Ethel noticed that individuals from the original, pure line of bearded goats als

Question

Ethel noticed that individuals from the original, pure line of bearded goats also had straight tails, whereas the pure beardless goats had curly tails, and all traits bred true (i.e., both traits were homozygous in the parental generation). She realized she was poised to do a classic Mendelian dihybrid cross. She crossed a pure bearded, straight-tail male (BBSS) to a beardless, curly-tailed female (bbss) and got 100% bearded, straight-tailed F1 offspring. She let the F1s breed amongst themselves. Assuming Mendel’s law of independent assortment...

What possible gametes are present in the F1 individuals?

a.

BBSS, bbss

d.

BBSS, BbSs, bbss

b.

BS, Bs, bS, bs

e.

all options are correct.

c.

BS, bs

I understand this answer but what would the question have to ask, for the answer to be BS, bs?

These are the possible gametes the indivudual could pass on, but what would we answer if we were asked for the individual's gametes present in his/her genotype?

What proportion of individuals in the F2 population have beards AND curly tails?

a. 56% d. 75%
b. 19% e. 66%
c. 6%

The answer is B.19%, but I don't understand why you need to multiply the Bearded (12/16) by the curly (4/16). Why don't we just look at our punnett square and find out the percentage of indivuals that are bearded and curly?(This would give a different percentage) Isn't the question asking for individuals with BOTH together?

a.

BBSS, bbss

d.

BBSS, BbSs, bbss

b.

BS, Bs, bS, bs

e.

all options are correct.

c.

BS, bs

Explanation / Answer

The only way BS and bs would be the options would be if the genes were completely linked together (i.e. segregate together).

The Punnett square would have to be written out with every combination of possible parent alleles on each side and you would see that indeed 3/16 (or 18.75%) have a bearded and curly phenotype. It is most likely that you are drawing the Punnett Square incorrectly since it is difficult with multiple genes to get every combination and hence we multiply the probabilities instead!

Hope this helps!

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