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The allele resulting in straight hair (H) and the allele resulting in curly hair (H) show incomplete dominance. Therefore a heterozygote has -you probably guessed it -wavy hair! (HH'). A man who has straight hair and a woman who has curly hair marry. What are the expected phenotypes and genotypes of their children?

Sagot :

Tessyaelvidn

Answer:

All their children will have the genotype HH' and they will all have wavy hair.

Explanation:

The man must have the genotype HH because only the genotype HH will be expressed as straight hair, since people with the genotype HH' (heterozygotes) have wavy hair instead of either straight or curly. By this logic, the woman must have the genotype H'H'.

Using the Punnet square, we can see that all four boxes have the outcome of HH', meaning that the couple have a 100% chance of having children with the genotype HH', and all their children will have wavy hair.

View image Tessyaelv
TerryMarthaidn

All their children are expected to have wavy hair (Hh) phenotypes.

Let's represent straight hair allele as H and curly hair allele as h.

Given that straight hair (H) and curly hair (h) alleles show incomplete dominance:

    • Straight hair (HH) is homozygous dominant.

    • Curly hair (hh) is homozygous recessive.

    • Wavy hair (Hh) is the heterozygote, which shows the intermediate phenotype.

Now, let's consider the offspring of a man with straight hair (HH) and a woman with curly hair (hh):

The man's genotype: HH (straight hair)

The woman's genotype: hh (curly hair)

Their possible combinations in the offspring:

    1. H from the father (HH) and h from the mother (hh) -> Hh (wavy hair)

    2. H from the father (HH) and h from the mother (hh) -> Hh (wavy hair)

    3. H from the father (HH) and h from the mother (hh) -> Hh (wavy hair)

    4. H from the father (HH) and h from the mother (hh) -> Hh (wavy hair)

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