Final answer:
A homozygous dominant brown mouse crossed with a heterozygous brown mouse will yield offspring with a genotypic ratio of 1:1 (1 BB:1 Bb) using a Punnett square, while all presenting the brown phenotype since 'B' is dominant.
Step-by-step explanation:
The student's question pertains to an exercise involving a genetic cross using a Punnett square. In this cross, a homozygous dominant brown mouse (presumably with the genotype BB for the trait in question, where 'B' is the dominant brown allele and 'b' would be the recessive tan allele) is crossed with a heterozygous brown mouse (with the genotype Bb).
To set up a Punnett square for this cross, we list the alleles from one parent across the top and the alleles from the other parent down the side. In this case, the top row will have 'B' from the homozygous parent, while the side will have 'B' and 'b' from the heterozygous parent. When we fill the interior squares, it shows us the possible combinations of alleles that the offspring could have. Since one parent is homozygous dominant, all of the offspring will receive a 'B' allele and thus will have a brown phenotype. The genotypic ratio would be 1 BB:1 Bb, reflecting the fact that 50% of the offspring will be homozygous dominant (BB) and the other 50% heterozygous (Bb), but all will show the dominant brown phenotype.