AP Biology – Genetics Practice Answer Key
Note: anywhere that allele letter is not specified, you may choose your own. Which side of a square holds maternal vs. paternal gametes is also up to you.
MONOHYBRID AND POLYHYBRID CROSSES
1.
F / fF / FF / Ff
f / Ff / ff
The trees must both be heterozygous. Some offspring are ff; therefore, both parents must have at least one f allele. Both parents are fuzzy, therefore, they also have at least one F allele.
2. Yellow must be dominant. Were yellow recessive and red dominant, two red plants could produce yellow offspring:
R / RR / RR / Rr
r / Rr / rr
But this problem says that yellow tulips must have a yellow parent. If yellow is dominant and red is recessive, then red tulips are invariably homozygous recessive, producing this cross with no yellow offspring:
r / rr / rr / rr
r / rr / rr
All other crosses involve at least one yellow tulip, and at least some dominant offspring.
3a. AbC, abC
b. ABD, ABd, aBD, aBd
c. ABDE, AbDE, aBDE, abDE
4a. 50%
b. 50%
5. We seek progeny that will be dominant for A, B, and D, but recessive for C and E.
P dominant phenotype for A: .75
P dominant phenotype for B: .5
P recessive phenotype for C: .5
P dominant phenotype for D: .75
P recessive phenotype for E: 1.0
.75*.5*.5*.75*1 = .14 or 14% chance.
6. (3/8 = .375, 1/8 = .125)
A good way to start is with the recessive/recessive puppies. A .125 frequency could only be obtained via .5*.25 operation. .25 only occurs in heterozygote vs heterozygote crosses, so the parents are both heterozygous for one of the two genes. For the other gene, one parent must be heterozygous and the other must be homozygous recessive in order to have a .5 chance of homozygous recessive puppies. Our options are therefore either BbSs x Bbss or BbSs x bbSs.
The former produces a homozygous dominant chance of (.75*.5) = .375, which fits the data. But so does the latter, so that isn’t helpful.
However, we know that there is a .375 chance of black spotted, and a .125 chance of chestnut solid. If the parents are BbSs x Bbss, the chance of black spotted is .75*.5 = .375, and the chance of chestnut solid is .25*.5 = .125, which fits the data.
If the parents are BbSs x bbSs, the chance of black spotted is .5 * .25 = .125, which is does not match the data.
The parents’ genotypes are BbSs x Bbss.
7a. .5*.25*.25*.25 = .0078 or .78%
b. .5*.5*.5*.5 = .0625 or 6.25%
c. .5*.25*.25*.25 = .0078 or .78%
INCOMPLETE DOMINANCE, CODOMINANCE, MULTIPLE ALLELES, LETHALITY
8. Purple = RW, white = WW
P / WW / PW / WW
W / PW / WW
1:1 Purple:White, 1:1 PW:WW
9. Color exhibits incomplete dominance between the black and white alleles. A gray rooster and black hen would have a 50% chance of gray and 50% chance of black offspring.
10. Red must be RR, White WW, Roan RW. Mating RW x RW:
R / WR / RR / RW
W / RW / WW
25% RR, 50% RW, 25% WW
25% Red, 50% Roan, 25% White
11a. The parents are C? x ChCh or Chc. If parent one is CC, all offspring in F1 would be black, so that’s not correct. Possible parents are therefore:
CCh x ChCh =1:1 black:chinchilla, fits the data
Cc x ChCh = 1:1 black:chinchilla, also fits
CCh x Chc = 1:1 black:chinchilla, also fits
Cc x Chc = 2:1:1 black:chinchilla:white, no fit
So the parents can be narrowed down to three pairs, but not definitively determined.
b. This time, the parents must definitely be Cc x Chc for reasons outlined in part a.
12. Baby 1 cannot belong to Lucy/Josh. Baby 2 cannot belong to Jane/Bob, Lucy/Josh, or Mary/John. Baby 3 cannot belong to Mary/John or Doris & Dan. Baby 4 cannot belong to Jane/Bob or Lucy/Josh.
Parents / Possible BabiesMary & John / 1,4
Doris & Dan / 1,2,4
Jane & Bob / 1, 3
Lucy & Josh / 3
So Baby 1: Jane&Bob
Baby 2: Doris & Dan
Baby 3: Duke & Betty
Baby 4: Mary & John
Jane&Bob
o / oA / Ao / Ao
? / ?o / ?o
Lucy & Josh
o / oo / oo / oo
o / oo / oo
Mary & John
o / oA / Ao / Ao
B / Bo / Bo
Doris & Dan
B / ?A / AB / A?
B / BB / B?
13. Hairlessness must be the heterozygous condition, normal hair must be the homozygous dominant condition, and homozygous recessive is lethal. The crosses are:
B / bB / BB / Bb
B / BB / Bb
½ normal hair, ½ hairless, matching the data.
B / bB / BB / Bb
b / Bb / bb
Because the bb zygote dies, 2/3 of offspring of the second cross are hairless, 1/3 are normal hair.
