homework 5 ans
5.01 In an experiment designed to study the inheritance of flower color in four-o'clocks, two plants with pink flowers were crossed. In the progeny from this cross, there were 42 plants with red flowers, 86 with pink flowers, and 39 with white flowers. Using a chi-square test, determine whether those numbers are consistent with the single-gene, incomplete dominance inheritance pattern.
5.01 In an experiment designed to study the inheritance of flower color in four-o'clocks, two plants with pink flowers were crossed. In the progeny from this cross, there were 42 plants with red flowers, 86 with pink flowers, and 39 with white flowers. Using a chi-square test, determine whether those numbers are consistent with the single-gene, incomplete dominance inheritance pattern.
Answer:
Answer:
• Using df = 2, 0.258 is less than the table value of 5.99, so the data are consistent with incomplete dominance.
• Using df = 2, 0.258 is less than the table value of 5.99, so the data are consistent with incomplete dominance.
5.02 Trying to understand the inheritance of the dominant yellow gene in mice, researchers mated two yellow heterozygous mice. A typical result was 56 yellow progeny to 31 wild-type.
5.02 Trying to understand the inheritance of the dominant yellow gene in mice, researchers mated two yellow heterozygous mice. A typical result was 56 yellow progeny to 31 wild-type.
• (a) Use a chi-square test to determine if the outcome of this cross is consistent with the usual 3:1 ratio predicted by Mendelian inheritance for a dominant gene.
• (a) Use a chi-square test to determine if the outcome of this cross is consistent with the usual 3:1 ratio predicted by Mendelian inheritance for a dominant gene.
Answer:
Answer:
• Chi-square = 5.25 which is greater than the table value of 3.84 (df=1). Inconsistent
• Chi-square = 5.25 which is greater than the table value of 3.84 (df=1). Inconsistent
• (b) You will find that the chi-square test done in part (a) indicates the data are not consistent. Now try the hypothesis that the dominant allele is lethal in the homozygous condition. Repeat the chi-square test.
• (b) You will find that the chi-square test done in part (a) indicates the data are not consistent. Now try the hypothesis that the dominant allele is lethal in the homozygous condition. Repeat the chi-square test.
Answer:
Answer:
• Chi-square = 0.21 which is less than the table value of 3.84 (df=1). Consistent
• Chi-square = 0.21 which is less than the table value of 3.84 (df=1). Consistent
5.03 Which blood group type of the ABO system is known as the universal donor? Which type is the universal recipient? Explain your answers. Note: this is something we don't usually talk about in lecture, so you may have to look this up. It is not directly related to the inheritance pattern, but is worth knowing about because it is related to the properties of the red blood cells.
5.03 Which blood group type of the ABO system is known as the universal donor? Which type is the universal recipient? Explain your answers. Note: this is something we don't usually talk about in lecture, so you may have to look this up. It is not directly related to the inheritance pattern, but is worth knowing about because it is related to the properties of the red blood cells.
Answer: Type O is the universal donor. Type O red blood cells have neither A nor B type antigen and so do not stimulate antibody production when given to a recipient with any blood type. Type AB is the universal recipient. AB red blood cells have both A and B antigens, so an AB recipient would not produce antibodies when given any blood type.
Answer: Type O is the universal donor. Type O red blood cells have neither A nor B type antigen and so do not stimulate antibody production when given to a recipient with any blood type. Type AB is the universal recipient. AB red blood cells have both A and B antigens, so an AB recipient would not produce antibodies when given any blood type.
5.04 A woman with blood type A marries a man with blood type O.
5.04 A woman with blood type A marries a man with blood type O.
• (a) Given only this information, determine all the blood types that are possible for their children?
• (a) Given only this information, determine all the blood types that are possible for their children?
Answer: The woman could have genotypes IAIA or IAi. The man must be genotype ii. Therefore, these parents could produce children with blood types A or O.
Answer: The woman could have genotypes IAIA or IAi. The man must be genotype ii. Therefore, these parents could produce children with blood types A or O.
