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In humans, the heterozygous state of the gene (Bb), homozygous dominant (BB), or homozygous state of the gene (aa) leads to the appearance of a type of blindness. Only the genotype (Aabb) is normal. A blind woman, whose parents are sighted, is married to a person with genotype AaBb. What will be the genotypic and phenotypic constitution of the offspring?

Respuesta :

Answer:

To determine the genotypic and phenotypic constitution of the offspring, we need to consider the possible combinations of alleles from the parents.

The blind woman has the genotype aa, which means she carries two recessive alleles for the blindness gene.

Her husband has the genotype AaBb, which means he carries one dominant allele and one recessive allele for the blindness gene.

Let's consider the possible combinations of alleles that the offspring can inherit from each parent:

From the blind woman (aa):

- Each child will inherit one allele from the blind woman, so they will all receive an 'a' allele.

From the husband (AaBb):

- Each child has a 50% chance of inheriting the dominant 'A' allele and a 50% chance of inheriting the recessive 'a' allele.

- Each child also has a 50% chance of inheriting the dominant 'B' allele and a 50% chance of inheriting the recessive 'b' allele.

Now, let's consider the genotypic and phenotypic constitution of the offspring:

Genotypic Constitution:

- Each child will have the genotype Aa, inheriting one dominant 'A' allele and one recessive 'a' allele from their father.

- Each child will also have the genotype aa, inheriting one recessive 'a' allele from their mother.

Phenotypic Constitution:

- Since the blindness gene is recessive, the presence of at least one dominant 'A' allele will prevent the expression of blindness.

- Therefore, all offspring will be phenotypically normal and will not exhibit blindness.

In summary, the genotypic constitution of the offspring will be Aa and aa, while the phenotypic constitution of all the offspring will be normal, without any signs of blindness.