Question

1. in humans, assume that spotted skin (d) is dominant over non - spotted skin (d) and that wooly hair (h) is dominant over non - wooly hair (h). cross a marriage between a heterozygous spotted, non - wooly man with a heterozygous wooly - haired, non - spotted woman. give genotypic and phenotypic ratios of offspring.
2. in horses, black is dependent upon a dominant gene (b) and chestnut upon its recessive allele (b). the trotting gait is due to a dominant gene (t) and the pacing gait to its recessive allele (t). a homozygous black pacer is mated to a homozygous chestnut trotter. give the genotypic and phenotypic ratios.

Explanation:

Problem 1 (Human Traits Cross)

Step1: Determine Parental Genotypes

• Man: Heterozygous spotted (Dd), non - wooly (hh) → Genotype: \(Ddhh\)
• Woman: Heterozygous wooly - haired (Hh), non - spotted (dd) → Genotype: \(ddHh\)

Step2: Determine Parental Gametes

• Man (\(Ddhh\)): Produces gametes by segregating alleles. For D/d: D or d; for H/h: only h (since hh). So gametes: \(Dh\), \(dh\) (each with frequency 1/2)
• Woman (\(ddHh\)): For D/d: only d; for H/h: H or h. So gametes: \(dH\), \(dh\) (each with frequency 1/2)

Step3: Construct Punnett Square

	\(Dh\)	\(dh\)
\(dh\)	\(Ddhh\)	\(ddhh\)

Step4: Count Genotypes and Phenotypes

• Genotypes:
• \(DdHh\): 1
• \(ddHh\): 1
• \(Ddhh\): 1
• \(ddhh\): 1
• Phenotypes:
• Spotted, wooly (\(DdHh\)): 1
• Non - spotted, wooly (\(ddHh\)): 1
• Spotted, non - wooly (\(Ddhh\)): 1
• Non - spotted, non - wooly (\(ddhh\)): 1

Step5: Calculate Ratios

• Genotypic Ratio: \(DdHh:ddHh:Ddhh:ddhh = 1:1:1:1\)
• Phenotypic Ratio: Spotted wooly : Non - spotted wooly : Spotted non - wooly : Non - spotted non - wooly \(= 1:1:1:1\)

Problem 2 (Horse Traits Cross)

Step1: Determine Parental Genotypes

• P1 (homozygous black pacer): Black is dominant (B), pacer is recessive (t). So genotype: \(BBtt\)
• P2 (homozygous chestnut trotter): Chestnut is recessive (b), trotter is dominant (T). So genotype: \(bbTT\)

Step2: Determine Parental Gametes

• P1 (\(BBtt\)): For B/b: only B; for T/t: only t. So gametes: \(Bt\) (all gametes are \(Bt\))
• P2 (\(bbTT\)): For B/b: only b; for T/t: only T. So gametes: \(bT\) (all gametes are \(bT\))

Step3: Construct Punnett Square

	\(Bt\)

Step4: Count Genotypes and Phenotypes

• Genotype: \(BbTt\) (all offspring)
• Phenotype: Black trotter (since B is dominant for black, T is dominant for trotter)

Step5: Calculate Ratios

• Genotypic Ratio: \(BbTt: - - - = 1:0\) (since all are \(BbTt\))
• Phenotypic Ratio: Black trotter : - - - \(= 1:0\) (all are black trotters)

Answer:

(Problem 1):

• Genotypic Ratio: \(DdHh:ddHh:Ddhh:ddhh = 1:1:1:1\)
• Phenotypic Ratio: Spotted wooly : Non - spotted wooly : Spotted non - wooly : Non - spotted non - wooly \(= 1:1:1:1\)