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problem 2: parents: two brown mice, one with homozygous dominant coat c…

Question

problem 2:
parents: two brown mice, one with homozygous dominant coat color and the other heterozygous for coat color.
create a punnett square and determine the possible genotypes of the offspring.
what are the phenotypes and percentage chance of each phenotype?
genotypes
phenotypes
percentage chance

Explanation:

Step1: Define allele notation

Let $B$ = dominant brown allele, $b$ = recessive non-brown allele.
Homozygous dominant parent: $BB$; Heterozygous parent: $Bb$.

Step2: Set up Punnett square

$B$$B$
$b$$Bb$$Bb$

Step3: Identify offspring genotypes

From the Punnett square, the genotypes are $BB$ and $Bb$.

Step4: Determine phenotypes and percentages

Both $BB$ and $Bb$ express the dominant brown phenotype.
Total offspring squares = 4.
$BB$ count = 2, so percentage: $\frac{2}{4} \times 100\% = 50\%$
$Bb$ count = 2, so percentage: $\frac{2}{4} \times 100\% = 50\%$
All offspring have brown coat phenotype, so 100% brown.

Answer:

Punnett Square:
$B$$B$
$b$$Bb$$Bb$
Genotypes, Phenotypes, and Percentage Chance:
Genotypes$BB$$Bb$
Percentage Chance50%50%

Overall phenotype percentage: 100% chance of brown coat offspring.