Answer sheet attached for punnett square questions
1. Striped = TT, Speckled = PP, Striped and speckled = PT
2. Punnett square shown in attachment
a) 100% PT and therefore 100% striped and speckled.
b) 0%
c) 0%
d) 100%
3. Punnett square shown in attachment
a) PP, PT, TT, striped, speckled, or striped and speckled
b) 25%
c) 25%
d) 50%
4. Punnett square shown in attachment
a) PT and TT, striped and speckled and striped
b) Striped - 50%, speckled - 0%, striped and speckled 50%.
5. Punnett square shown in attachment
a) PP, YY and PY. Pink, yellow and spotted.
b) 25%
c) 25%
d) 50%
6. Punnett square shown in attachment
a) The possible genotypes are YY and PY, yellow and spotted or "juju".
b) 50%
c) 0%
d) 50%
7. Punnett square attached, Yellow - 0%, Pink 0%, Juju - 100%
8. Punnett square attached, Yellow - 0%, Pink 50%, Juju - 50%
Explanation:
1. Since the traits are codominant, the expression of one doesn't mask the expression of the other. Therefore to be purely striped, the snail would need to have 2 copies of the striped allele (TT). To be speckled, would need two copies of the speckled allele (PP). To be both, would have one of each (PT)
2. a-d) Gary is striped, so must have the phenotype TT. A speckled snail would have the genotype PP. The punnett square (attached) shows all the offspring would have the genotype PT, and would therefore all be striped and speckled.
3. a -d). The punnett square showing the cross is attached. The resulting genotypes are 1 TT: 2 TP: 1 PP. That means 25% of the offspring will be TT, (striped), 25% of the offspring will be PP (speckled), and 50% of the offspring will be TP (speckled and striped).
4. a -b). The punnett square showing the cross is attached (TT x TP). The possible genotypes are TT or TP, with the ratios being 50:50. This means 50% of the offspring will be striped and 50% will be striped and speckled.
5. a-d) Spotted jellyfish have the genotype PY. Therefore, the cross between 2 spotted jellyfish is PY x PY. The punnett square is attached. The possible genotypes are PP, PY and YY, and therefore the possible genotypes are pink, spotted, and yellow. The ratios are 1 PP: 2 PY: 1 YY. Therefore the percentages are 50% PY (spotted), 25% PP (pink) and 25% yellow (YY).
6. a-d) A yellow jellyfish has the genotype YY. A "Juju" (spotted) has the phenotype PY. The cross is shown in the attached punnett square. The cross would produce 50% YY and 50% PY genotypes, and therefore 50% yellow and 50% spotted or "Juju" phenotypes.
7. The punnett square for this problem is attached. A pink jellyfish has the genotype PP, a yellow jellyfish has the genotype YY. The punnett square shows that all the jellyfish are PY. If 100 jellyfish were produced, you would get 100% jujus (PY)
8. The punnett square for this problem is attached. A pink jellyfish has the genotype PP. A Juju has the genotype PY. The resulting offspring are PP or PY, with 50% having the PP genotype and 50% the PY. Therefore 0% would be yellow, 50% would be pink (PP), and 50% would be juju (PY).
