Question

Many textbooks and informational websites that discuss the topic of genetic inheritance provide a list of human traits that are purportedly controlled by a single locus. If the genes that control the inheritance of these traits have one recessive and one dominant allele, dominant phenotypes will be more common.

Hand clasping is thought to be one such trait. Clasping one’s hands with the right thumb over the left is thought to be the recessive phenotype. If the frequencies of the dominant and recessive alleles are equal, and if mating is at random with respect to the trait[1], then we expect the frequency of recessive phenotypes to be 0.25.

I discussed this in another class I teach (BIOL 463) and we found that 5 people had the R over L phenotype, and 9 people had the L over R phenotype. Based on this sample, test the hypothesis that the frequency of recessive phenotypes is 0.25.

[1] In other words, we assume that people do not choose who to mate with based on how they clasp their hands

1. Several lectures ago, we did the same hand-clasping exercise in BIOL 260. This provided a much bigger sample. By my in-class count, we had 47 people with the R over L phenotype and 89 with the L over R phenotype.

Based on this larger sample, calculate the probability of obtaining the observed data, or data more extreme, if the null hypothesis were true. Explain your procedure like you did in part b), but no need to show the actual calculations: just give the probability (3pts).

2. Based on the larger sample, what do you conclude? (1pt)

3. Compare your results from part b and part f. Explain any similarities and differences in the results (1pt)

Answer 1

Solution

Let X = Number of persons with R over L phenotype in a sample of n persons. Then,

X ~ B(n, p), where p = probability a person is R over L phenotype, which is also the same as the frequency of recessive phenotypes, that is given to be 0.25.

Thus, X ~ B(n, 0.25) ……………………………………………………………………….. (1)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then

probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n …………………..(2)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution: BINOMDIST(Number_s:Trials:Probability_s:Cumulative), what is within brackets is (x:n:p:True)] ………………………………………………………………….(2a)

Now to work out the solution,

Part (?)

5 people had the R over L phenotype, and 9 people had the L over R phenotype.

=> in a sample of 14, 5 people had the R over L phenotype. This in conjunction with (1) and (2), would mean:

P(X = 5) = (¹⁴C₅)(0.25⁵)(0.75)⁹

= 0.1468.

The above probability being significant (more than 0.05), we accept the hypothesis that the frequency of recessive phenotypes is 0.25 ANSWER

Part (1)

Here n = (47 + 89) = 136.

Since np = 34 and np(1 - p) = 25.5 are both greater than 5 binomial probability can be approximated by Normal Distribution.

We want P(X ≥ 47)

= P[Z ≥ {(47 – 34)/√(34 x 0.75)}], where Z ~ N(0, 1)

= P(Z ≥ 2.6)

= 0.004 ANSWER

Part (2)

When the sample size increased to almost 10-fold, the probability reduced considerably. While, in the former case the hypothesis is accepted, in the latter case it is rejected.

Since larger sample enhances reliability, it is only reasonable to conclude that the frequency of recessive phenotypes is not 0.25 ANSWER

DONE