• (b) Which blood types are impossible?
• (b) Which blood types are impossible?
Answer: These parents could not produce children with blood types B or AB.
Answer: These parents could not produce children with blood types B or AB.
• (c) Would the answers to a and b change if it was known that both of the woman's parents had blood type AB? Explain.
• (c) Would the answers to a and b change if it was known that both of the woman's parents had blood type AB? Explain.
Answer: Yes. If the woman's parents were both type AB, it would mean that her genotype could only be IAIA, because neither of her parents would carry the i allele. In this case, all her children would be blood type A because they would always inherit a IA allele from their mother.
Answer: Yes. If the woman's parents were both type AB, it would mean that her genotype could only be IAIA, because neither of her parents would carry the i allele. In this case, all her children would be blood type A because they would always inherit a IA allele from their mother.
5.05 A woman with type O blood, whose father has type A and whose mother has type B has a child with type O. There is a dispute over the identity of the child's father. Two men are possible fathers. One is type AB and the other is type A.
5.05 A woman with type O blood, whose father has type A and whose mother has type B has a child with type O. There is a dispute over the identity of the child's father. Two men are possible fathers. One is type AB and the other is type A.
• (a) What is the mother's genotype?
• (a) What is the mother's genotype?
Answer: The mother must have genotype ii because that is the only genotype that produces type O blood.
Answer: The mother must have genotype ii because that is the only genotype that produces type O blood.
• (b) Which man could be the father?
• (b) Which man could be the father?
Answer: Only the type A man could be the father. The man with type AB could not carry an i allele and so could not produce type O children.
Answer: Only the type A man could be the father. The man with type AB could not carry an i allele and so could not produce type O children.
• (c) If this man is the father, what is his genotype?
• (c) If this man is the father, what is his genotype?
Answer: He would have to be heterozygous (IAi). If he were homozygous, he would not be able to produce type O children.
Answer: He would have to be heterozygous (IAi). If he were homozygous, he would not be able to produce type O children.
• (d) What are the genotypes of the woman's parents?
• (d) What are the genotypes of the woman's parents?
Answer: The woman's parents would have genotypes IAi and IBi. This is the only way they could have a daughter who is type O.
Answer: The woman's parents would have genotypes IAi and IBi. This is the only way they could have a daughter who is type O.
5.06 In the peppered moth, Biston betularia, there are three alleles that determine body color. These alleles are all at the same locus. The allele for pale color (m) is recessive. A second allele (M’), which is dominant to m, produces a mottled color called insularia. The third allele (M), which is dominant to both of the other two, produces a melanic moth (very dark colored). A female moth having the typical pale color is mated to a male melanic. If half the progeny are melanic and half are insularia, what were the genotypes of the two parents?
5.06 In the peppered moth, Biston betularia, there are three alleles that determine body color. These alleles are all at the same locus. The allele for pale color (m) is recessive. A second allele (M’), which is dominant to m, produces a mottled color called insularia. The third allele (M), which is dominant to both of the other two, produces a melanic moth (very dark colored). A female moth having the typical pale color is mated to a male melanic. If half the progeny are melanic and half are insularia, what were the genotypes of the two parents?
Answer: The female must be mm because pale color is recessive. The male must have at least one M allele because his phenotype is melanic. From his phenotype alone, his genotype could be MM, MM', or Mm. But because some of the offspring are insularia (M'm), he must be carrying a M' allele. Thus the male parent's genotype is MM'.
Answer: The female must be mm because pale color is recessive. The male must have at least one M allele because his phenotype is melanic. From his phenotype alone, his genotype could be MM, MM', or Mm. But because some of the offspring are insularia (M'm), he must be carrying a M' allele. Thus the male parent's genotype is MM'.
5.07 In a particular plant there is a gene with five alleles, B1, B2, B3, B4, and B5.
5.07 In a particular plant there is a gene with five alleles, B1, B2, B3, B4, and B5.
• (a) Given that two plants with genotypes B2B4 and B1B5 are mated, what types of progeny, and in what proportions, would you expect?
