Question

A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring.
Black appears dominant to albino
B - black (both BB and Bb genotypes)
b - albino (bb genotype)
This is a problem showing that 1 phenotype (black color) can be the result of two different genotypes (BB and Bb). These two crosses would be the equivalent of test crosses to determine genotype.
A black guinea pig crossed with an albino guinea pig produced 12 black offspring. This is the result (all black offspring) you would expect when breeding two true breeders (Two homozygotes BB x bb).
B B
b Bb Bb
b Bb Bb
When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained. This looks like the results you would expect between a heterozygote and an albino (Bb x bb).
B b
b Bb bb
b Bb bb

Answer 1

Answer:

A black guinea pig crossed with an albino guinea pig producing 12 black offspring

         b         b

B     Bb       Bb

B     Bb        Bb

The genotype of the black gunea pig will be BB in this situation and the genotype for the albino pig would be bb.

The gametes formed will be B, B,b, b.

The offsprings will have genotype Bb. They will be heterozygous black.

When the albino was crossed with a second black one, 7 blacks and 5 albinos were obtained

         b            b

B      Bb           Bb

b       bb          bb

The genotype of the black parent will be heterozygous i.e Bb in this condition. The genotype of the albino parent will be bb.

The gametes formed will be B,b,b,b.

50% of the offsprings will have the probability of having heterozygous Bb genotype. 50% of the offsprings will have the probability of having albino bb genotype.