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m_a_m_a [10]
2 years ago
7

Tongue-curling in humans is a dominant genetic trait (T). Suppose a man who is heterozygous for tongue-curling marries a woman w

ho is also heterozygous for this trait. Create a Punnett Square. What percentage of the offspring will be phenotypic tongue-curlers? and What is the genotypic ratio (different genotype possibility)

Biology
2 answers:
Monica [59]2 years ago
6 0

Explanation:

genotypic ratio:-1:2:1

TT: 25%

Tt: 50%

tt: 25%

the percentage of phenotypic tongue curlers is: 75%

pav-90 [236]2 years ago
5 0

Answer: The percentage of the offspring will be phenotypic tongue-curlers is 75% and the genotypic ratio is 1TT : 2Tt : 1tt

Explanation: Since tongue-curling in humans is a dominant genetic trait (T), then non tongue-curling is recessive trait and is represented as (t).

A man and woman who are heterozygous for tongue-curling will have genotype Tt. A cross between two heterozygous individuals will produce four offsprings, Tt x Tt = TT, Tt, Tt and tt.

The genotype ratio is 1TT : 2Tt : 1tt.

Since (T) is dominant, genotypes TT and Tt will manifest phenotypically as tongue-curling, while tt will manifest phenotypically as non tongue-curling.

Since three out of four offsprings are phenotypic tongue-curlers, the percentage of the offspring that are tongue-curlers is 3/4 x 100 = 75%

See the attached punnet square for more information

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Answer:

p = 0.34

Explanation:

The green allele is recessive, meaning two copies of q (qq) are required to be green. Conversely, animals that are either pp or pq will be blue.

If 44 organisms are green, that means 44 are qq.

For genotype frequencies, the equation is:

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Which is denoted as

p² + 2pq + q² = 1

We know that q² = 44/100 = 0.44

To work out q, we can do \sqrt{0.44} = 0.66

For allele frequencies, the total must add up to 100%, so

p + q =1

We know that q= 0.66

So p = 0.34, because 0.66 + 0.34 = 1

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