Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

Question

Harley Tran

Biology

20 November, 05:00

Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

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Maleah Washington · Answer 1

The independent assortment of genes is a principle of Mendel and of genetics. Since humans are diploid organisms (they have 2 copies of their genetic information), we have that each gamete carries one of the two available alleles for each feature. These gametes are created in equal proportions.

We have that both parents are heterozygous for a feature, lets say H. Thus, their genotype is Hh. Hence, the child will get with probability 50%=1/2 H from the father and 50%=1/2 h from the father; same from the mother. By doing a Punnett square, we get that there is 1/4 chance that the child is HH, 1/2 chance that it is Hh and 1/4 that is hh. Since the disease is recessive (so both alleles are needed), we have that the chance that the kid has the disease is 1/4.