• (a) Given that two plants with genotypes B2B4 and B1B5 are mated, what types of progeny, and in what proportions, would you expect?
Answer:
Answer:
• (b) For the same gene, if a progeny from a single mating has equal numbers of B1B2 and B2B4 individuals (and no other genotypes), what are the parents' genotypes?
• (b) For the same gene, if a progeny from a single mating has equal numbers of B1B2 and B2B4 individuals (and no other genotypes), what are the parents' genotypes?
Answer: The only way to account for the observed offspring is that one parent has genotype B2B2 and the other has genotype B1B4 . (Just construct a 2-box Punnett square to see what the parent gametes must be.)
Answer: The only way to account for the observed offspring is that one parent has genotype B2B2 and the other has genotype B1B4 . (Just construct a 2-box Punnett square to see what the parent gametes must be.)
5.08 What is the difference between dominance and epistasis? Why didn't any pairs of genes used by Mendel show epistasis?
5.08 What is the difference between dominance and epistasis? Why didn't any pairs of genes used by Mendel show epistasis?
Answer: Dominance means that, for a single gene locus, one allele's expression dominates over another. Epistasis means that an allele at one locus (one gene) influences the expression of another allele at a different locus (two genes are involved). The genes that Mendel chose to use in dihybrid crosses were for such unrelated traits that no gene interaction was observed. Epistasis occurs when more than one gene affects the same trait.
Answer: Dominance means that, for a single gene locus, one allele's expression dominates over another. Epistasis means that an allele at one locus (one gene) influences the expression of another allele at a different locus (two genes are involved). The genes that Mendel chose to use in dihybrid crosses were for such unrelated traits that no gene interaction was observed. Epistasis occurs when more than one gene affects the same trait.
5.09 In chickens, comb shape is determined by two interacting genes (R/r and P/p). A walnut comb is produced when at least one dominant R allele and at least one dominant P allele are present (R- P-). A rose comb is produced when at least one dominantR allele is present and the second gene is homozygous recessive (R- pp). A pea comb is produced when the first gene is homozygous recessive and at least one dominant R allele is present (rr P-). When both genes are homozygous recessive, a single comb is produced. Predict the phenotype ratios for each of the following matings:
5.09 In chickens, comb shape is determined by two interacting genes (R/r and P/p). A walnut comb is produced when at least one dominant R allele and at least one dominant P allele are present (R- P-). A rose comb is produced when at least one dominantR allele is present and the second gene is homozygous recessive (R- pp). A pea comb is produced when the first gene is homozygous recessive and at least one dominant R allele is present (rr P-). When both genes are homozygous recessive, a single comb is produced. Predict the phenotype ratios for each of the following matings:
• Each of the following answers can be calculated from a forked-line analysis:
• Each of the following answers can be calculated from a forked-line analysis:
• (a) RR PP X rr pp Answer: All are walnut (Rr Pp)
• (a) RR PP X rr pp Answer: All are walnut (Rr Pp)
• (b) Rr pp X Rr pp Answer: 3/4 rose (R- pp), 1/4 single (rr pp)
• (b) Rr pp X Rr pp Answer: 3/4 rose (R- pp), 1/4 single (rr pp)
• (c) Rr Pp X rr pp Answer: 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp)
• (c) Rr Pp X rr pp Answer: 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp)
• (d) Rr pp X rr Pp Answer: 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp)
• (d) Rr pp X rr Pp Answer: 1/4 walnut (Rr Pp), 1/4 rose (Rr pp), 1/4 pea (rr Pp), 1/4 single (rr pp)
• (e) Rr Pp X Rr Pp Answer: 9/16 walnut (R-P-), 3/16 rose (R-pp), 3/16 pea (rr P-), 1/16 single (rr pp)
• (e) Rr Pp X Rr Pp Answer: 9/16 walnut (R-P-), 3/16 rose (R-pp), 3/16 pea (rr P-), 1/16 single (rr pp)
• (f) rr pp X rr Pp Answer: 1/2 pea (rr Pp), 1/2 single (rr pp)
• (f) rr pp X rr Pp Answer: 1/2 pea (rr Pp), 1/2 single (rr pp)
5.10 As discussed in class, one form of fur color in mice is controlled by the interaction of two gene resulting in three phenotypes: agouti, black, and albino. Two agouti mice are crossed repeatedly, and the following offspring are produced: 46 agouti, 16 black, and 23 albino.
