Can someone me, i will give you brainliest.
there is a pdf attachment with the questions
i really need your

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).

1. b.

2. d.

3. a.

5. d.

6. c.

these are the only ones that i know, i hope this , and good luck.