Question

rryy x rryy (round and yellow are dominant) considering the p1 above, what are the chances that the f1 generation will have yellow, wrinkly seeds? hint: use a dihybrid cross.
7/8
1/8
1/16
3/16

Explanation:

Step1: Determine the gametes

For \(RrYy\), the possible gametes are \(RY\), \(Ry\), \(rY\), \(ry\) (using the principle of independent assortment).

Step2: Set up the dihybrid cross

We can use a Punnett - square for a dihybrid cross. The cross \(RrYy\times RrYy\) is equivalent to \((Rr\times Rr)\) and \((Yy\times Yy)\) simultaneously.
For the seed - shape gene (\(Rr\times Rr\)):
The cross \(Rr\times Rr\) gives the genotypic ratio \(RR:Rr:rr = 1:2:1\). The probability of getting \(rr\) (wrinkled, recessive) is \(\frac{1}{4}\).
For the seed - color gene (\(Yy\times Yy\)):
The cross \(Yy\times Yy\) gives the genotypic ratio \(YY:Yy:yy=1:2:1\). The probability of getting \(Y-\) (yellow, since \(Y\) is dominant) is \(\frac{3}{4}\) (\(YY + Yy=\frac{1 + 2}{4}\)).

Step3: Use the multiplication rule

Since the two traits (seed - shape and seed - color) are independent (Mendel's law of independent assortment), the probability of getting yellow (\(Y-\)) and wrinkled (\(rr\)) seeds is the product of the probabilities of each individual trait.
Let \(P(Y - rr)\) be the probability of yellow and wrinkled seeds. Using the formula \(P(A\cap B)=P(A)\times P(B)\) (where \(A\) is the event of yellow seeds and \(B\) is the event of wrinkled seeds), we have \(P(Y - rr)=\frac{3}{4}\times\frac{1}{4}\).

Answer:

\(\frac{3}{16}\)