Question

in the cross bb x bb (two heterozygous mice), what percentage of the offspring would you predict to have the same phenotype as the parents? 0% 25% 50% 75% 100% next, you perform a cross with a mouse with black fur and a mouse with white fur. which result(s) for the offspring would confirm that the mouse with black fur was heterozygous? 4 mice with black fur 8 mice with black fur, 1 mouse with white fur 5 mice with black fur, 6 mice with white fur 6 mice with black fur, 2 mice with white fur

Explanation:

Step1: Set up Punnett - square

The cross is Bb x Bb. The possible gametes from each parent are B and b. The Punnett - square has 4 cells: BB, Bb, Bb, bb.

Step2: Determine phenotypes

Assume B is for black fur (dominant) and b is for white fur (recessive). BB and Bb have the black - fur phenotype, while bb has the white - fur phenotype.

Step3: Calculate percentage

There are 3 out of 4 genotypes (BB, Bb, Bb) that result in the black - fur phenotype (same as parents). So the percentage is $\frac{3}{4}\times100\% = 75\%$.

For the second part:
A cross between a black - fur mouse (B_) and a white - fur mouse (bb). If the black - fur mouse is heterozygous (Bb), the cross is Bb x bb. The expected ratio of offspring is 1 black - fur (Bb) : 1 white - fur (bb).

• Option A (4 mice with black fur) does not confirm heterozygosity as all could be from a homozygous dominant cross (BB x bb).
• Option B (8 mice with black fur, 1 mouse with white fur) is not a typical ratio for a Bb x bb cross.
• Option C (5 mice with black fur, 6 mice with white fur) is close to a 1:1 ratio, which would confirm heterozygosity.
• Option D (6 mice with black fur, 2 mice with white fur) is a 3:1 ratio, which would suggest the black - fur parent is homozygous dominant (BB).

Answer:

First part: D. 75%
Second part: C. 5 mice with black fur, 6 mice with white fur