Punnett squares predict the probability of different allele combinations based on the parent genotypes. Thus, a Punnett square would predict that two heterozygous parents will produce 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive child. But what if they only have 1 child? What if they have 6 children? If they have 1 homozygous dominant and 1 homozygous recessive, does that mean that the next two children would be heterozygotes? Explain your answer.

Question

Biology

Precious

12 November, 18:08

Punnett squares predict the probability of different allele combinations based on the parent genotypes. Thus, a Punnett square would predict that two heterozygous parents will produce 1 homozygous dominant, 2 heterozygous, and 1 homozygous recessive child. But what if they only have 1 child? What if they have 6 children? If they have 1 homozygous dominant and 1 homozygous recessive, does that mean that the next two children would be heterozygotes? Explain your answer.

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Case Webster · Answer 1

If they have only one child, there is a 50% chance of being heterozygous, 25% chance of being homozygous dominant and 25% chance of being homozygous recessive. If they have 6 children, you should consider each birth separately. If they have 1 homozygous dominant and 1 homozygous recessive, the next child the next child has a 50% chance of being heterozygous.

Explanation:

We are talking about probability/proportions. Two heterozygous parents (Aa) will produce 50% of heterozygous (Aa and aA), 25% of homozygous dominant (AA) and 25% of homozygous recessive (aa). The odds are independent. The odds are independent if we consider the simplest case. If there is gene linkage and we take it into account, the situation becomes more complicated, and the odds change.