Question

two brown eyed parents (bb) have a baby. what is the chance the baby is blue eyed? 50% 25% 75% 0 %

Explanation:

Step1: Recall Mendelian Genetics

In Mendelian genetics, for a single - gene trait with two alleles (here, \(B\) for brown eyes, dominant; \(b\) for blue eyes, recessive), when two heterozygous parents (\(Bb\)) are crossed, we use a Punnett square. The possible gametes from each parent are \(B\) and \(b\).

Step2: Construct Punnett Square

The Punnett square for \(Bb\times Bb\) is:

	\(B\)	\(b\)
\(b\)	\(Bb\)	\(bb\)

From the Punnett square, the genotypes are \(BB\) (1/4), \(Bb\) (2/4), and \(bb\) (1/4). The phenotype for \(bb\) is blue eyes (since blue eyes are recessive, only homozygous recessive shows the blue - eye trait).

Step3: Calculate Probability

The probability of the baby having the \(bb\) genotype (and thus blue eyes) is \(\frac{1}{4}=25\%\).

Answer:

25% (corresponding to the option with 25%)