Jamal and Martina both have freckles, but their son Maddox does not. Show how this is possible in a Punnett square. If Jamal and Martina have more children, what is the probability that each will have freckles?

This is a Biology question.

Probability that each additional child that Jamal and Martina will have freckles is 3/4.

Step-by-step explanation:

The possession of freckles in humans is a dominant trait (FF or Ff) while no freckles is a recessive trait (ff). This means that in the gene responsible for freckles, an individual must possess the dominant allele (F) to possess it. First of all, it is important to note that Jamal and Martina produced a son without freckles (recessive trait) because both parents are heterozygous for the trait i.e. they possess a combination of the dominant (freckles) allele and the recessive (no freckles) allele.

In that case, when Jamal and Martina come together who are both heterozygous (Ff), they produce gametes that have alleles F and f. Using a punnet square (see attachment), four possible offsprings will be produced with genotypes: FF, Ff, Ff, and ff.

The genotypes FF, Ff and Ff will all possess freckles phenotypically while tje genotype ff will not possess freckle. Hence, Maddox by a 1/4 chance must have fallen into this category with genotype (ff).

According to the illustration above, the probability that Jamal and Martina will produce offsprings with freckles is 3/4 (FF, Ff, Ff) while the probability that they will produce offsprings with no freckles is 1/4 (ff). Hence, if they have more children, the probability that each child will possess freckles is 3/4.

See attachment for the punnet square.

step-by-step explanation:

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when the line is dashed:

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about the problem:

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