Pls help with 14-18 I have no idea what to do thx in advance

14a. 1 short hair and 3 long hair. 14b. 3 short hair and 1 long hair. 15. Genotype: Hh. Phenotype: hornless. 16. Genotype: Rr. Phenotype: red.

17. Father's phenotype: non-roller. Mother's phenotype: roller. Father's genotype: rr. Mother's genotype: Rr. Probability of tongue-rolling child: 1/2 or 50%.

18. 1/4 or 25%

Explanation: 14a. Set up your Punnett square with Ss on one side and ss on the other. The children will be Ss, ss, ss, and ss. 3 are long hair (ss) and 1 is short hair (Ss).

14b. Heterozygous means they are Ss. Set up the Punnett square with Ss on both sides. The children are SS, Ss, Ss, and ss. 1 is long hair (ss) and 3 are short hair (SS and Ss).

15. The parents are HH (homozygous hornless) and hh (homozygous horned). Set up a Punnett square with HH on one side and hh on the other. The offspring are all Hh. The genotype is Hh, showing the phenotype (trait) of hornless.

16. The parents are RR (homozygous for red) and rr (yellow can only be rr). Set up the Punnett square with RR and rr. The offspring are all Rr. The genotype is Rr, with a phenotype of red.

17. The father's phenotype is non-roller. His genotype is recessive (rr). The mother's phenotype is roller with a heterozygous (Rr) genotype. Set up the Punnett square with rr and Rr. The offspring are Rr, Rr, rr, and rr. Two of these are tongue rollers (Rr), making the probability 2/4=1/2=50%.

18. If a brown-eyed person had a blue-eyed parent, they have the genotype of Bb. Set up the Punnett square with Bb on both sides. The offspring are BB, Bb, Bb, and bb. Only one of these is blue-eyed (bb), making the probability 1/4=25%.

Hopefully this was helpful. I'm not sure if my explanations made complete sense.

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