Question

1. make a punnett square to cross the following: a male hamster with black fur and a long tail with a female hamster with white fur and a short tail. the male is heterozygous for both traits. white fur and a short tail are recessive traits. what are the phenotypic ratios and genotypic ratios of this dihybrid cross? (25 pts)

Explanation:

Step1: Define Alleles

Let \( B \) = black fur (dominant), \( b \) = white fur (recessive); \( L \) = long tail (dominant), \( l \) = short tail (recessive). Male: heterozygous for both, so genotype \( BbLl \). Female: white ( \( bb \)) and short tail ( \( ll \)), so genotype \( bbll \).

Step2: Determine Gametes

Male (\( BbLl \)) produces gametes: \( BL, Bl, bL, bl \) (each with equal probability). Female (\( bbll \)) produces gametes: \( bl \) (only one type).

Step3: Construct Punnett Square

	\( BL \)	\( Bl \)	\( bL \)	\( bl \)
\( bl \)	\( BbLl \)	\( Bbll \)	\( bbLl \)	\( bbll \)
\( bl \)	\( BbLl \)	\( Bbll \)	\( bbLl \)	\( bbll \)
\( bl \)	\( BbLl \)	\( Bbll \)	\( bbLl \)	\( bbll \)

Step4: Analyze Genotypes

Count each genotype:

• \( BbLl \): 4 times
• \( Bbll \): 4 times
• \( bbLl \): 4 times
• \( bbll \): 4 times

Genotypic ratio: \( BbLl : Bbll : bbLl : bbll = 1:1:1:1 \)

Step5: Analyze Phenotypes

• \( BbLl \): black, long (dominant for both)
• \( Bbll \): black, short (dominant fur, recessive tail)
• \( bbLl \): white, long (recessive fur, dominant tail)
• \( bbll \): white, short (recessive for both)

Each phenotype has 4 individuals. Phenotypic ratio: black - long : black - short : white - long : white - short \( = 1:1:1:1 \)

Answer:

Genotypic ratio: \( BbLl : Bbll : bbLl : bbll = 1:1:1:1 \)
Phenotypic ratio: black - long : black - short : white - long : white - short \( = 1:1:1:1 \)