in a dihybrid cross for round and yellow seeds (rryy x rryy), what is the probability of having green and…

in a dihybrid cross for round and yellow seeds (rryy x rryy), what is the probability of having green and wrinkled seeds?\nkey: r = round\nr = wrinkled\ny = yellow\ny = green\n1:16\n3:16\n8:16\n9:16

in a dihybrid cross for round and yellow seeds (rryy x rryy), what is the probability of having green and wrinkled seeds?\nkey: r = round\nr = wrinkled\ny = yellow\ny = green\n1:16\n3:16\n8:16\n9:16

Answer

Explanation:

Step1: Analyze the cross for each trait separately

For the seed shape trait ((Rr\times Rr)): The Punnett - square gives genotypes (RR:Rr:rr = 1:2:1). The probability of getting (rr) (wrinkled) is (\frac{1}{4}). For the seed color trait ((Yy\times Yy)): The Punnett - square gives genotypes (YY:Yy:yy=1:2:1). The probability of getting (yy) (green) is (\frac{1}{4}).

Step2: Use the multiplication rule for independent events

Since the two traits (seed shape and seed color) are independent (Mendel's law of independent assortment), the probability of getting both (rr) and (yy) is the product of their individual probabilities. Let (P(rr)) be the probability of wrinkled seeds and (P(yy)) be the probability of green seeds. Then (P(rr\cap yy)=P(rr)\times P(yy)) Substitute (P(rr)=\frac{1}{4}) and (P(yy)=\frac{1}{4}) into the formula: (P(rr\cap yy)=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16})

Answer:

1:16