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if a man is homozygous dominant for bent fifth finger and a woman is homozygous recessive. What is the probability that any offspring will have a bent fifth finger?

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Dgzhearts

The probability of the offspring having a bent finger is 100% we can calculate this using a punnet square like one attached, where B is the dominant bent fifth finger and b is the recessive bent fifth finger, the man's genes are along the top and the woman's genes are along the side.

Ver imagen Dgzhearts

Answer:

100 %

All the offsprings will have bent fifth finger if dominant trait is expressed

Explanation:

Let "F" represents the dominant bent fifth finger trait while

"f" represents the recessive trait.

Now , the genotype of parents is as follows -

a) Man with homozygous dominant  bent fifth finger "FF"

b) Woman with homozygous recessive trait "ff"

When a cross is carried out between these two parents, the following offsprings will result -

FF x ff

Ff, Ff, Ff, Ff

All the offsprings are heterpzygous with two paired allele of which one is dominant and the other is recessive.

Thus, all the offsprings will have bent fifth finger if dominant trait is expressed.

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