mdP

Hint 3. What are the recombinant genotypes between genes m and d? You must first determine which offspring have recombinant genotypes for each pair of genes in order to calculate their recombination frequencies. Which of the following genotypes are recombinant genotypes between genes m and d? Select all that apply. ANSWER:

MmDdPp MmDdpp MmddPp Mmddpp mmDdPp mmDdpp mmddPp mmddpp

ANSWER:

Correct The recombination frequencies between genes can be used to construct a linkage map, as you have just done. The closer two genes are to each other on the same chromosome, the less likely they are to be separated by a crossover event, resulting in a lower recombination frequency.

To calculate the distance between m and p, you would add the map distances between m and d and between d and p (12 cM + 5 cM = 17 cM). You probably noticed that the recombination frequency between m and p is slightly less than that sum. This is because of double crossover eventsthe times that crossovers occur both between m and d and between d and p. The second crossover effectively “cancels out” the first, reducing the number of recombinants between m and p that are observed, while contributing to the number of recombinants between each of the other two pairs of genes. Therefore, adding the smaller map distances to calculate larger distances avoids inaccuracies due to doublecrossover events that produce nonrecombinant genotypes.

Activity: Polyploid Plants


Part A Humans are diploid and have 46 chromosomes (or two sets). How many sets of chromosomes are found in each human gamete? ANSWER: 1 2 3 4 5

Correct Meiosis cuts the number of chromosomal sets in half.

Part B Humans are diploid and have 46 chromosomes. How many chromosomes are found in each human gamete? ANSWER: 12 23 36 45 92

Correct 23 is onehalf of 46.

Part C _____ is the process by which haploid gametes form a diploid zygote. ANSWER: Embryogenesis Meiosis Gastrulation Fertilization Mitosis

Correct The ing of haploid gametes produces a diploid zygote.

Part D A particular diploid plant species has 48 chromosomes, or two sets. A mutation occurs and gametes with 48 chromosomes are produced. If self fertilization occurs, the zygote will have _____ set(s) of chromosomes. ANSWER: 1 2 3 4 5

Correct If selffertilization occurs, and two gametes (each having 48 chromosomes) unite, the zygote will have 96 chromosomes (or 4 sets).

Part E Which of these  applies to an organism with extra sets of chromosomes? ANSWER: monosomy haploid trisomy polyploid diploid

Correct Polyploid individuals have more than two complete sets of chromosomes.

Part F Mutant tetraploid plants _____. ANSWER: are usually sickly are able to interbreed with their parents have an odd number of chromosomes are unable to interbreed with a diploid plant unable to selffertilize

Correct This is why polyploidy is a mechanisms of plant speciation.

Part G Most polyploid plants arise as a result of _____. ANSWER:

selffertilization a mutation of gamete formation meiosis mitosis hybridization

Correct Most polyploid plants are the result of the hybridization of two different parental species.

Activity: Mistakes in Meiosis

Click here to view this animation. Then answer the questions.

Part A What is a nondisjunction?

Hint 1. Think about the literal meaning of the word nondisjunction. ANSWER: An error in cell division that causes homologous chromosomes or sister chromatids to move to the same side of the dividing cell An error in which a diploid cell or organism has an extra chromosome of one type, producing a chromosome number of 2n + 1 An error in which a diploid cell or organism lacks a chromosome of one type, producing a chromosome number of 2n 1 None of the above

Correct Nondisjunction refers to the failure of pairs of chromosomes or sister chromatids to separate during meiosis or mitosis.

Part B When can nondisjunction occur? Choose the best answer.

Hint 1. Nondisjunction can occur in cells of the germ line (cells that produce gametes) or in other body cells. ANSWER:

In mitosis, when sister chromatids fail to separate In meiosis, when sister chromatids fail to separate In meiosis, when homologous chromosomes fail to separate All three answers are correct.

Correct Nondisjunction errors can occur in meiosis I, when homologous chromosomes fail to separate, or in either mitosis or meiosis II, when sister chromatids fail to separate.