5.10 As discussed in class, one form of fur color in mice is controlled by the interaction of two gene resulting in three phenotypes: agouti, black, and albino. Two agouti mice are crossed repeatedly, and the following offspring are produced: 46 agouti, 16 black, and 23 albino.
• (a) Based on this cross, which common multiple gene inheritance pattern is operating here?
• (a) Based on this cross, which common multiple gene inheritance pattern is operating here?
Answer: The observed ratio seems to fit the 9:3:4 ratio very closely. This is indicative of the gene interaction pattern recessive epistasis.
Answer: The observed ratio seems to fit the 9:3:4 ratio very closely. This is indicative of the gene interaction pattern recessive epistasis.
• (b) Given your answer to part a, what were the genotypes of the parents?
• (b) Given your answer to part a, what were the genotypes of the parents?
Answer: 9:3:4 is the expected ratio for recessive epistasis when both parents are heterozygous for each gene, so the parents must be Aa Bb x Aa Bb.
Answer: 9:3:4 is the expected ratio for recessive epistasis when both parents are heterozygous for each gene, so the parents must be Aa Bb x Aa Bb.
• (c) Show a forked-line diagram for this cross that makes it clear how each genotype category corresponds to each phenotype.
• (c) Show a forked-line diagram for this cross that makes it clear how each genotype category corresponds to each phenotype.
Answer:
Answer:
• (d) For each of the following crosses, give the possible genotypes of the parents. Agouti x Agouti produces 37 agouti and 12 black
• (d) For each of the following crosses, give the possible genotypes of the parents. Agouti x Agouti produces 37 agouti and 12 black
Answer: A- Bb x A- Bb. The B gene must be heterozygous in both parents, because some offspring are black (bb). The exact genotype for the A gene cannot be determined completely, but both parents must have at least one copy of the dominant A allele. Additionally, we know that one or both parents must be homozygous (AA), because there were no albino (aa) offspring.
Answer: A- Bb x A- Bb. The B gene must be heterozygous in both parents, because some offspring are black (bb). The exact genotype for the A gene cannot be determined completely, but both parents must have at least one copy of the dominant A allele. Additionally, we know that one or both parents must be homozygous (AA), because there were no albino (aa) offspring.
Black x Albino produces 27 agouti and 24 albino
Black x Albino produces 27 agouti and 24 albino
Answer: Aa bb x aa BB. The black parent must be heterozygous for the A gene because some offspring were albino(aa). The albino parent must be homozygous for the B gene, otherwise some offspring would have been black (bb).
Answer: Aa bb x aa BB. The black parent must be heterozygous for the A gene because some offspring were albino(aa). The albino parent must be homozygous for the B gene, otherwise some offspring would have been black (bb).
Black x Black produces 28 black and 10 albino
Black x Black produces 28 black and 10 albino
Answer: Aa bb x Aa bb. Both parents must be heterozygous for the A gene, otherwise there could be no albino offspring.
Answer: Aa bb x Aa bb. Both parents must be heterozygous for the A gene, otherwise there could be no albino offspring.
5.11 The agouti fur color in mice actually results from alternating dark and light bands on each hair. Given this information and what you know about the gene interaction from the previous problem, propose a mechanism that could explain how each of the two genes is actually affecting the overall fur color (agouti, black, or albino).
5.11 The agouti fur color in mice actually results from alternating dark and light bands on each hair. Given this information and what you know about the gene interaction from the previous problem, propose a mechanism that could explain how each of the two genes is actually affecting the overall fur color (agouti, black, or albino).
