13. Professor Dropkin has been collecting data from couples known to be carriers
ID: 23170 • Letter: 1
Question
13. Professor Dropkin has been collecting data from couples known to be carriers or heterozygotes for the STR gene wowzer. The data collected look like this:CHILDREN'S PHENOTYPES
WW Ww ww
121 249 130
The expected number of heterzygotes is:
a. 250 b. 400 c.125 d. 249 e. 15
14. Using his hypothesis the expected number of ww homozygotes is:
a. 18 b. 36 c. 41.25 d. 18.33 e. 125
15. and the expected number of WW homozygotes is
a. 125 b. 19 c. 13.75 d. 36.67 e. 18.33
16. The chi-square value for goodness of fit of this data is Chi-square = SUM (obs-exp)2/exp
a. 0.0 b. 3.84 c. 0.332 d. 55 e. 0.152
PROBABILITY
df 0.95 0.90 0.50 0.25 0.05
1 .004 .016 .455 1.32 3.84
2 .103 .211 1.39 2.77 5.99
3 .352 .584 2.37 4.11 7.81
Explanation / Answer
Let's back up a little from Chi squared for a minute. Professor Dropkin has been looking at people who are carriers (heterozygous) for a condition. This means they have different alleles (versions of a gene), one "W" and one "w". So when a couple who were carries had children, the cross would be Ww x Ww. This would produce an EXPECTED genotype ratio of 1:2:1 .......|..W..|..w..| ..W..| ..w...| So if there were 500 children born, how would this fit into that ratio? Divide 500 by 4, and you get 125. So the "expected" ratio for 500 children would be 125 WW : 250 Ww : 125 ww (so far, this answers 13-15). Obs...|.Exp.. 121....| 125 249....|.250 130....|.125 (obs-exp)2/exp ((-4) x (-4))/125 = 0.128 ((-1) x (-1))/250 = 0.004 ((5) x (5))/125 = 0.2 0.128 + 0.004 + 0.2 = 0.332 You have two degrees of freedom (df) because you have three different groups of genotypes (WW, Ww, ww) and df is # groups -1. When deciding if a hypothesis should be rejected or accepted, the level of 0.95 is generally the cutoff point. So in the row for df = 2, find where 0.332 would occur. If it's to the "left" (on your chart, on other charts, the order of the numbers are reversed) of the column for 0.95 (0.95 to 1.00), then the hypothesis should be accepted because the data was consistent with the expected values. If it's to the "right" (0.94 to 0.05) the hypothesis should be rejected.
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