Question

1. a male guinea pig that was homozygous dominant for black fur and homozygous recessive for a smooth coat was bred with a female that was heterozygous for both black fur and a rough coat. select the correct genotypes in this cross: male bbrr × female bbrr male (bbrr) female (bbrr) phenotype probability list the expected number out of 16 black fur, rough coat: 8/16 black fur, smooth coat: 8/16 white fur, rough coat: 0/16 white fur, smooth coat: 0/16

Explanation:

Step1: Analyze male's gametes

The male has genotype \( BBrr \). Using the principle of segregation and independent assortment, for the \( B/b \) gene (since it is \( BB \), only \( B \) can be passed) and for the \( R/r \) gene (since it is \( rr \), only \( r \) can be passed). So male's gametes are all \( Br \).

Step2: Analyze female's gametes

The female has genotype \( BbRr \). For the \( B/b \) gene, it can produce \( B \) and \( b \) gametes; for the \( R/r \) gene, it can produce \( R \) and \( r \) gametes. Using the principle of independent assortment, the female's gametes are \( BR:Br:bR:br = 1:1:1:1 \).

Step3: Do the cross

When \( Br \) (male gamete) combines with \( BR \) (female gamete), the offspring genotype is \( BBRr \) (black fur, rough coat). When \( Br \) combines with \( Br \), the offspring genotype is \( BBrr \) (black fur, smooth coat). When \( Br \) combines with \( bR \), the offspring genotype is \( BbRr \) (black fur, rough coat). When \( Br \) combines with \( br \), the offspring genotype is \( Bbrr \) (black fur, smooth coat).
The probability of getting a genotype with \( B- \) (black fur) is \( 1 \) (since male is \( BB \) and female has \( B \) and \( b \), but \( B\times B \) and \( B\times b \) both result in \( B- \)). For the coat texture, when combining \( r \) (from male) with \( R \) (from female in half of female's gametes), the probability of rough coat (\( Rr \)) is \( \frac{1}{2} \), and the probability of smooth coat (\( rr \)) is \( \frac{1}{2} \).
The probability of black fur, rough coat: \( 1\times\frac{1}{2}=\frac{8}{16} \). The probability of black fur, smooth coat: \( 1\times\frac{1}{2}=\frac{8}{16} \). Since there is no \( bb \) (from male \( BB \)) in the offspring, the probability of white fur (\( bb \)) is \( 0 \). So white fur, rough coat and white fur, smooth coat probabilities are \( 0 \).

Answer:

Black fur, rough coat: \( 8/16 \)
Black fur, smooth coat: \( 8/16 \)
White fur, rough coat: \( 0/16 \)
White fur, smooth coat: \( 0/16 \)