Two genes control color in corn snakes as follows:O– B– snakes are brown, O– bb are orange, oo B– areblack, and oo bb are albino. An orange snake wasmated to a black snake, and a large number of F1progeny were obtained, all of which were brown.When the F1 snakes were mated to one another, theyproduced 100 brown offspring, 25 orange, 22 black,and 13 albino.a. What are the genotypes of the F1 snakes?b. What proportions of the different colors wouldhave been expected among the F2 snakes if the twoloci assort independently?

Two genes control color in corn snakes as follows:O– B– snakes are brown, O– bb are orange, oo B– areblack, and oo bb are albino. An orange snake wasmated to a black snake, and a large number of F1progeny were obtained, all of which were brown.When the F1 snakes were mated to one another, theyproduced 100 brown offspring, 25 orange, 22 black,and 13 albino.a. What are the genotypes of the F1 snakes?b. What proportions of the different colors wouldhave been expected among the F2 snakes if the twoloci assort independently?

Name: Two genes control color in corn snakes as follows:O– B– snakes are bro
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Description: Two genes control color in corn snakes as follows:O– B– snakes are brown, O– bb are orange, oo B– areblack, and oo bb are albino. An orange snake wasmated to a black snake, and a large number of F1progeny were obtained, all of which were brown.When t

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 2TYU: The F1 flies described in question 1 were mated with brown-eyed flies from a true-breeding line....

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The F1 flies described in question 1 were mated with brown-eyed flies from a true-breeding line. What phenotypes would you expect the offspring to have? (a) all with red eyes (b) all with brown eyes (c) half with red eyes and half with brown eyes (d) red-eyed females and brown-eyed males (e) brown-eyed females and red-eyed males
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?
In humans, curly hair is dominant to straight hair. A woman heterozygous for curly hair marries a man with straight hair and they have children. a. What is the probability that the first child will have curly hair? b. What is the probability that in a family of four children 3 will be curly and 1 will be straight haired? c. What is the probability that the first three children will be curly-haired girls and the next will be a straight haired boy?
In humans, the mic2 gene (involved in antibody production) is found on sex chromosomes and has pseudoautosomal inheritance. If a female that is a mic2a is mated to a phenotypically mic2b male then what would be resulting ratios of the offspring? (note; the mic2a and 2b alleles are codominant) O 50% mic2a and 50% mic2b for both sexes O 100% mic2a/2b for both males and females O all females are mic2a/2b whereas males are all mic2a all females are mic2b whereas males are all mic2a
Earlobe attachment is controlled by a single gene. The free (F) condition is dominant over the attached (f) condition. A woman with free earlobes produced several children with a man having attached earlobes. The following results were obtained with the F1 offspring: 8 free, 7 attached. What are the parents’ genotypes? a. A heterozygous free-lobed man is mated to a heterozygous free-lobed woman. What are the genotypes and phenotypes of their possible offspring?
You perform a cross between two true-breeding strains of zucchini, One has green fruit and the other has yellow fruit. The F1 plants are all green, but when these are crossed, the F2 plants consist of 9 green: 7 yellow. A. Assuming that wild-type zucchini are green, how would you describe the phenomenon that occurred in the F1 of part (e) of the parents and progeny?.
You have been given a virgin Drosophila female. You notice that the bristles on her thorax are much shorter than normal. You mate her with a normal male (with long bristles) and obtain the following F1 progeny1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. A cross of the F1 long-bristled females with their brothers gives only long-bristled F2. A cross of short-bristled females with their brothers gives 1 3 short-bristled females, 1 3 long-bristled females, and 1 3 long-bristled males. Provide a genetic hypothesis to account for all these results, showing genotypes in every cross.
Forked bristles, miniature wings, and sable bodies are all homozygous recessive traits whose genes are located on the same chromosome. You cross female flies that have forked bristles (f) with male flies that have both miniature wings (m) and sable bodies (s). You then cross the F1 of this cross with flies that have sable bodies, forked bristles, and miniature wings, and obtain the following numbers of offspring with the following phenotypes out of a total of 1500 offspring: Sable bodies and miniature wings 463 Sable bodies and forked bristles 3 Miniature wings and forked bristles 49 Forked bristles, miniature wings, sable bodies 99 Miniature wings 5 Wild type 107 Sable bodies 55 Forked bristles…
Forked bristles, miniature wings, and sable bodies are all homozygous recessive traits whose genes are located on the same chromosome. You cross female flies that have forked bristles (f) with male flies that have both miniature wings (m) and sable bodies (s). You then cross the F1 of this cross with flies that have sable bodies, forked bristles, and miniature wings, and obtain the following numbers of offspring with the following phenotypes out of a total of 1500 offspring: Sable bodies and miniature wings 463 Sable bodies and forked bristles 3 Miniature wings and forked bristles 49 Forked bristles, miniature wings, sable bodies 99 Miniature wings 5 Wild type 107 Sable bodies 55 Forked bristles…
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 a certain type of pea plant, the gene for yellow color (Y) is dominant to the gene for green color (y), and the gene for round shape (R) is dominant to the gene for wrinkled shape (r). If a homozygous dominant plant is mated with a homozygous recessive plant (P-gen), what proportion in the F2 generation will be homozygous recessive? a) 1/4 b) 1/16 c) 3/4 d) 3/8
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.