Question

Use the completed Punnett square in Part B to answer the questions below about the F2 generation.

Answer 1

Complete 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.

Note: You will find the Punnet square in the attached files.  

Note that questions 3 and 4 require a more advanced evaluation of probabilities than do questions 1 and 2 because you have to consider all possible groupings and orders of three F2 seeds. If you need help with these calculations, use Hint 3.

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 along 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?  

Answer:

1) 3/4              

2) 1/3

3) 63/64

4) 27/64

Explanation:

• Y is the dominant allele and expresses yellow color

• y is the recessive allele and expresses green color
• Seeds with YY and Yy genotype are yellow
• Seeds with yy genotype are green

Parental) Yy   x   Yy

Gametes) Y   y   Y   y

Punnet Square)

                      (1/2) Y            (1/2) y

      (1/2) Y    (1/4) YY         (1/4) Yy

      (1/2) y     (1/4) yY          (1/4) yy  

F2)  1/4 YY

      2/4 = 1/2 Yy

      1/4 yy

1) There will be 3/4 of probabilities of getting a yellow F2 seed

   1/4 YY + 2/4 Yy = 3/4 Y-

   The symbol "-" means that in its position there might be either a Y or y allele.

2) As there are only three yellow possible genotypes for yellow seeds, the fourth genotype for green seeds is not considered. Our pool now is only yellow seeds. So the probability of getting a pure breeding yellow seed among all yellow seeds is 1/3.  

   1/3 YY

   1/3 Yy

   1/3 yY

In the pool of yellow seeds, there are three genotypes, and only one of them corresponds to pure breeding yellow seed.

3) To calculate the probability of getting at least one yellow seed among the three seeds randomly chosen, we need to multiply and then perform addition. This is, when we take three seeds from the pool, we have different possibilities of getting seeds with different genotypes. So:

• the three chosen seeds could be yellow. In this case, we multiply their probabilities. The probability of getting one yellow seed is 3/4, so, the probabilities of getting three yellow seeds are: 3/4 x 3/4 x 3/4 = 27/64

There might also be one green sees and two yellow seed, so we multiply their probabilities in the order in which we might get them. This is:

• Two yellow seeds and one green seed: 3/4 x 3/4 x 1/4 = 9/64
• One yellow seed, one green seed, and one yellow seed: 3/4 x 1/4 x 3/4 = 9/64
• one green seed, and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64

And we could also get two green seeds and only one yellow seed:

• Two green and one yellow seed: 1/4 x 1/4 x 3/4 = 3/64
• one green, one yellow and one green: 1/4 x 3/4x 1/4 = 3/64
• one yellow seed and two green seeds: 3/4 x 1/4 x 1/4 = 3/64

Now, having all the possibilities, we just need to add all the possibilities:

27/64 + 9/64 + 9/64 + 9/64 + 3/64 + 3/64 + 3/64= 63/64.

Note that we did not include the possibility of getting three green seeds (1/4 x 1/4 x 1/4). This is because we were asked to calculate the probabilities of getting AT LEAST one yellow seed.

4) To calculate the probability of getting two yellow seeds and one green seed among the three seeds randomly chosen, we should multiply probabilities and then sum them up.

The orders in which we could get the seeds are:

• Two yellows and one green: 3/4 x 3/4 x 1/4 = 9/64
• One yellow, one green, and one yellow: 3/4 x 1/4 x 3/4 = 9/64
• one green seed and two yellow seeds: 1/4 x 3/4 x 3/4 = 9/64

Now, having all the possibilities, we just need to add all the possibilities:

6/64 + 9/64 + 9/64 = 27/64.