Question

part c - using the punnett square to make predictions
use the completed punnett square in part b to answer the questions below about the f₂ generation.
drag the probabilities on the left to the blanks on the right to answer the questions. terms can be used once, more than once, or not at all.
view available hint(s)

1. what is the probability that an f₂ seed chosen at random will be yellow?
2. what is the probability that an f₂ seed chosen at random from among the yellow seeds will breed true when selfed?
3. what is the probability that three f₂ seeds chosen at random will include at least one yellow seed?
4. what is the probability that three f₂ seeds chosen at random will include one green seed and two yellow seeds?

probabilities (left): 0, 3/4, 1, 1/4, 1/2, 2/3, 27/64, 9/16, 9/64, 63/64, 1/64, 1/3

Explanation:

Step1: Recall Mendelian Genetics

In a typical Mendelian dihybrid cross (though here maybe monohybrid for seed color), the F₂ generation for a dominant (yellow) and recessive (green) trait has a phenotypic ratio of 3:1 (yellow:green). So the probability of a yellow seed ($P(Y)$) is $\frac{3}{4}$, green ($P(G)$) is $\frac{1}{4}$.

Step2: Solve Question 1 (Probability of yellow F₂ seed)

From the 3:1 ratio, yellow seeds (dominant) have probability $\frac{3}{4}$. Wait, but let's check the options. Wait, maybe it's a trihybrid? Wait, no, the options include $\frac{27}{64}$, $\frac{9}{16}$, etc. Wait, maybe the F₂ is from a trihybrid? No, seed color is monohybrid. Wait, maybe the first question: if we consider a monohybrid cross, F₂ yellow is $\frac{3}{4}$. But the options have $\frac{3}{4}$? Wait the left has 0, $\frac{3}{4}$, 1, $\frac{1}{4}$, $\frac{1}{2}$, $\frac{2}{3}$, $\frac{27}{64}$, $\frac{9}{16}$, $\frac{9}{64}$, $\frac{63}{64}$, $\frac{1}{64}$, $\frac{1}{3}$. Wait, maybe the problem is about a trihybrid? No, seed color is monohybrid. Wait, maybe the first question: probability of yellow F₂ seed. In a monohybrid cross (Yy x Yy), F₂ has YY (1/4), Yy (2/4), yy (1/4). Yellow is YY + Yy = 3/4. So answer for Q1 is $\frac{3}{4}$.

Step3: Solve Question 2 (Yellow seed breeds true when selfed)

A seed that breeds true when selfed is homozygous. Yellow homozygous is YY. In F₂, probability of YY is 1/4 (from monohybrid cross: YY:Yy:yy = 1:2:1). Wait, no: in F₂, homozygous yellow (YY) is 1/4, heterozygous (Yy) is 2/4, homozygous green (yy) is 1/4. So the probability that a yellow seed (which is either YY or Yy) is homozygous (YY) is (1/4) / (3/4) = 1/3? Wait no, wait the question is: "probability that an F₂ seed chosen at random from among the yellow seeds will breed true when selfed". So among yellow seeds (3/4 of total), the homozygous yellow (YY) is 1/4 of total. So conditional probability: P(YY | yellow) = (1/4) / (3/4) = 1/3? But the options have 1/3? Wait the left has 1/3. Wait, maybe I messed up. Wait, if it's a monohybrid cross, F₂ yellow seeds: YY (1/4), Yy (2/4). So proportion of YY among yellow is (1/4)/(3/4) = 1/3. So Q2 answer is 1/3? Wait no, wait "breeds true" means homozygous. So yellow seeds that are homozygous (YY) will breed true (selfed YY gives all YY, yellow). Yy selfed gives 3/4 yellow, 1/4 green, so doesn't breed true. So among yellow seeds (YY and Yy), the proportion that is YY is 1/3. So Q2: 1/3.

Step4: Solve Question 3 (At least one yellow in three F₂ seeds)

Probability of at least one yellow = 1 - probability of all green. Probability of green is 1/4. So P(all green) = $(\frac{1}{4})^3 = \frac{1}{64}$. Thus, P(at least one yellow) = 1 - $\frac{1}{64}$ = $\frac{63}{64}$.

Step5: Solve Question 4 (One green and two yellow in three F₂ seeds)

This is a binomial probability: $C(3,1) \times P(G) \times P(Y)^2$. $C(3,1) = 3$, $P(G) = \frac{1}{4}$, $P(Y) = \frac{3}{4}$. So $3 \times \frac{1}{4} \times (\frac{3}{4})^2 = 3 \times \frac{1}{4} \times \frac{9}{16} = \frac{27}{64}$.

Answer:

1. $\boldsymbol{\frac{3}{4}}$
2. $\boldsymbol{\frac{1}{3}}$
3. $\boldsymbol{\frac{63}{64}}$
4. $\boldsymbol{\frac{27}{64}}$

(Assuming the questions are 1 to 4 as labeled, with the respective solutions above. The first question's answer is $\frac{3}{4}$, second $\frac{1}{3}$, third $\frac{63}{64}$, fourth $\frac{27}{64}$.)