Answer: Given the allele we have been using in the previous problem, one explanation would be: One gene (A- or aa)determines whether any dark pigment is produce (A- : pigment produced, aa : albino). The second gene determines the distribution of the pigment on each hair (B- : alternating dark and light bands - agouti, bb : solid pigmented hair - black). When no pigment is produced (aa) it would not matter if the mouse was B- or bb, so albino is the result. In other words, although both genes affect the same trait (fur color), one actually controls the formation of pigment; the other controls the distribution of the pigment when it is produced.
Answer: Given the allele we have been using in the previous problem, one explanation would be: One gene (A- or aa)determines whether any dark pigment is produce (A- : pigment produced, aa : albino). The second gene determines the distribution of the pigment on each hair (B- : alternating dark and light bands - agouti, bb : solid pigmented hair - black). When no pigment is produced (aa) it would not matter if the mouse was B- or bb, so albino is the result. In other words, although both genes affect the same trait (fur color), one actually controls the formation of pigment; the other controls the distribution of the pigment when it is produced.
5.12 In summer squash, there are two pairs of alleles that determine fruit color. The two genes sort independently. Two white-fruited plants are crossed. Both parents are known to be heterozygous for both genes. The cross produces the following offspring: 20 green-fruited plants, 58 yellow-fruited plants, and 218 white-fruited plants.
5.12 In summer squash, there are two pairs of alleles that determine fruit color. The two genes sort independently. Two white-fruited plants are crossed. Both parents are known to be heterozygous for both genes. The cross produces the following offspring: 20 green-fruited plants, 58 yellow-fruited plants, and 218 white-fruited plants.
• (a) Based on the observed ratio, which common type of epistasis is operating here? That is, which kind of modified dihybrid ratio most closely fits these data?
• (a) Based on the observed ratio, which common type of epistasis is operating here? That is, which kind of modified dihybrid ratio most closely fits these data?
Answer: The ratio comes close to 12:3:1, which indicates dominant epistasis.
Answer: The ratio comes close to 12:3:1, which indicates dominant epistasis.
• (b) List the four genotype classes in the offspring and give the corresponding phenotype of each.
• (b) List the four genotype classes in the offspring and give the corresponding phenotype of each.
Answer:
Answer:
A-B- = white
A-B- = white
A-bb = white
A-bb = white
aaB- = yellow
aaB- = yellow
aabb = green
aabb = green
• (c) If a doubly heterozygous white plant is crossed with a green plant, what phenotype ratio would you expect in the progeny?
• (c) If a doubly heterozygous white plant is crossed with a green plant, what phenotype ratio would you expect in the progeny?
Answer:
Answer:
5.13 Two types of epistasis discussed were duplicate and complementary gene action. How would you distinguish between these types of epistasis in determining purple and white flower color if you could set up any crosses necessary?
5.13 Two types of epistasis discussed were duplicate and complementary gene action. How would you distinguish between these types of epistasis in determining purple and white flower color if you could set up any crosses necessary?
Answer: One way would be to do a test cross between a double heterozygous parent and a completely recessive parent (CcPp x ccpp). The expected ratios would be different depending on which mode of inheritance was acting. See the forked-line diagram below. Another way would be to cross a double heterozygous plant with itself (CcPp x CcPp). This cross should produce a 9:7 ratio of purple to white if the mechanism is complementary inheritance or a 15:1 ratio if it is duplicate inheritance.
Answer: One way would be to do a test cross between a double heterozygous parent and a completely recessive parent (CcPp x ccpp). The expected ratios would be different depending on which mode of inheritance was acting. See the forked-line diagram below. Another way would be to cross a double heterozygous plant with itself (CcPp x CcPp). This cross should produce a 9:7 ratio of purple to white if the mechanism is complementary inheritance or a 15:1 ratio if it is duplicate inheritance.
5.14 The pedigree shown depicts the inheritance of a certain rare human disease. Solid symbols indicate diseased individuals; open symbols indicate normal individuals.
5.14 The pedigree shown depicts the inheritance of a certain rare human disease. Solid symbols indicate diseased individuals; open symbols indicate normal individuals.