Part C Which syndrome is characterized by the XO chromosome abnormality? ANSWER: Turner syndrome Down syndrome Klinefelter syndrome Trisomy X

Correct Turner syndrome is characterized by a chromosome pattern of XO, or one X chromosome and no Y chromosome.

Part D What kind of cell results when a diploid and a haploid gamete fuse during fertilization?

Hint 1. Consider whether the  refer to entire sets of chromosomes or to a single chromosome. ANSWER: A monosomic cell A triploid cell A trisomic cell A monoploid cell

Correct A triploid cell has three sets of chromosomes: the two from the diploid gamete and the one from the haploid gamete.

Part E Of the following chromosomal abnormalities, which type is most likely to be viable in humans?

Hint 1. There are only a small number of aneuploid conditions that humans can tolerate; most changes in chromosome number are lethal. ANSWER:

Trisomy Monosomy Triploidy Haploidy

Correct Some trisomies are viable in humans, but they result in developmental abnormalities, such as Down syndrome (trisomy 21). Only humans with trisomies of the small chromosomes (13, 18, or 21) or the sex chromosomes survive past birth.

Part F If a diploid cell undergoes meiosis and produces two gametes that are normal, and one with n − 1 chromosomes, and one with n + 1 chromosomes, what type of error occurred?

Hint 1. Note that half of the gametes are affected—this is a clue to the stage of meiosis at which the error occurred. ANSWER: No error occurred; these are normal gametes. A nondisjunction error occurred in meiosis II, in which both sister chromatids of a chromosome migrated to the same pole of the cell. A nondisjunction error occurred in meiosis I, in which both  of all homologous pairs migrated to the same pole of the cell. A nondisjunction error occurred in meiosis I, in which both  of a homologous pair migrated to the same pole of the cell.

Correct When this error occurs in meiosis II, only half of the gametes are affected.

Part G If a diploid cell undergoes meiosis and produces two gametes with n + 1 chromosomes and two gametes with n− 1 chromosomes, what type of error occurred?

Hint 1. Note that all of the gametes are affected—this is a clue to the stage of meiosis at which the error occurred. ANSWER: A nondisjunction error occurred in meiosis II, in which both sister chromatids of a chromosome migrated to the same pole of the cell. A nondisjunction error occurred in meiosis I, in which both  of all homologous pairs migrated to the same pole of the cell. No error occurred; these are normal gametes. A nondisjunction error occurred in meiosis I, in which both  of a homologous pair migrated to the same pole of the cell.

Correct When this type of error occurs, half of the gametes receive one too many chromosomes, and the other half receive one too few. When all of the gametes are affected, the error probably occurred in meiosis I.

Chapter 15 PreTest Question 6 Part A Which of the following results in a situation in which the chromosome number is either 2n+1 or 2n1 ?

Hint 1. Aneuploidy can be detected by examining a karyotype. ANSWER: polyploidy methylation inversion gene linkage aneuploidy

Correct An aneuploid individual has either too many or too few chromosomes, usually as a result of nondisjunction.

Chapter 15 PreTest Question 7 Part A What results if a fragment of a chromosome breaks off and then reattaches to the original chromosome at the same place but in the reverse direction?

Hint 1. Review the terminology for altered chromosome structure. ANSWER: a translocation a nondisjunction an inversion polyploidy a deletion

Correct The question describes an inversion.


Part A Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual? ANSWER: 47, XYY 47, XXY 45, X 47, +21 47, XXX

Correct


Part A A phenotypically normal prospective couple seeks genetic counseling because the man knows that he has a translocation of a portion of his chromosome 4 that has been exchanged with a portion of his chromosome 12. Although he is normal because his translocation is balanced, he and his wife want to know the probability that his sperm will be abnormal. What is your prognosis regarding his sperm? ANSWER: His sperm will be sterile and the couple might consider adoption. 1/4 will be normal, 1/4 will have the translocation, and 1/2 will have duplications and deletions. 1/2 will be normal and the rest will have the father's translocation. All will carry the same translocation as the father. None will carry the translocation since abnormal sperm will die.

