Widow's peak is a dominant trait (W). Tongue-rolling is a dominant trait (T). Two people who are heterozygous for both traits (widow's peak, tong-roller) get married and have children. Complete the Punnett square to show the parents the probabilities for each child, then answer the question that follows. What is the probability that the parents will have a child without a widow's peak and unable to roll his tongue?

Group of answer choices

3/16

9/16

0/16

1/16

The correct answers are -

A. 1)  WWTt    2) WWTt   3)WwTt   4)wwTT

B. 1/16

Explanation:

We know that each gene have minimum two alleles or forms of the gene one is dominant and the other is recessive. So, it is given that W is dominant trait that represents widow's peak so the recessive allele would be w, similarly for the tongue-rolling trait dominant is T n recessive would be t.

So the dihybrid cross of heterozygous parents for both is shown in the punnet square given which lack few possibilities that can easily be carried out by the gametes mention there that are

1. WWTt  by WT and Wt gamets

2. WWTt  by WT and Wt

3)WwTt   by Wt and wT

4)wwTT by wT and wT

And the possibility of a child without a widow's peak and unable to roll his tongue would be possible only if the alleles are recessive for the both trait which is one out of 16 in punnet square thus the possibility would be - 1/16

they both go up but then go back down. they both end lower than they started. lastly, they both are competitive/the second on.

explanation:

the answer is: reproductively

i think i spelled it wrong lol

anyways hope i

-lavira: )