A man heterozygous for albino gene marries a woman who is also heterozygous for the gene. Both have normal skin colour. The probability that they will have ...

Answer Details

The probability that a man heterozygous for albino gene marries a woman who is also heterozygous for the gene and they will have an albino child is "1/4."

The man is heterozygous for the albino gene, which means he has one dominant allele for normal skin color and one recessive allele for albino skin color. The woman is also heterozygous for the same gene. Therefore, both have a 50% chance of passing on the dominant allele and a 50% chance of passing on the recessive allele to their offspring.

To determine the probability of having an albino child, we need to use a Punnett square.

      A     a
   --------------
A | AA  |  Aa |
  |-------|------|
a | Aa  |  aa |
   --------------

In the Punnett square above, we have listed all possible allele combinations that can be passed on from the parents to their offspring. The uppercase A represents the dominant allele for normal skin color, and the lowercase a represents the recessive allele for albino skin color.

From the Punnett square, we can see that there is a 25% chance (1/4) of the couple having an albino child (represented by the aa genotype), and a 75% chance (3/4) of having a child with normal skin color.

Therefore, the probability that the man and woman will have an albino child is "1/4."