Mastering Biology Chp. 14 HW

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 PART A – Identifying the genotype How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous? Cross the green pod plant with the yellow pod plant [A cross between a plant of unknown genotype and one that is known to be homozygous recessive is called a test cross because the recessive homozygote tests whether there are any recessive alleles in the unknown. Because the recessive homozygote will contribute an allele for the recessive characteristic to each offspring, the second allele (from the unknown genotype) will determine the offspring’s phenotype.] PART B – Diagramming a cross using a Punnett square Punnett squares can be used to predict the two possible outcomes of the botanist’s test cross. The Punnett square on the left shows the predicted result if the unknown plant is homozygous (GG); the Punnett square on the right shows the predicted result if the unknown plant is heterozygous (Gg). Drag the labels to the correct locations on the Punnett squares. (G is the symbol for the green-pod allele and g is the symbol for the yellow-pod allele.) You can use a label once, more than once, or not at all. The genotypes in a Punnett square show all the possible combinations of alleles in offspring that could result from the particular cross. A Punnett square reveals the expected probabilities of each genotype among the offspring. For example, the Punnett square on the right reveals that there is a 50% chance that each offspring will have green pods and a 50% chance that each offspring will have yellow pods. PART C – Relationship with Mendel’s findings Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel’s findings does her test cross illustrate? law of segregation [The law of segregation states that the two alleles for a gene separate during gamete formation, and end up in different gametes. In the case of the heterozygous green-pod plant (Gg), one gamete will receive the dominant allele (G), and the other gamete will receive the recessive allele (g). The law of segregation accounts for the prediction that 50% of the offspring of the test cross will have green pods and 50% will have yellow pods.] PART D – Relationship of allele behavior to meiosis During which part of meiosis (meiosis I or meiosis II) do the two alleles of a gene separate? During which phase does the separation occur? State your answer as meiosis I or meiosis II followed by a comma and the name of the phase (for example, if your answer is meiosis II and metaphase, enter meiosis II, metaphase). meiosis I, anaphase [Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.] PART A – Deducing phenotypes and genotypes of selfed parents Mendel studied pea plants dihybrid for seed shape (round versus wrinkled) and seed color (yellow versus green). Recall that -the round allele (R) is dominant to the wrinkled allele (r) and -the yellow allele (Y) is dominant to the green allele (y). The table below shows the F1 progeny that result from selfing four different parent pea plants. Use the phenotypes of the F1 progeny to deduce the genotype and phenotype of each parent plant. Complete the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all. The ability to deduce an organism’s genotype from the phenotype(s) of its progeny is an important skill in solving genetics problems. In this example, the logic was simplified because the parent plants were selfed, and therefore only one parental genotype was involved. PART B – Deducing genotypes of crossed parents A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F1: [round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green]. Use this information to deduce the genotypes of the parent plants. Indicate the genotypes by dragging the correct label to the appropriate location. Answering this question requires two logical steps: -First, eliminate those genotypes that are inconsistent with the phenotypes of the parents. -Then examine the remaining possibilities: Which are consistent with the phenotypes of the progeny? Because the cross produces green progeny (and both parents are yellow), both parents must be Yy: If either parent were YY, only yellow progeny would result. The cross also produces wrinkled progeny, so the [round, yellow] parent must be heterozygous (Rr): If it were RR, all of the progeny would be Rr and have a round phenotype. PART C – Predicting genotypes and phenotypic frequencies of progeny For the cross in Part B, predict the frequencies of each of the phenotypes in the F1 progeny, and determine the genotype(s) present in each phenotypic class. Complete the diagram by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all. The frequencies of the four phenotypic classes are determined by applying the multiplication rule. The green progeny, for example, make up ¼ of the total progeny because the probability of their formation depends on receiving a y allele from both parents (probability = ½ x ½ = ¼). Similarly, the wrinkled progeny make up ½ of the total progeny because the probability of their formation depends on receiving an r allele from both parents (probability = 1 x ½ = ½). Because the [wrinkled, green] progeny are simultaneously green and wrinkled, and these events are independent, apply the multiplication rule to obtain their frequency: ¼ x ½ = 1/8. PART A – Determining relationships between alleles You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure wild-type 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 wild-type 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. An allele is never intrinsically "dominant" or "recessive." Instead, these terms 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 wild-type 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. Remember that p (the pale mutant allele) and P (the wild-type 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. Remember that F (the forked mutant allele) is dominant to f + (the wild-type allele). Labels may be used once, more than once, or not at all. For help getting started, see the hints. 