In humans, black hair is dominant over blonde hair. Using letters B and b, show using a genetic diagram, why a male parent with black hair had some children with black hair, and some children with blonde hair with his wife who had blonde hair.

Here are the solutions to your genetics problems. 1. Human Hair Color Step 1: Determine parental genotypes. Black hair (B) is dominant over blonde hair (b). The wife has blonde hair, so her genotype must be homozygous recessive: bb. The male parent has black hair, but since they have children with blonde hair (genotype bb), he must carry the recessive allele. Therefore, his genotype is heterozygous: Bb. Step 2: Construct a genetic diagram (Punnett square). Parents (P): Male (Bb) × Female (bb) Gametes: Male (B, b), Female (b, b) |c|c|c| Gametes & B & b \\ b & Bb & bb \\ b & Bb & bb \\ Step 3: Explain the offspring phenotypes. The Punnett square shows that the possible genotypes for the offspring are Bb and bb. Bb results in black hair*. bb results in blonde hair*. This demonstrates that a male parent with black hair (Bb) and a wife with blonde hair (bb) can have children with both black hair and blonde hair. 2. Cattle Coat Color Alleles: B for black coat color, W for white coat color. Pure breeding black bull: BB Pure breeding white cow: WW F1 offspring: Roan (a mix of black and white hairs, indicating codominance). a) (i) Show the genotype of the offspring (F1 generation). Step 1: Determine parental genotypes. Parents (P): Black bull (BB) × White cow (WW) Gametes: Bull (B), Cow (W) Step 2: Construct a genetic diagram (Punnett square). |c|c| Gametes & B \\ W & BW \\ Step 3: State the F1 genotype. All F1 offspring will have the genotype BW. a) (ii) Explain why neither the black bull nor the cow had their color expressed in the phenotype of the offspring. This phenomenon is called codominance. In codominance, both alleles (B for black and W for white) are fully and equally expressed in the heterozygous offspring (BW). This results in a phenotype (roan) that displays both black and white hairs, rather than a blend or one allele masking the other. b) Using a genetic diagram, predict the chances of the Second Filial (F2) offspring having black coat color if an F1 roan bull was crossed with one of the roan F1 offspring. Step 1: Determine parental genotypes for the F2 cross. Parents (F1): Roan bull (BW) × Roan cow (BW) Gametes: Bull (B, W), Cow (B, W) Step 2: Construct a genetic diagram (Punnett square). |c|c|c| Gametes & B & W \\ B & BB & BW \\ W & BW & WW \\ Step 3: Determine F2 genotypes and phenotypes. Genotypes: 1 BB : 2 BW : 1 WW Phenotypes: 1 black : 2 roan : 1 white Step 4: Predict the chances of F2 offspring having black coat color. Black coat color corresponds to the genotype BB. From the Punnett square, 1 out of 4 offspring have the BB genotype. The chance of an F2 offspring having black coat color is (1)/(4) or 25\%. Send me the next one 📸