Question

Suppose there are 50 couples with the same blood type and hemoglobin genotypes. They live on a small, isolated Pacific island on which very few mosquitoes have been identified. All the individuals are heterozygous for type A blood and have sickle cell trait. The 50 couples had 224 children over the years. The children were all tested for blood type and for the presence of the sickle cell allele.

Here are the results. Testing Results Blood Test Results Number of Children Type A, normal RBCs 48 Type O, normal RBCs 18 Type A, sickle cell trait 92 Type O, sickle cell trait 33 Type A, sickle cell disease 27 Type O, sickle cell disease 6 Are these data significant? Explain using a chi-square statistical analysis test.

a. What is χ2 ?

b. Calculate df.

c. Using the critical values table, determine the P value.

d. Interpret the P value as it relates to these data. Explain the significance.

From what you know about hemoglobin, sickle cell disease, and blood type, what selection pressure is acting on this population of children and causing the null hypothesis to be rejected? Explain your answer. (Hint: Look at the actual differences between the observed and expected numbers.)

Answer 1

x2= 13.58

Step-by-step explanation:

χ2 =sum ( o− e) ^2 /e

χ2 = 13.58

degree of freedom = 6 – 1 = 5

using the critical values table,

P value.

P < 0.05 (0.025 < P < 0.01)

As P < 0.05, the null hypothesis is rejected,

This indicates that there is a statistically significant difference between the observed and expected data.

Accept------ Alternative hypothesis :

================

null hypothesis :

there was NO significant difference between the observed and expected data.

Alternative hypothesis :

there was significant difference between the observed and expected data.

Answer 2

Step-by-step explanation:

Observed phenotype(o)Expected(e) (o-e) (o-e)²/ e

Type A, normal RBCs 48 42 6 0.86

Type O, normal RBCs 18 14 4 1.14

Type A,sickle cell trait 92 84 8 0.76

Type O,sickle cell trait 33 28 5 0.89

Type A,sickle cell Dx 27 42 -15 5.36

Type O, sickle cell Dx 6 14 -8 4.57

a. chi-square value x² = 13.58

b. the degrees of freedom (df).

df = 6 – 1 = 5

c. Using the Critical Values Table, we determine the p value. p < 0.05 (0.025 < p <0.01)

d. To Interpret the p value as it relates to these data and explain the significance.

Since p < 0.05, the null hypothesis is rejected, this suggests that there is a statistically significant

difference between the observed and expected data. Therefore, the difference between the observed and expected data is not solely due to chance.

e. From what you know about hemoglobin, sickle cell disease, and blood type, what selection pressure is acting on this population of children causing the null hypothesis to be rejected? Explain your answer. (Hint: Look at the actual

differences between the observed and expected numbers).

This population of individuals is isolated on a small Pacific island on which very little or no quality healthcare is available. Thus, there is selection against the children with sickle cell disease. The presence of this selection pressure distort the observed numbers from the expected values, causing the null hypothesis to be rejected, which suggests that something other than chance is acting on the population which would be the selection against the SS genotype in this case.