Name____________________________
AP Biology-Krock
Genetics Problems #5
1. A child has type AB blood. If one parent is blood type A, what are the possible genotypes of the
other parent? AB, BB, BO
2. Mrs. Doe and Mrs. Roe had babies at the same hospital at the same time. Mrs. Roe brought
home a girl (Nancy), and Mrs. Doe brought home a son (Richard). Mrs. Doe was certain she
had delivered a daughter and filed a lawsuit charging that the hospital mixed up the children.
Blood types are as follows:
Mrs. Doe: Type O
Mrs. Roe: Type B
Mr. Doe: Type AB
Mr. Roe: Type B
Nancy:
Type O
Richard: Type B
Could Mrs. Doe be correct in her allegation? NO Can you be certain? Explain. Mr. Doe is AB
so he cannot have a child with Type O blood.
3. In a cross involving linked genes in the fruit fly, Drosophila, a male homozygous for the
recessive linked genes purple eye (p) and black body (b) was crossed with a female
heterozygous at both loci. The following offspring resulted:
151 wild type (normal phenotype for both characteristics)
8
purple eyes and normal body color
10
normal eyes and black body color
131 purple eyes and black body color
a. Are these two genes linked? How do you know? YES. THE RATIOS DO NOT MATCH THE
NON-LINKED PHENOTYPE RATIOS.
PARENTAL CROSS = ppbb X PpBb
b. Calculate the distance between the eye color and body color loci in map units.
8 + 10 = 18/300 = 6% = 6 map units
4a. In Drosophila, the allele for miniature wing (m) is recessive to the allele for normal wing (M),
and the gene for vermillion eye color (v) is recessive to the allele for normal eye color (V). A
female heterozygous for eye color and wing size was mated to a vermillion-eyed, miniaturewinged male. The following offspring (F1) were collected:
140 normal wing, normal eyes
3
normal wing, vermillion eyes
6
miniature wing, normal eyes
151 miniature wing, vermillion eyes
PARENTAL CROSS = VvMm X vvmm
1. What were the linkage groups of the female parent? VM and vm
2. What is the distance (crossover value) between the two loci, v and m?
3 + 6 = 9/300 = 3% = 3 map units
1
4b. A female Drosophila heterozygous for the recessive alleles sable body (s) and miniature wing
(m) was mated with a sable-bodied, miniature-winged male with the following progeny:
250 wild type
15
normal body, miniature wings
20
sable body, normal wings
215 sable body, miniature wings
1. Diagram the linkage map of the female parent. PARENTAL CROSS = SsMm X ssmm
35/500 = 7% = 7 map units
4c. Draw the relative positions of the loci of the v, s, and m alleles on a chromosome. This cannot
be conclusively answered-show all possibilities.
4d. What further cross would provide a conclusive answer? What single cross would have allowed
mapping of all 3 loci? SsVv X ssvv; SsVvMm X ssvvmm
5. The crossover frequency between linked genes A and B is 40%; between B and C is 20%;
between C and D is 10%; between C and A is 20%; and, between D and B is 10%. Draw the
arrangement of the sequence of genes on the chromosome, labeling the distance between
each locus.
B D C
A
10
10
10
10
6. In a series of breeding experiments among frogs, a linkage group composed of genes A, B, C,
and D was found to show the following crossover frequencies. Map the chromosome.
A
B
C
D
A
B
-10%
10%
---
4%
6%
9%
19%
C
4%
6%
---
13%
D
9%
19%
13%
---
D
A
9
C
4
B
6
7. Suppose two AaBbCcDd individuals are mated and further suppose that each gene assorts
independently. What is the probability of an offspring with:
a. the genotype of the parent?
½ X ½ X ½ X ½ = 1/16
b. A_B_C_D_ phenotype?
¾ X ¾ X ¾ X ¾ = 81/256
c. aabbccdd genotype?
¼ X ¼ X ¼ X ¼ = 1/256
2