chi square question for biology

Question

Chi Square

Consider the scenario where one parent is type AB blood and the other parent is heterozygous for type B blood. A sample of 150 couples collectively produced a total of 38 children with type A blood, 117 children with type B blood, and 38 children with type AB blood. 

1. Determine the expected ratio of blood types for offspring produced by the above cross.
2. Calculate a chi-square value for this example in the space below. 
3. Could the difference in the actual outcome relative to the expected ratio be due to chance variation or is it a result of some other factor? 
4. What conclusion(s) could you draw from the data? 

Studies show that the following are the approximate frequencies of the various ABO blood types in the U.S.: 39% A, 12% B, 4% AB, and 45% O. Do the data given in the following table represent a satisfactory sample of the U.S. population? Complete Table 6, calculate the X2, and answer the questions based on your calculations. Calculate E values to the nearest one-hundredth (0.01).

1. Calculate a chi-square value for this example in the space below.
2. What conclusion(s) could you draw from the data?

Answer 1

a) According to the use of a Punnett square, the blood type probabilities for each offspring are as follows:

B= 50%, A=25%, AB=25% Below I drew up a simplified version of the Punnettt square in this situation. Usually when talking about blood type, you would use I^A, I ^B, and i to distinguish between the antigens A, B and O(no antigen). On the top of the square is the AB parent. Along the left side is the parent that is heterozygous for B. This means that the 2nd parent has the blood type BO. Remember that the O blood type is recessive, so you need two i alleles for the person to have an O blood type. Each quadrant represents a 1/4 chance of receiving that specific combination. We notice that 2/4 of the possibilities include obtaining a B blood type. Therefore, we can say that there is a 50% chance the offspring will be B blood type. Also notice that it is impossible to have an O type offspring with this combination of parents. If this set of parents does have an O type child they may want to go on the Maury show because "you are NOT the father (or mother)".

A B

B (AB) (BB)

O (AO) (BO)

b)The X²= sum (Observed value -expected value)²

-------------------------------------------------------

expected value

B= [(117-75)^2] /75 = 23.52

A= [(38-37.5)^2]/37.5 = 0.006

AB= [(38-37.5)^2]/37.5 = 0.006

X²= 23.52+ 0.006 + 0.006 = 23.53

c/d) This is a very high number. Analyzing statistics always depends on what P value you are using for the conclusion of your hypothesis. We hypothesized that the offspring results would follow our Punnett square. With a p < 0.05, we would have to reject our hypothesis. This means that the difference between expected and observed results is significant. There may be another factor playing a role here other than chance. Perhaps the sample size is too small?

I think I am missing the table for the second part.