In minks, the gene for brown fur (B) is dominant to the gene for silver fur (b). Which set of genotypes represents a cross that could produce offspring with silver fur from parents that both have brown fur?
A. Bb × Bb
B. BB × Bb
C. BB × bb
D. Bb × bb
E. none of the above
Answer (3)
The set of **Bb × Bb **genotypes represents a cross that could produce offspring with silver fur from parents that both have brown fur. The correct option is A .
What is genotype?
An organism's genotype is made up of all of its genetic components. The alleles or variations that an individual carries in a specific gene or genetic region are also referred to as the genotype.
The clinical presentation of a patient is the observable expression of this genotype, or phenotype .
The brown fur gene ( B ) is dominant over the silver fur gene in minks ( b ).
The parents' phenotypic is brown , which indicates that they have at least one allele for brown fur (B).
While both parents have brown fur, the " bb " gene produces silver fur. The genotype " Bb " is shared by both parents , who are heterozygous .
Thus, the correct option is A .
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The set of genotypes that could produce offspring with silver fur from parents that both have brown fur is (A) Bb imes Bb. This cross can result in offspring with a genotype of bb, resulting in silver fur, with a 25% probability using a Punnett square analysis.
In minks, the gene for brown fur (B) is dominant to the gene for silver fur (b). To produce offspring with silver fur from parents that both have brown fur, at least one parent must carry the recessive (b) allele. Since silver fur is recessive, both alleles in the offspring must be silver (bb) to express this trait.
The only set of genotypes from the options that represents a cross that could produce offspring with silver fur from parents that both have brown fur is (A) Bb imes Bb. When both parents are heterozygous for the brown fur gene (Bb), there is a possibility for the offspring to receive one recessive allele (b) from each parent, resulting in the genotype bb, which is silver fur. A Punnett square analysis of this cross would show that there is a 25% chance for offspring to be homozygous recessive (bb) and thus have silver fur.
To produce silver fur offspring (bb) from parents both with brown fur (B), the genotype Bb x Bb is suitable as it includes the possibility of getting bb offspring. The correct answer from the options provided is A (Bb × Bb) .
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