Correct


Part A An inversion in a human chromosome often results in no demonstrable phenotypic effect in the individual. What else may occur? ANSWER: Some abnormal gametes may be formed. The individual is more likely to get cancer. There is an increased frequency of mutation. All inverted chromosomes are deleted. There may be deletions later in life.

Correct


Part A Down syndrome has a frequency in the U.S. population of ~ 1/700 live births. In which of the following groups would you expect this frequency to be significantly higher? ANSWER: people living in equatorial areas of the world very small population groups people in Latin or South America the Inuit and other peoples in very cold habitats No groups have such higher frequency.

Correct


Part A A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition? ANSWER: One member of the couple underwent nondisjunction in somatic cell production. The woman inherited this tendency from her parents. One member of the couple carried a translocation. The mother had a chromosomal duplication. One member of the couple underwent nondisjunction in gamete production.

Correct


Part A At what point in cell division is a chromosome lost so that, after fertilization with a normal gamete, the result is an embryo with 45, X? I. an error in anaphase I II. an error in anaphase II III. an error of the first postfertilization mitosis IV. an error in pairing ANSWER: I, II, or III only I or II only I, II, III, or IV III or IV only II or IV only

Correct


Part A Which of the following is true of aneuploidies in general? ANSWER: An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication. Of all human aneuploidies, only Down syndrome is associated with mental retardation. A monosomy is more frequent than a trisomy. 45 X is the only known human liveborn monosomy. Some human aneuploidies have selective advantage in some environments.

Correct

Chapter 15 PreTest Question 8 Part A What phenomenon occurs when a particular allele will either be expressed or silenced, depending on whether it is inherited from a male or a female?

Hint 1. The phenomenon is associated with the methylation of certain genes during gametogenesis. ANSWER: polyploidy sex linkage aneuploidy extranuclear inheritance genomic imprinting

Correct The sex of the parent transmitting a given chromosome can affect the chromosome's expression in the offspring.

Chapter 15 PreTest Question 9 Part A How are human mitochondria inherited?

Hint 1. Mitochondria are not found within a cell's nucleus. ANSWER: as an Xlinked trait without DNA from the mother only from the father only as linear DNA

Correct An individual's mitochondria are all derived from those found in the ovum.


Part A A gene is considered to be Mendelian if its inheritance pattern follows Mendel's laws. Which of the following would be considered Mendelian? ANSWER:

a gene transmitted from mothers to sons and from fathers to daughters a gene derived solely from maternal inheritance a gene whose expression varies depending on the gender of the transmitting parent a gene transmitted via the cytoplasm or cytoplasmic structures a gene transmitted by a virus to eggproducing cells

Correct


Part A Mitochondrial DNA is primarily involved in coding for proteins needed for electron transport. Therefore, in which body systems would you expect most mitochondrial gene mutations to be exhibited? ANSWER: the immune system and the blood the nervous and muscular systems the skin and senses the circulation system the excretory and respiratory systems

Correct


Part A A certain kind of snail can have a righthanded direction of shell coiling (DD or Dd) or lefthanded coiling (dd). If direction of coiling is due to a protein deposited by the mother in the egg cytoplasm, then a Dd eggproducing snail and a dd spermproducing snail will have offspring of which genotype(s) and phenotype(s)? ANSWER: 1/2 Dd:1/2 dd; half right and half left coiling 1/2 Dd:1/2 dd; all right coiling all Dd; all right coiling all Dd; half right and half left coiling all Dd; all left coiling

Correct


Part A

The pedigree in the figure shows the transmission of a trait in a particular family. Based on this pattern of transmission, the trait is most likely ANSWER: sexlinked recessive. autosomal recessive. mitochondrial. autosomal dominant. sexlinked dominant.

