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.

p1 genotype____ p1 gametes , , , __
p2 genotype____ p2 gametes , , , __

Explanation:

Step1: Determine Parental Genotypes

The man is heterozygous spotted (Dd) and non - wooly (hh), so his genotype is $Ddhh$. The woman is heterozygous wooly - haired (Hh) and non - spotted (dd), so her genotype is $ddHh$.

Step2: Determine Parental Gametes

For the man ($Ddhh$), using the law of segregation, the possible gametes are formed by combining the alleles for each trait. For the D/d trait: D and d; for the H/h trait: only h (since it's hh). So the gametes are $Dh$ and $dh$.
For the woman ($ddHh$), for the D/d trait: only d (since it's dd); for the H/h trait: H and h. So the gametes are $dH$ and $dh$.

Step3: Create Punnett Square

	$Dh$	$dh$
$dh$	$Ddhh$	$ddhh$

Step4: Analyze Genotypes and Phenotypes

Genotypes:

• $DdHh$: 1
• $ddHh$: 1
• $Ddhh$: 1
• $ddhh$: 1

So the genotypic ratio is $DdHh:ddHh:Ddhh:ddhh = 1:1:1:1$.

Phenotypes:

• $DdHh$: Spotted, Wooly - haired
• $ddHh$: Non - spotted, Wooly - haired
• $Ddhh$: Spotted, Non - wooly
• $ddhh$: Non - spotted, Non - wooly

Each phenotype has a count of 1, so the phenotypic ratio is Spotted Wooly: Non - spotted Wooly: Spotted Non - wooly: Non - spotted Non - wooly $= 1:1:1:1$.

Answer:

Genotypic ratio: $DdHh:ddHh:Ddhh:ddhh = 1:1:1:1$; Phenotypic ratio: Spotted Wooly: Non - spotted Wooly: Spotted Non - wooly: Non - spotted Non - wooly $= 1:1:1:1$