Question

1. cross a heterozygous running, homozygous brown mouse with a waltzing albino mouse.

a. what is the probability that an offspring will be running brown?
b. what is the probability that a mouse will be running albino?
c. what is the probability that a mouse will be waltzing brown?
d. what is the probability that a mouse will be waltzing albino?

Explanation:

Step1: Define alleles

Let \(R\) represent the dominant allele for running (waltzing is \(r\)), and \(B\) represent the dominant allele for brown (albino is \(b\)). The heterozygous running, homozygous brown mouse has the genotype \(RrBB\), and the waltzing albino mouse has the genotype \(rrbb\).

Step2: Determine gametes

The \(RrBB\) mouse can produce two types of gametes: \(RB\) and \(rB\). The \(rrbb\) mouse can produce only one type of gamete: \(rb\).

Step3: Set up Punnett - square

When we set up the Punnett - square, we have the following combinations:

	\(RB\)	\(rB\)
\(rb\)	\(RrBb\)	\(rrBb\)
\(rb\)	\(RrBb\)	\(rrBb\)
\(rb\)	\(RrBb\)	\(rrBb\)

Step4: Calculate probabilities

• a. For running brown (\(R - B-\)): The genotypes \(RrBb\) represent running brown mice. There are 4 \(RrBb\) out of 8 total offspring, so the probability is \(\frac{1}{2}\).
• b. For running albino (\(R - bb\)): Since there are no \(bb\) genotypes in the offspring, the probability is \(0\).
• c. For waltzing brown (\(rrB-\)): The genotypes \(rrBb\) represent waltzing brown mice. There are 4 \(rrBb\) out of 8 total offspring, so the probability is \(\frac{1}{2}\).
• d. For waltzing albino (\(rrbb\)): Since there are no \(rrbb\) genotypes in the offspring, the probability is \(0\).

Answer:

a. \(\frac{1}{2}\)
b. \(0\)
c. \(\frac{1}{2}\)
d. \(0\)