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In mice, the waltzing allele (w) that causes the mouse to run in circles due to an inner ear defect is recessive to the non- waltzing allele(W). If a heterozygous mouse mated wit ha walting mouse. If a heterozygous mouse mated with a waltzing mouse, a). 100% of the offspring would have a copy of the recessive allele. b). 75% of the offspring would have the dominant trait. c).100% of the offspring would have the same copy of the dominant allele. d). 75% of the offspring would have the recessive trait.

 

 

 

3 years ago

Answered By Emily D

Howdy! This is a Punnett square problem but I'm going to go over some definitions before we draw our square.

Dominant - this allele will "dominate" the phenotype (displayed trait) even if you only have one copy

Recessive - this allele will be hidden by the dominant allele; you need TWO copies of it to display its phenotype

Heterozygous - the two alleles are not the same. Usually this means one dominant and one recessive copy

Homozygous - the two alleles are the same (either both dominant or both recessive)

Okay so now that we have that out of the way, let's break down the given information: The waltzing allele, w, is recessive. The non-waltzing allele, W, is dominant. Two heterozygous (Ww x Ww) are mated.

Let's draw our punnett square now:

___| W  | w  |

.W_|WW|Ww|

_w..|Ww|.ww.|

Our offspring are:

WW

Ww

Ww

ww

So we have: 75% carry the recessive allele (ww and 2Ww), 25% display the recessive trait (only ww), 75% carry the dominant allele AND display the dominant trait (WW and 2Ww), and only one has two copies of the dominant allele (WW).

The only answer this matches with is B. 75% of the offspring would have the dominant trait