There is recessive epistasis for these phenotypes, with ALBINO epistatic to AGOUTI and BLACK. True-breeding ALBINO mice were crossed to true-breeding BLACK mice. All the F1 from this cross had AGOUTI-colored hair. For the F2 progeny, 89 were AGOUTI, 32 were BLACK, and 39 were ALBINO. What is the expected number of F2 ALBINO mice that have the genotype AA cc? a) All of them (39) b) 10

There is recessive epistasis for these phenotypes, with ALBINO epistatic to AGOUTI and BLACK. True-breeding ALBINO mice were crossed to true-breeding BLACK mice. All the F1 from this cross had AGOUTI-colored hair. For the F2 progeny, 89 were AGOUTI, 32 were BLACK, and 39 were ALBINO. What is the expected number of F2 ALBINO mice that have the genotype AA cc? a) All of them (39) b) 10

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 5QP: As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women....

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Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?
As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
When a white guinea pig is crossed to a yellow guinea pig, all the progeny are cream-colored. From a cross of two such F1’s, the F2 showed 32 white, 66 cream, and 30 yellow guinea pigs. How is this trait inherited? _______________ Identify the genotype of the white guinea pig. ________ Identify the genotype of the yellow guinea pig. ________ Identify the genotype of the cream-colored guinea pig. ________ Identify the genotypes and the corresponding phenotypes of the F2 progeny. Give the genotypic ratio. __________________ Give the phenotypic ratio. ___________________
In Drosophila, a cross was made between a yellowbodied male with vestigial (not fully developed)wings and a wild-type female (brown body). The F1generation consisted of wild-type males and wild-typefemales. F1 males and females were crossed, and theF2 progeny consisted of 16 yellow-bodied males withvestigial wings, 48 yellow-bodied males with normalwings, 15 males with brown bodies and vestigialwings, 49 wild-type males, 31 brown-bodied femaleswith vestigial wings, and 97 wild-type females.Explain the inheritance of the two genes in questionbased on these results.
In dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.
A rooster with a particular comb morphology calledwalnut was crossed to a hen with a type of comb morphology known as single. The F1 progeny all had walnut combs. When F1 males and females were crossedto each other, 93 walnut and 11 single combs wereseen among the F2 progeny, but there were also29 birds with a new kind of comb called rose and32 birds with another new comb type called pea.a. Explain how comb morphology is inherited.b. What progeny would result from crossing a homozygous rose-combed hen with a homozygouspea-combed rooster? What phenotypes and ratioswould be seen in the F2 progeny?c. A particular walnut rooster was crossed to a peahen, and the progeny consisted of 12 walnut,11 pea, 3 rose, and 4 single chickens. What are thelikely genotypes of the parents?d. A different walnut rooster was crossed to a rosehen, and all the progeny were walnut. What are thepossible genotypes of the parents?
The Labrador Retriever’s coat colours are black, yellow, and chocolate.Yellow is produced by the presence of a recessive epistatic gene, which has the effect of masking the black and chocolate colour genes. Black (B) is dominant to chocolate (b).Therefore, the possible genotypes and corresponding phenotypes are: Phenotype Possible Genotypes Black G__B__ Brown G__bb Yellow gg__ Predict the percentage of offspring that would be black in colour for each of the following crosses. Record your answer as a whole number. Cross: GgBbGgBB GGBbGGBb Percentage ofOffspring: Answer% Answer%
When true-breeding mice with brown fur and short tails (BBtt)were crossed to true-breeding mice with white fur and long tails(bbTT), all of the F1 offspring had brown fur and long tails. TheF1 offspring were crossed to mice with white fur and short tails.What are the possible phenotypes of the F2 offspring? Which F2offspring are recombinant, and which are nonrecombinant? Whatare the ratios of phenotypes of the F2 offspring if independentassortment is taking place? How are the ratios affected by linkage?
What is a reciprocal cross? Suppose that a gene is found as a wildtype (functional) allele and a recessive mutant (nonfunctional)allele. What would be the expected outcomes of reciprocal crossesif a true-breeding normal individual was crossed to a true-breedingindividual carrying the mutant allele? What would be the results ifthe gene is maternally inherited?
Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.If we assume that no double crossovers occur, what percentageof F2 offspring is likely to be homozygous for all three genes?
Radishes may be round, oval and lomg. Aside from this, they may be red, purple or white in color. Red x white produces progeny, all of which are purple. If purple oval will becrossed with purple oval, how many types of phenotypes will appear and If purple oval are crossed with white, long radishes, then what genotypic and phenotypicratios exist in the progeny? Indicate the phenotypes of the offspring.