In the genetics sample problem 'Feather Color and Beak Shape in Birds', where red feathers (R) are dominant over blue (r) and sharp beaks (B) are dominant over blunt (b), what percentage of offspring from the cross RrBb × rrBb will have blue feathers and a blunt beak?

What you are looking for in this cross

You want offspring that show both recessive phenotypes: blue feathers and blunt beak. That means each offspring must be $rr$ (blue) and $bb$ (blunt), so the target genotype is $rrbb$.

Split it into two separate monohybrid crosses

Because feather color and beak shape are controlled by different genes, you can find each probability separately and then multiply.

Feather color gene ($R/r$)

Cross: $Rr \times rr$

Possible offspring at the $R$ locus:

• $Rr$ (red)
• $rr$ (blue)

So, $$P(rr)=\frac{1}{2}$$

Beak shape gene ($B/b$)

Cross: $Bb \times Bb$

Genotype ratio at the $B$ locus:

• $BB$: $Bb$: $bb$ = $1:2:1$

So, $$P(bb)=\frac{1}{4}$$

Combine the probabilities (both traits at once)

To have blue feathers and a blunt beak, the offspring must be both $rr$ and $bb$: $$P(rrbb)=P(rr)\cdot P(bb)=\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}=0.125$$

So the percentage is: $$0.125 \times 100\% = 12.5\%$$

Quick check using 8 equally likely outcomes

You can also think of 2 possible outcomes for feather color (from $Rr \times rr$) and 4 for beak shape (from $Bb \times Bb$), giving $2 \times 4 = 8$ combined outcomes. Only 1 of those 8 is $rrbb$, so $1/8 = 12.5\%$.