In humans, the gene for curly hair (H) is incompletely dominant to the gene for straight hair (h). Individuals that are heterozygous (Hh) have wavy hair. Two wavy-haired heterozygous parents have a child. What is the likeliness that the child will have wavy hair?

The question is describing a autosomal type of inheritance pattern, being the dominant trait curly hair (allele H), and the recessive (non-dominant) trait straight hair (allele h). There are three phenotypes one can obtain from this allele: HH ⇒ Curly hair, Hh ⇒ Wavy hair, and hh ⇒straight hair. We are informed that two wavy-haired heterozygous parents have a child, meaning that both parents present a genotype of Hh. To obtain the probability of type of offspring, it is necessary to perform a Punnet square:

H | h

———————————

H | HH | Hh

———————————

h | Hh | hh

One can conclude that the amount of offspring with wavy-hair phenotype, will be 2/4 = 1/2 = 50%

akposevictor akposevictor · Answer 2 · 2021-11-03T21:12:29+01:00

The likeliness that the child that the two wavy-haired heterozygous parents will have will possess wavy hair is: 50%

Given the following:

gene for curly hair --> (H) (exhibits incomplete dominance)
gene for straight hair --> (h)
gene for heterozygous individuals exhibiting wavy hair --> (Hh)

Two parents that are heterozygous are: (Hh) x (Hh)

The cross between both parents will produce the following offspring as shown in the Punnett Square attached below:

1 HH - Curly hair - 25%
2 Hh - wavy hair - 50%
1 hh - straight hair - 25%

Therefore, the likeliness that the child that the two wavy-haired heterozygous parents will have will possess wavy hair is: 50%

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