April Quijano
12/30/07
Answers to Genetics Packet
2) Flower position, stem length, and seed shape were three characters that Mendel studied. Each is
controlled by an independently assorting gene and has dominant and recessive expression. If a plant
that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring
would you expect to be as follows?
a) homozygous for the three dominant traits
AATTRR = ¼ x ¼ x ¼ = 1/64
b) homozygous for the three recessive traits
aattrr= ¼ x ¼ x ¼ = 1/64
c) heterozygous for all three characters
AaTtRr = ½ x ½ x ½ = 1/8
d) homozygous for axial and tall, heterozygous for seed shape
AATTRr = ¼ x ¼ x ½ = 1/32
3) A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino is
crossed with a second black one, 7 blacks and 5 albinos are obtained. What is the best explanation for
this genetic situation? Write genotypes for the parents, gametes, and offspring.
The best explanation is that at the first mating both were homozygous of their alleles. But in the
second mating, the black one was heterozygous of their alleles. At the first mating, the black one’s
genotype was BB, while the albino was bb. The gametes were B and b, and the offspring had the
genotypes Bb. On the second mating, while the albino was bb, the black one was Bb. The gametes were
b and B, and the offspring created both Bb and bb.
4) In sesame plants, the one-pod condition (P) is dominant to the three pod condition (p), and normal
leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently. Determine
the genotype for the two parents for all possible matings producing the following offspring.
a) 318 one pod, normal leaf: 98 one pod, winkled leaf.
The one pod gene must be either homozygous dominant or heterozygous. Therefore the
parents can both be PPLl, or one can be PpLl and the other be PPLl.
b) 323 three pod, normal leaf: 106 three pod, wrinkled leaf.
The pod number allele has to be homozygous recessive in both parents. Also, to make sure most
of the offspring is has three pod, normal leaf alleles, the parents have to be both ppLl.
c) 401 one pod, normal leaf.
To make that all offspring have at least the dominant allele for both traits and not have the
recessive allele showing, the parents can have be heterozygous for only one trait. Therefore, the parents
can be both PPLL, or PPLL with either PpLL or PPLl. PpLL and PPLL cannot cross with each other.
d) 150 one pod, normal leaf: 147 one pod, wrinkled leaf: 51 three pod, normal leaf: 48 three pod,
wrinkled leaf
Since the majority of the offspring have the dominant allele for the pod number, the parents
should be heterozygous for that allele (Pp). However, since a lot of the offspring have the recessive
allele for the leaf type, it seems that one of the parents must be homozygous for the leaf type. So the
parents must be PpLl and Ppll.
e) 223 one pod, normal leaf: 72 one pod, wrinkled leaf: 51 three pod normal leaf: 27 three pod,
wrinkled leaf.
Since most of the offspring has both dominant alleles, the parents can be only both PpLl. Doing
other combinations usually leaves the other three kinds having very small numbers.