In sheep, the allele for belly fur (A) is dominant to the allele for no belly fur (a). A mother with the genotype Aa and a fathe
r with the genotype Aa produce an offspring.
2 answers:
Answer:
THE COMPLETE QUESTION IS:
In sheep, the allele for belly fur (A) is dominant to the allele for no belly fur (a). A mother with the genotype Aa and a father with the genotype Aa produce an offspring. What is the percent chance that the offspring will have no belly fur?
A) 0% B) 100% C) 50% D) 75% E) 25%
Correct answer is E (25%)
Explanation:
This cross is a typical monohybrid cross because it involves a single gene coding for belly fur in sheep. (A)
According to the question, the allele coding for presence of belly fur (A) is dominant over the allele coding for no belly fur (a) i.e. According to Mendel's law of dominance, The allele A is masking the expression of allele a, hence allele A is the dominant allele while allele a is the recessive allele.
Both parents in this question are heterozygous i.e. they possess a combination of two different alleles in their genotype (Aa), but since the allele A will mask allele a, the parent sheeps will be phenotypically belly-furred.
In a cross between a Father (Aa) and Mother (Aa), each allele will segregate into gametes, according to Mendel's law of segregation. (See attached image).
Four possible offsprings will be produced with the following genotypes:
1 AA (homozygous dominant) phenotypically with belly fur
2 Aa (heterozygous dominant) phenotypically with belly fur
1 aa (homozygous recessive) phenotypically with no belly fur
Hence, according to the question, the percent chance that an offspring will have no belly fur (aa) is 1/4 × 100= 25%
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