Correct


Part A Imagine a human disorder that is inherited as a dominant, Xlinked trait. How would the frequency of this disorder vary between males and females? ANSWER: Males would display this disorder with greater frequency than females. Males and females would display this disorder with equal frequency. Females would display this disorder with greater frequency than males.

Correct Men have only one X chromosome, whereas women have two. Thus, women have twice the chance of inheriting the dominant allele that causes this disorder.  that the disorder is caused by a dominant allele, so an individual who inherits a single dominant allele will display the disorder. Read about the inheritance of Xlinked genes.


Part A Select the correct statement(s) about sex determination in animals. Select all that apply. ANSWER: The mechanism of sex determination varies with different animal species. In all animals, males are XY and females are XX. In all animals, males have the SRY gene and females lack this gene.

Correct There are various chromosomal systems for sex determination found in animals.


Part A Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel’s law of independent assortment? ANSWER: Sister chromatids separate during anaphase II. The arrangement of each pair of homologous chromosomes on the metaphase plate during metaphase I is random with respect to the arrangements of other pairs. The two alleles for each gene separate as homologous chromosomes move apart during anaphase I.

Correct Alleles of genes on nonhomologous chromosomes assort independently during gamete formation.


Part A Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel’s law of segregation? ANSWER: Sister chromatids separate during anaphase II. The arrangement of each pair of homologous chromosomes on the metaphase plate during metaphase I is random with respect to the arrangements of other pairs. The two alleles for each gene separate as homologous chromosomes move apart during anaphase I.

Correct The chromosomal basis of Mendel’s law of segregation state that the two alleles for each gene separate during gamete formation.


Part A Gregor Mendel set up a dihybrid cross with one pea plant from the parental generation (P) producing round yellow peas and the other pea plant producing wrinkled green peas. The F2 generation included 315 plants producing round yellow peas, 108 with round green peas, 101 with wrinkled yellow peas, and 32 with wrinkled green peas. How would these results have differed if pea shape and pea color had been linked genes, located close together on the same chromosome? ANSWER: The F2 generation would have included a higher percentage of pea plants producing yellow peas. The F2 generation would have included a lower percentage of pea plants producing wrinkled, green peas. The F2 generation would have included a higher percentage of pea plants producing round, yellow peas.

Correct If pea shape and pea color were determined by closely linked genes, they would not show independent assortment. Chromosomes would carry the combination of alleles seen in the P generation: yellow + round or green + wrinkled. Approximately ¾ or 75% of the F2 generation would inherit at least one chromosome carrying the yellow + round alleles. That is higher than Mendel’s observed percentage of 315/556 or 56%. Read how linkage affects inheritance.

Current Events: In Choosing a Sperm Donor, a Roll of the Genetic Dice (New York Times, 5/14/2012) Read this New York Times article and then answer the questions. In Choosing a Sperm Donor, a Roll of the Genetic Dice (New York Times, 5/14/2012) Registration with The New York Times provides instant access to breaking news with NYTimes.com. To , go to http://www.nytimes.com/. Visit http://www.nytimes.com/content/help/rights//ofservice.html to review the current NYT  of Service.

Part A Cystic fibrosis is an inherited disorder that affects which of the following? ANSWER: lungs heart liver All of the above.

Correct

Part B How long can sperm be cryogenically stored and still be viable? ANSWER: for more than 20 years less than 2 years only 6 months about 10 years

Correct

Part C Which federal agency ensures that sperm donors are tested for communicable diseases? ANSWER: Department of Health and Human Services Food and Drug istration National Institutes of Health Centers for Disease Control

Correct

Part D Which of the following is true?

ANSWER: Only men can be carriers for genetic disorders. Neither women nor men can be carriers for genetic disorders. Both women and men can be carriers for genetic disorders. Only women can be carriers for genetic disorders

Correct

Part E Which of the following is a leading cause of sudden death among young athletes? ANSWER: NF1 HCM Fragile X syndrome cystic fibrosis

Correct Score Summary: Your score on this assignment is 99.0%. You received 95.05 out of a possible total of 96 points.

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