SEX-LINKAGE, EPISTASIS, PEDIGREES
14a. The mother must be a carrier, XH Xh. The son receives his X chromosome from her, and his X chromosome bears the hemophilia allele, therefore she has a recessive allele. She also has a dominant allele because she is not a hemophiliac.
b.
Xh / YXH / XH Xh / XHY
Xh / Xh Xh / XhY
Their odds are 50% for either a hemophiliac son, a hemophiliac daughter, or a hemophiliac child in general.
15.
Xr / YXR / XR Xr / XRY
X? / X? Xr / X?Y
They will certainly have some red-eyed females (XR Xr) and some red-eyed males (XRY). However, without knowing the mother’s full genotype, we don’t know whether they could also have white-eyed females and males (Xr Xr and XrY). They will if she’s heterozygous.
16. When not otherwise specified, assume that sex-linked means X-linked. Regardless, we know that the calico cat is the female, because males cannot be heterozygous for sex-linked traits. Furthermore, a Punnett Square is actually unnecessary, but here it is anyways:
Xh / YXH / XH Xh / XHY
Xh / Xh Xh / XhY
As was just stated “males cannot be heterozygous for sex-linked traits,” therefore the odds of a calico male kitten are 0%.
17. Girl: He must also be colorblind; she received a colorblind X-chromosome from him, and he only has the one. Boy: Nothing at all.
18a. To start, we know E??? x E???. Of nine offspring, seven are golden, which is closest to 75%. The parents are most likely heterozygous for E. Then, of the ee puppies, half are brown and half are black. The parents are most likely Bb x bb, although they could be Bb x Bb because a sample size of two is difficult to draw a good conclusion from. So the answer is most likely EeBb x Eebb.
b. To start, we know eeB? X E???. Not all offspring are golden, so the male must be Ee. Of non-golden offspring, about ¾ are dominant black and about ¼ are recessive brown, so both parents are most likely heterozygous (eeBb x EeBb). But the male could be homozygous recessive instead (Eebb).
19. Either autosomal dominant or X-linked recessive. All affected people have an affected parent, suggesting dominance if there is no sex-linkage. If there is sex-linkage, a woman is affected, so it’s not Y-linked. There are two girls whose father is affected though they are not; if it were X-linked dominant, they would be affected due to receiving his affected X chromosome. It could be X-linked recessive, even though it does not skip generations; sons of an affected mother are both affected, and all others on the tree are consistent with X-linked recessiveness if person I-4 is heterozygous.
20. Most likely autosomal recessive. It skips a generation (person IV-6 is affected, but no affected parent), so it can’t be dominant or Y-linked. X-linked recessive is possible but requires assuming several heterozygous women marrying into the family, so it’s less likely.
21. Autosomal dominant. All affected people have an affected parent, suggesting dominance if there is no sex-linkage. If there is sex-linkage, women are affected, so it’s not Y-linked. It can’t be X-linked dominant, girl III-9 is unaffected though her father is affected. It also can’t be recessive; examining that same girl, it would mean that her parents were XrY and XrXr; she could not be dominant and unaffected.
22. Deaf-mutism:
23.
If the parents are fully heterozygous, their genotypes are (using R and Y but other letters are permitted) RrYy x RrYy . The odds of offspring can be calculated using a dihybrid cross, or two multiplied monohybrid crosses:
RY / Ry / rY / ryRY / RRYY / RRYy / RrYY / RrYy
Ry / RRYy / RRyy / RrYy / Rryy
rY / RrYY / RrYy / rrYY / rrYy
ry / RrYy / Rryy / rrYy / rryy
Round+Yellow = 9/16 Round+Green = 3/16 Wrinkled+Yellow = 3/16 Wrinkled+Green = 1/16
OR
R / rR / RR / Rr
r / Rr / rr
Y / y
Y / YY / Yy
y / Yy / yy
Round = ¾, Wrinkled = ¼ Yellow = ¾, Green = ¼
So:
Round+Yellow = ¾ * ¾ = 9/16
Round+Green = ¾* ¼ = 3/16
Wrinkled+Yellow = ¾* ¼ = 3/16
Wrinkled+Green = ¼ * ¼ = 1/16
The expected number of offspring of each type is found by multiplying the chance of that phenotype by offspring totals, providing the following expected values (alongside the actual values given in the problem)
Phenotype / Expected / ActualRound+Yellow / 9/16 * 556 = 313 / 315
Round+Green / 3/16 * 556 =104 / 108
Wrinkled+Yellow / 3/16 * 556 =104 / 101
Wrinkled+Green / 1/16 * 556 =35 / 32
Chi-square analysis proceeds using the equation (o-e)2e
Round+Yellow = (315-313)2 / 313 = .0128
Round+Green = (108-104)2 / 104 = .154
Wrinkled+Yellow = (101-104)2 / 104 = .0866
Wrinkled+Green = (32-35)2 / 35 = .257
.0128, .154, .0866, .257 = .510
Degrees of freedom = 4-1 = 3 and p .05
The value in the chi square table for 3 degrees of freedom, p .05 is 7.82
.5104 < 7.82, so the null hypothesis is supported: it is most likely that the deviations from the values predicted by the Punnett Square are due to chance/random assortment.