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Oksi-84 [34.3K]
4 years ago
14

In a diploid individual, one chromosome carries A and B genes, and the homologous chromosome carries different forms (alleles) o

f these same genes, a and b. If there is a single crossover between these two genes involving non-sister chromatids during metaphase I of meiosis, the resulting four gametes are:
A. AB, AB, ab, ab.
B. AB, ab, AB, ab.
C. AaBb, AaBb, AaBb, AaBb.
D. AB, Ab, aB, ab.
E. Ab, Ab, aB, aB.
Biology
1 answer:
Helen [10]4 years ago
4 0

Answer:

D

Explanation:

This involves a dihybrid inheritance I.e. two genes are being passed on. During meiosis, specifically, the Prophase stage, homologous chromosomes (similar but non-identical chromosomes received from each parent) line side by side. According to the question, one chromosome contains A and B alleles and its homologue, received by the other parent carries a and b alleles. This means that the diploid individual has a genotype AaBb for that gene.

According to Mendel's law of independent assortment, the alleles separate independently of one another into gametes. I.e. allele A and a separates into the gametes without affecting alleles B and b of the other gene.

Crossing-over, which is the exchange of chromosomal segment occurs between the two homologues. Hence, the exchange of chromosomal segments containing alleles in the individual will possibly produce four gametes with the genotypes: AB, Ab, aB, ab.

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