Question

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

albinism is also inherited.
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 \(R\) - running (dominant), \(r\) - waltzing (recessive), \(B\) - brown (dominant), \(b\) - albino (recessive). The first mouse is \(RrBb\) and the second is \(rrBb\).

Step2: Determine gametes

The first mouse (\(RrBb\)) can produce \(RB\), \(Rb\), \(rB\), \(rb\) gametes. The second mouse (\(rrBb\)) can produce \(rB\), \(rb\) gametes.

Step3: Create Punnett - square

	\(RB\)	\(Rb\)	\(rB\)	\(rb\)
\(rb\)	\(RrBb\)	\(Rrbb\)	\(rrBb\)	\(rrbb\)

Step4: Calculate probabilities

• a. For running brown (\(R - B-\)): There are 3 out of 8 offspring with the \(R - B-\) genotype (\(RrBB\), \(RrBb\), \(RrBb\)), so the probability is \(\frac{3}{8}\).
• b. For running albino (\(R - bb\)): There is 1 out of 8 offspring with the \(R - bb\) genotype (\(Rrbb\)), so the probability is \(\frac{1}{8}\).
• c. For waltzing brown (\(rrB-\)): There are 3 out of 8 offspring with the \(rrB-\) genotype (\(rrBB\), \(rrBb\), \(rrBb\)), so the probability is \(\frac{3}{8}\).
• d. For waltzing albino (\(rrbb\)): There is 1 out of 8 offspring with the \(rrbb\) genotype, so the probability is \(\frac{1}{8}\).

Answer:

a. \(\frac{3}{8}\)
b. \(\frac{1}{8}\)
c. \(\frac{3}{8}\)
d. \(\frac{1}{8}\)