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Difficulty: HardGenetics and Heredity

In guinea pigs, black coat color (BB) is dominant over white coat color (bb). Two black guinea pigs produce a white offspring. What is the probability that their next offspring will have a heterozygous genotype?

  1. A
    25%
  2. 50%Answer
  3. C
    75%
  4. D
    100%

Answer

50%
Because the two black parents produced a white offspring with genotype bbbb, each parent must carry and pass on a recessive bb allele alongside their dominant BB allele, making both parents heterozygous (BbBb). A Punnett square of Bb×BbBb \times Bb produces genotypes in the ratio 1 BB:2 Bb:1 bb1\ BB : 2\ Bb : 1\ bb. The probability of producing a heterozygous offspring (BbBb) is 2 out of 4, or 50%.

Step-by-Step Solution

1
Deduce parental genotypes from offspring phenotype
Both parents have the genotype BbBb
Since white coat color is recessive (bbbb), the white offspring must inherit one recessive allele (bb) from each parent. Because both parents display the dominant black coat, both must be heterozygous carriers (BbBb).
2
Set up a monohybrid cross for Bb×BbBb \times Bb
Genotypic distribution is 1 BB:2 Bb:1 bb1\ BB : 2\ Bb : 1\ bb
Crossing two heterozygous parents produces four equally likely gamete combinations: BBBB (1/4), BbBb (2/4), and bbbb (1/4).
3
Calculate the probability of obtaining a heterozygous genotype
2 out of 4, which equals 50%
The question specifically asks for the heterozygous genotype (BbBb), which accounts for 2 of the 4 outcomes (50%).

Key Concept

Determination of parental genotypes and calculation of genotypic ratios in a monohybrid cross
Estimated Time:1m 0s
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Topics

Genetics and Heredity

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