AP Genetics #1

Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive. What percentage of the offspring will show recombinant phenotypes resulting from crossovers?

• 50%

• 25%

• 75%

• 100%

A wild type fruit fly (heterozygous for gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are:

wild-type 721

black-purple 751

gray-purple 49

black-red 45

What is the combination frequency between the genes for body color and eye color? Express your answer as a percentage.

• 8%

• 6%

• 15%

• 20%

Two genes of a flower, one controlling blue (B) versus white (b) petals and the other controlling round (R) versus oval (r) stamens, are linked and are 10 map units apart. You cross a homozygous blue-oval plant with a homozygous white round plant. The resulting F₁ progeny are crossed with homozygous white oval parents and 1,000 F₂ progeny are obtained. How many F₂ plants of each of the four phenotypes do you expect?

_____ blue-oval

_____ white-round

_____blue-round

_____white-oval

• 50; 450; 450:50

• 50;50;450;450

• 450;450;50;50

• 400;400;100;100

A man with hemophilia (a recessive, sex linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait.

What is the probability that a daughter of this mating will be a hemophilic?

• 0/8

• 1/8

• 1/4

• 1/16

A man with hemophilia (a recessive, sex linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait.

What is the probability that a son will be a hemophilic?

• 1/4

• 1/8

• 1/2

• 1/16

A man with hemophilia (a recessive, sex linked condition) has a daughter of normal phenotype. She marries a man who is normal for the trait.

If the couple has four sons, what is the probability that all four will be born with hemophilia?

• 1/2

• 1/4

• 1/8

• 1/16

A space probe discovers a planet inhabited by creatures that reproduce with the same heredity patterns seen in humans. Three phenotypic characters are: height (T=tall, t=dwarf); head appendages (A= antennae, a=no antennae); nose morphology (S=upturned snout, s= downturned snout).

For tall heterozygotes with antennae, the offspring are:

tall-antennae-46

dwarf-antennae- 7

dwarf no antennae- 42

tall-no antennae- 5

Calculate the recombination frequency between T and A.

• 20%

• 33%

• 12%

• 17%

A space probe discovers a planet inhabited by creatures that reproduce with the same heredity patterns seen in humans. Three phenotypic characters are: height (T=tall, t=dwarf); head appendages (A= antennae, a=no antennae); nose morphology (S=upturned snout, s= downturned snout).

antennae-upturned snout: 47

antennae-downturned snout: 2

no antennae- downturned snout: 48

no antennae-upturned snout: 3

Calculate the recombination frequency between A and S

• 5%

• 10%

• 12%

• 20%

A space probe discovers a planet inhabited by creatures that reproduce with the same heredity patterns seen in humans. Three phenotypic characters are: height (T=tall, t=dwarf); head appendages (A= antennae, a=no antennae); nose morphology (S=upturned snout, s= downturned snout).

Scientists do a further testcross using a heterozygote for height and nose morphology. The offspring are:

tall-upturned snout: 40

dwarf-upturned snout: 9

dwarf-downturned snout: 42

tall-downturned snout: 9

Calculate the recombination frequency between T and S.

• 5%

• 12%

• 16%

• 20%

A space probe discovers a planet inhabited by creatures that reproduce with the same heredity patterns seen in humans. Three phenotypic characters are: height (T=tall, t=dwarf); head appendages (A= antennae, a=no antennae); nose morphology (S=upturned snout, s= downturned snout).

What is the correct sequence of the three linked genes?

• A-T-S

• T-S-A

• T-A-S

• none of these are correct