27. Part A. Make a punnett square (as best as you can in this online format) and cross parents both with the thumb genes Aa (A=normal thumb, a=hitchhikers thumb) and show the results. (1 pt)
A a
A AA Aa
a Aa aa
27 Part B. (1 pt) Approximately what percentage of their children would have a hitchhikers thumb?
Approximately 25% percent of their children would have a hitchhikers thumb.
27 Part C. (1 pt) What percentage of their offspring shown in your punnett square COULD have a child with a hitchhikers thumb assuming they reproduced with someone carrying the "a" allele? Note: EC may be awarded for well detailed answers!
Please check my punnet square
A a
A AA Aa
a Aa aa
and if the percentage I cam up with is correct for part b. I really don't understand part c so please explain that to me!!! Thank you!
a, and b parts are ok
part c
aa x Aa
offspring then aA,aa, aA, aa or half will have aa
To answer part C, we need to consider the genotypes of the offspring shown in the punnett square.
From the punnett square, we can see that there are four possible combinations of genotypes: AA, Aa, Aa, and aa.
Out of these, only the individuals with the genotype Aa can pass on the "a" allele to their offspring, which is necessary for the child to have a hitchhiker's thumb.
So, out of the four possible genotypes, two (AA and aa) cannot have a child with a hitchhiker's thumb.
Therefore, 50% of the offspring shown in the punnett square have the potential to have a child with a hitchhiker's thumb if they reproduce with someone carrying the "a" allele.