Lab 9: Grammar 2: Syntactic Structures & Syntactic Ambiguities

Lab 13: Genetics and Evolution of Language

1. Twin study of dyslexia. Bakwin (1973) studied 31 pairs of monozygotic (MZ) twins and 31 pairs of diazygotic (DZ) twins, in which at least 1 twin in each twin pair was dyslexic. He found that in 26 of the MZ twin pairs, both twins were dyslexic, and in 5 of the MZ twin pairs, only one of the twins was dyslexic. In 9 of the DZ twin pairs, both twins were dyslexic and in 22 of the DZ twin pairs, only one of the twins was dyslexic.

The pairwise concordance rates for twins can be calculated using the following formula:

number of concordant twin pairs

total number of twin pairs

A. In this study, what is the pair-wise concordance rates for MZ twins? 83%

B. In this study, what is the pair-wise concordance rates for DZ twins? 29%

C. One can use the chi-square statistic to determine whether the difference in concordance rates for MZ and DZ twins is statistically significant. Using the formula for the chi-square statistic given at the end of this handout and chi-square table provided by your TA, calculate the chi-square statistic and determine the probability that this outcome is the result of chance alone.

2. Twin study of SLI. Tombin & Buckwalter (1994) studied 56 pairs of monozygotic (MZ) twins and 26 pairs of diazygotic (DZ) twins, in which at least 1 twin in each twin pair was diagnosed as having SLI. They found that in 45 of the MZ twin pairs, both twins were SLI, and in 11 of the MZ twin pairs, only one of the twins was SLI. In 10 of the DZ twin pairs, both twins were SLI and in 16 of the DZ twin pairs, only one of the twins was SLI.

A. In this study, what is the pair-wise concordance rates for MZ twins? 83%

B. In this study, what is the pair-wise concordance rates for DZ twins? 29%

C. Calculate the chi-square statistic and determine the probability that this outcome is the result of chance alone.

3. Reproductive success and evolutionary fitness.

A. Leslie Loudmouth has 3 children, and each of these children in turn have 3 children, who in turn each have 3 children, etc. Robin Rarelytalk has 2 children and each of these children in turn have 2 children, who in turn each have 2 children, etc..

1. Counting Leslie and Robin as the first generation, what is the least number of generations of Loudmouths before Loudmouths outnumber Rarelytalks by 2 to 1?

2. Counting Leslie and Robin as the first generation, what is the least number of generations of Loudmouths before Loudmouths outnumber Rarelytalks by 10 to 1?

3. Assuming 25 years between generations for both Loudmouths and Rarelytalks, how many years will it be before the number of Loudmouth relatives outnumber the number of Rarelytalk relatives by 2 to 1?

4. Assuming 25 years between generations for both Loudmouths and Rarelytalks, how many years will it be before the number of Loudmouth relatives outnumber the number of Rarelytalk relatives by 10 to 1?

5. If Leslie and Robin (and their children and their children's children etc.) each have 3 children, but the average number of years between generations is 20 years for the Loudmouths and 25 years for the Rarelytalks, how many years will it be before the number of Loudmouths outnumber the Rarelytalks by 2 to 1?

B. Sandy Slicktalker and Terry Taciturn have large families. Terry has 5 children, and each of these children in turn have 5 children, who in turn each have 5 children, etc. Sandy has 6 children, and each of these children in turn have 6 children, who in turn have 6 children, etc. .

1. Counting Sandy and Terry as the first generation, what is the least number of generations of Slicktalkers before Slicktalkers outnumber Taciturns by 2 to 1?

2. Assuming 25 years between generations for both Slicktalkers and Taciturns, how many years will it be before the number of Slicktalker relatives outnumber the number of Taciturn relatives by 2 to 1?

3. If Sandy and Terry (and their children and their children's children etc.) each have 3 children, but the average number of years between generations is 20 years for the Slicktalkers and 25 years for the Taciturns, how many years will it be before the number of Slicktalkers outnumber the Taciturns by 2 to 1?

WARNING: the numbers involved are very big. You may not want to attempt the following if you haven't figured out a formula for calculating the number of generations.

Extra credit 1: Counting Sandy and Terry as the first generation, what is the least number of generations of Slicktalkers before Slicktalkers outnumber Taciturns by 10 to 1?

Extra credit 2. Assuming 25 years between generations for both Slicktalkers and Taciturns, how many years will it be before the number of Slicktalker relatives outnumber the number of Taciturn relatives by 10 to 1?

Chi-Square Statistic Formula

The computation of c2 for a table of any size is most easily accomplished by

where O = the number of instances observed and E = the number of instances expected given the null hypothesis. This quantity, when basedon independent observations and expected frequencies that are not too small, tends to be distributed as c2 on degrees of freedom (df) = (number of rows-1) x (number of columns -1).

Here's a simple example. Let's test the null hypothesis that sophomores and juniors are equally likely to make the Dean's List. We observe that of 100 sophomores, 20 make the Dean's List. Of 80 juniors, 25 make the Dean's List. The observed frequencies are:

	Dean's List	Not Dean's List	TOTAL
Sophomore	20	80	100
Junior	25	55	80
TOTAL	45	135	180

The expected frequencies are:

	Dean's List	Not Dean's List	TOTAL
Sophomore	25	75	100
Junior	20	60	80
TOTAL	45	135	180

So the c2 = 1+ .333 + 1.25 + .417 = 3.00. The df is 1, so if we look it up in the table, we find that there is greater than a 5% chance that these result were due to chance alone, and we cannot confidently reject the null hypothesis.