A black guinea pig crossed with an albino guinea pig produced twelve black offspring. When the albino was crossed with a second black animal, six blacks and six albinos were obtained. What is the best explanation for this genetic situation?
A- Albino is dominant; black is incompletely dominant.
B- Albino is recessive; black is codominant.
C- Albino is recessive; black is dominant.
D- Albino and black are codominant.

C- Albino is recessive; black is dominant

Explanation:

The gene involved in this question codes for pigment color in guinea pig. One allele codes for black color (B) and the other codes for albino color (b). According to the question, a black guinea pig when crossed with an albino guinea pig produced all black offsprings. This is because the black allele (B) is dominant over the albino allele (b) i.e. The black allele will always mask the expression of the white allele in a heterozygous state.

This is the case of the first cross. The black and albino guinea pigs were both homozygous i.e. had genotypes BB and bb respectively. Hence, they produced offsprings with a Bb genotype, which is heterozygous. Since B allele (black) is dominant over the albino allele (b), it will cover up the expression of albino color. Hence, all offsprings will appear black.

In the second cross between the same albino guinea pig and another black guinea pig, 6 black and 6 albino guinea pigs were produced. This is because the black guinea pig was not homozygous (BB), it was rather genotypically heterozygous (Bb). Hence it will produce offsprings in a 1:1 ratio consisting of 1 black: 1 albino, which is accurate in this case i.e 6:6 = 1:1

the principle of segregation describes how pairs of gene variants are separated into reproductive cells. from his data, mendel formulated the principle of segregation. we now know that the segregation of genes occurs during meiosis in eukaryotes, which is a process that produces reproductive cells called gametes.