Question

A geneticist is studying the inheritance of two traits, let's just call them G and H. Each has two alleles and displays simple dominant/recessive phenotypes. They think that the two traits are independently assorting. The possible phenotypes of the offspring are dominant for G and H, dominant for G and recessive for H, recessive for G and dominant for H, and recessive for both G and H.

1. What is the predicted ratio of phenotypes of the offspring if these two traits are independent? input your answer as #:#:etc.

They cross two individuals who are heterozygous for both traits a number of times and end up producing 463 total offspring.  

If the traits are independent how many offspring would be expected to be:(remember you can have parts of offspring so round each to the nearest whole number, they should add up to 463)

2. Dominant for both traits?

3. Dominant for G and recessive for H?

4. Dominant for G and recessive for H?

5. Recessive for both traits?

What they observed was that 225 of the offspring have are dominant for G and H, 93 dominant for G and recessive for H, 106 recessive for G and dominant for H and 39 are recessive for both G and H. They ask you to use a chi-square test to determine if these results support their hypothesis. Remember , the null hypothesis is that the two traits are inherited independently.

6. How many classes (outcomes) are there?

7. Calculate the chi-square. X² =

8. How many degrees of freedom are there?

9. Use the table below. What is the critical value if we use a p of .05?

10. Based on your calculated X² value and the critical value you determined would you reject the null hypothesis? (yes or no)

11. Are the two traits independent? (yes or no)

Answer 1

There are two tratis here G and H. And as mentioned there are four possible phenotypes. Let the dominant alleles be G and H and the recessive alleles be g and h.

Cross was performed between individuals who were heterozygous for both the traits. Hence, these two individuals were GgHh. Therefore, they will give rise to four types of gametes: GH, Gh, gH and gh.

Their cross can be performed using a Punnett square as:

	GH	Gh	gH	gh
GH	GGHH	GGHh	GgHH	GgHh
Gh	GGHg	GGhh	GgHh	Gghh
gH	GgHH	GgHh	ggHH	ggHh
gh	GgHh	Gghh	ggHh	gghh

There are four phenotypes as mentioned. These are:

1. Dominant for G and dominant for H: This will happen for the genotypes GGHH, GGHh, GgHH, GgHh. As can be seen from the Punnett square, the proportion of this is: 9/16

2. Dominant for G and recessive for H: This will happen for the genotypes GGhh and Gghh. As can be seen from the Punnett square, the proportion of this is: 3/16

3. Recessive for G and dominant for H: This will happen for the genotypes ggHH and ggHh. As can be seen from the Punnett square, the proportion of this is: 3/16

4. Recessive for G and recessive for H: This will happen for the genotype gghh. As can be seen from the Punnett square, the proportion of this is: 1/16

The Punnett square is generated assuming that the two traits assort independently in the parents.

Hence, we get an expected ratio of phenotypes as

Dominant for G and H: Dominant for G and recessive for H: Recessive for G and dominant for H: Recessive for G and H = 9:3:3:1

There are a total of 463 offsprings sampled. Using the expected frequencies, we can calculate the expected number of offsprings of each phenotype:

Summarizing this in a table, we get:

Phenotype	Expected
Dominant for G and H	260
Dominant for G and recessive for H	87
Recessive for G and dominant for H	87
Recessive for G and H	29
TOTAL	463

Adding the observed values from the question to the table, we get:

Phenotype	Expected	Observed
Dominant for G and H	260	225
Dominant for G and recessive for H	87	93
Recessive for G and dominant for H	87	106
Recessive for G and H	29	39
TOTAL	463	463

Now, to check if the observed values are statistically different from the expected values, we shall perform a chi-square test.

To do so, we first calculate:

Doing so, we get:

Phenotype	Expected	Observed
Dominant for G and H	260	225	4.7115
Dominant for G and recessive for H	87	93	0.4138
Recessive for G and dominant for H	87	106	4.1494
Recessive for G and H	29	39	3.4483

Now, the chi-square statistic is given using the formula:

Summing the last column, we get:

The number of categories here are n = 4. Hence, the degrees of freedom becomes df = n - 1 = 4 - 1 = 3.

Using a chi-square table, and a significance level of 0.05, we get a critical value of 7.815.

Since the chi-square statistic is greater than the critical value, we can reject the null hypothesis i.e. reject the claim that the observed values are the same as the expected values.

Therefore, we conclude that the observed values are different from the expected values and therefore the genes are not assorting independently.