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? The forked mutation and the twist mutation are codominant alleles of the same locus. [The absence of wild-type 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 wild-type genotype for both loci and therefore a wild-type phenotype. In this case, the F1 expresses both mutant phenotypes, indicating that the forked and twist alleles are codominant.] PART D – Assigning 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. 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. 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? 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. [Read about Mendel’s experiments and about quantitative characters.] Look at the Punnett square, which shows the predicted offspring of the F2 generation from a cross between a plant with yellow-round seeds (YYRR) and a plant with green-wrinkled seeds (yyrr). Select the correct statement about wrinkled yellow seeds in the F2 generation. 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%. [Read about the Law of Independent Assortment and about how to solve complex genetics problems using the rules of probability.] Each chromosome in this homologous pair possesses a different allele for flower color. Which statement about this homologous pair of chromosomes is correct? These homologous chromosomes represent a maternal and a paternal chromosome. [Homologous chromosomes initially come together during fertilization with the contribution of a set of chromosomes from the sperm and the egg.] When a dominant allele coexists with a recessive allele in a heterozygote individual, how do they interact with each other? They do not interact at all. [An allele is called dominant because it appears in the phenotype of a heterozygote, not because it subdues a recessive allele. Read about dominance.] Refer to the drawings in the figure of a single pair of homologous chromosomes as they might appear during various stages of either mitosis or meiosis. Which diagram represents anaphase I of meiosis? I (1) Refer to the drawings in the figure of a single pair of homologous chromosomes as they might appear during various stages of either mitosis or meiosis. Which diagram(s) represent(s) anaphase II of meiosis? V (5) You have isolated DNA from three different cell types of an organism, determined the relative DNA content for each type, and plotted the results on the graph shown in the figure. Which sample(s) of DNA might be from a skin cell arrested in G0 of the cell cycle? I (1) PART A What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.) bbgg [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 _____. 1 black eyes, green skin : 1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin [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. bbgg [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. dihybrid [The cross is examining two characters.] PART E A phenotypic ratio of 9:3:3:1 in the offspring of a cross indicates that _____. both parents are heterozygous for both genes [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 _____. Mendel’s laws of segregation and independent assortment [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] 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 [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? Mendel’s law of independent assortment is being violated. [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] BBg [All of the offspring have black eyes, and there is a 3:1 ratio of skin color.] PAERT J In a situation in which genes assort independently, what is the ratio of the gametes produced by an AaBB individual? 1 AB : 1 aB [This is the correct ratio of gametes.] PART A – A trait governed by two or more genes In mice, agouti fur is a dominant trait resulting in individual hairs having a light band of pigment on an otherwise dark hair shaft. A mouse with agouti fur is shown here, along with a mouse with solid color fur, which is the recessive phenotype (A = agouti; a = solid color). A separate gene, which is not linked to the agouti gene, can result in either a dominant black pigment or a recessive brown pigment (B = black; b = brown). A litter of mice from the mating of two agouti black parents includes offspring with the following fur colors: -solid color, black -solid color, brown (sometimes called chocolate) -agouti black -agouti brown (sometimes called cinnamon) What would be the expected frequency of agouti brown offspring in the litter? 