• (a) How can you explain the observed pattern of inheritance? You can assume that each gene involved has a penetrance value of 1.0. That is, every individual with the appropriate genotype will exhibit the disease.
• (a) How can you explain the observed pattern of inheritance? You can assume that each gene involved has a penetrance value of 1.0. That is, every individual with the appropriate genotype will exhibit the disease.
Answer: This trait cannot be explained by simple single-gene inheritance. (Using a one-gene explanation, try to determine if it is dominant or recessive. You will see a contradiction.) It does fit the pattern of 2-gene epistasis with complimentary inheritance. II-3 is A-bb, II-4 is aaB-. Both have the disease, but they can have a child without the disease. In this case, the child would have genotype AaBb.
Answer: This trait cannot be explained by simple single-gene inheritance. (Using a one-gene explanation, try to determine if it is dominant or recessive. You will see a contradiction.) It does fit the pattern of 2-gene epistasis with complimentary inheritance. II-3 is A-bb, II-4 is aaB-. Both have the disease, but they can have a child without the disease. In this case, the child would have genotype AaBb.
• (b) Based on your explanation in part a, what are the genotypes of II-3 and II-4?
• (b) Based on your explanation in part a, what are the genotypes of II-3 and II-4?
Answer: II-3 is AAbb, II-4 is aaBB. (This assumes that the recessive alleles are rare, which the problem states.)
Answer: II-3 is AAbb, II-4 is aaBB. (This assumes that the recessive alleles are rare, which the problem states.)
• (c) If II-3 and II-4 have another child, what is the probability that he or she will have the disease?
• (c) If II-3 and II-4 have another child, what is the probability that he or she will have the disease?
Answer: If the disease is rare, there is very little chance that they will have diseased children. If we are correct about the genotypes of the parents (see part b above), then all their children would be AaBb.
Answer: If the disease is rare, there is very little chance that they will have diseased children. If we are correct about the genotypes of the parents (see part b above), then all their children would be AaBb.
• (d) Assuming that this disease is very rare, what are the most likely genotypes of I-1, I-2, I-3, and I-4?
• (d) Assuming that this disease is very rare, what are the most likely genotypes of I-1, I-2, I-3, and I-4?
Answer: I-1 is AABb, I-2 is AABb, I-3 is AaBB, and I-4 is AaBB.
Answer: I-1 is AABb, I-2 is AABb, I-3 is AaBB, and I-4 is AaBB.
5.15 In wheat kernel color is determined by a pair of genes in a quantitative way. Each of the two genes can have two alleles (A1, A2, B1, B2). Kernel color ranges from red, when four type 1 alleles (either A1 or B1) are present, to white, when four type 2 alleles (either A2 or B2) are present. Three intermediate colors (dark pink, medium pink, and light pink) can occur depending on the relative numbers type 1 and type 2 alleles (this example was discussed in lecture). Using the forked-line approach, calculate the expected proportions of the five phenotypes that would be produced by a cross between two wheat plants with medium pink kernels that are heterozygous for both genes: A1A2 B1B2 x A1A2 B1B2 .
5.15 In wheat kernel color is determined by a pair of genes in a quantitative way. Each of the two genes can have two alleles (A1, A2, B1, B2). Kernel color ranges from red, when four type 1 alleles (either A1 or B1) are present, to white, when four type 2 alleles (either A2 or B2) are present. Three intermediate colors (dark pink, medium pink, and light pink) can occur depending on the relative numbers type 1 and type 2 alleles (this example was discussed in lecture). Using the forked-line approach, calculate the expected proportions of the five phenotypes that would be produced by a cross between two wheat plants with medium pink kernels that are heterozygous for both genes: A1A2 B1B2 x A1A2 B1B2 .
Answer:
Answer:
•  Phenotype ratio 1/16 red, 1/4 dark pink, 3/8 pink, 1/4 light pink, 1/16 white
•  Phenotype ratio 1/16 red, 1/4 dark pink, 3/8 pink, 1/4 light pink, 1/16 white