3/16 [Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/4, and the probability of bb offspring is 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.] PART B – Lethal alleles and epistasis In addition to A and a, the "agouti" gene has a third allele, AY . Here is some information about the inheritance of the AY allele. -The AY allele is dominant to both A and a. -The homozygous genotype (AYAY ) results in lethality before birth. -The heterozygous genotypes (AYA or AYa) result in yellow fur color, regardless of which alleles are present for the B/b gene. (This effect exhibited by the AY allele is known as epistasis–when the expression of one gene masks the expression of a second gene.) Suppose you mate two mice with the genotypes AYaBb x AYaBb . Considering only the live-born offspring, what would be the expected frequency of mice with yellow fur? (For help getting started, see Hint 1.) Express your answer as a fraction using the slash symbol and no spaces (for example, 1/16). 2/3 [Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.] PART C – The effect of a third gene on fur color In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse. Match the phenotypes on the labels at left to the genotypes listed below. Labels can be used once, more than once, or not at all. AaBbcc – albino AaBBCC – aqouti black Aabbcc – albino AAbbCc – aqouti brown aaBbCc – solid color, black AABBcc – albino PART D – The effect of a fourth gene on fur color In the same mouse species, a fourth unlinked gene (gene P/p) also affects fur color. -For mice that are either homozygous dominant (PP) or heterozygous (Pp), the organism’s fur color is dictated by the other three genes (A/a, B/b, and C/c). -For mice that are homozygous recessive (pp), large patches of the organism’s fur are white. This condition is called piebaldism. In a cross between two mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur? (Hint: Consider each gene separately; then use the multiplication rule. For more help getting started, see Hint 1.) Express your answer as a fraction using the slash symbol and no spaces (for example, 3/16). 9/256 [Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BBor Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.] PART A – Determining the mode of inheritance The pedigrees below show the inheritance of three separate, rare autosomal conditions in different families. For each pedigree, decide if the condition is better explained as recessive or dominant. Drag the correct label to the appropriate location. Labels can be used once, more than once, or not at all. If an individual has a genetic condition that neither parent has, then that condition must be recessive. Dominant conditions require that every affected individual have at least one affected parent. In situations where the inheritance mode of a rare condition cannot be definitely determined, the most likely mode is the one that requires the fewest unrelated individuals to have the condition-causing allele. PART B – Determining genotypes in autosomal dominant pedigrees Pedigree 2 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal dominant condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all. You can deduce the genotype of an individual in a pedigree based on two types of information: the individual’s phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children. For autosomal dominant conditions: 1. Unaffected individuals are homozygous for the recessive, wild-type allele. 2. Affected individuals with only one affected parent are heterozygous. 3. Affected individuals with any unaffected children are heterozygous. 4. Affected individuals with two affected parents may be homozygous dominant or heterozygous. PART C – Determining genotypes in autosomal recessive pedigrees Pedigree 3 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal recessive condition. Note that individual II-3 has no family history of this rare condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all. You can deduce the genotype of an individual in a pedigree based on two types of information: the individual’s phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children. For autosomal recessive conditions: 1. Affected individuals are always homozygous recessive. 2. Unaffected children of an affected parent are always carriers (heterozygous). 3. Both parents of affected individuals must have at least one recessive allele. 4. If both parents are carriers, their unaffected children may be carriers or homozygous for the dominant, wild-type allele. 5. For rare conditions, you can assume that individuals marrying into a family do not carry the recessive allele if there is no evidence that they are carriers. PART D – Calculating probabilities in pedigrees The pedigree from Part C is shown below. Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition. Complete each statement by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all. You can calculate probabilities in pedigrees by considering the requirements for a specific outcome. In this case, the outcome that individual IV-1 will be affected has four requirements: 1. Individual III-3 is a carrier (probability = 2/3). 2. Individual III-4 is a carrier (probability = 1/2). 3. Individual III-3 passes the r allele to his child (probability = 1/2, assuming III-3 is a carrier, which is accounted for in requirement 1). 4. Individual III-4 passes the r allele to her child (probability = 1/2, assuming III-4 is a carrier, which is accounted for in requirement 2). Because requirements 1 AND 2 AND 3 AND 4 must be met for the outcome in question (individual IV-1 to be affected), you calculate the answer by applying the multiplication rule (as implied by "AND"): The probability that IV-1 will be affected (rr) = 2/3 x 1/2 x 1/2 x 1/2 = 1/12.

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