Question

Complete the following chi-square problems. (Show all work!!) In the cross AaBb x AaBb O-E (O-E) (O-E)?/E 311 105 1. Phenotype A-B- A-bb aaB- aabb Totals 25 calculated X2= mith, Beatty and Ware 7/14 revised by for Fall 2015, BIOL 251 Genetics Lab Manual, Benedictine Univ page 3

use problem above to answer

Answer 1

From the given data, the observed frequencies are

	B	bb	Total
A	311	105	416
aa	91	25	116
Total	402	130	532

The expected frequencies are

The expected values are computed in terms of row and column totals. In fact, the formula is Ejj = RXC where R; corresponds to the total sum of elements in row i, C; corresponds to the total sum of elements in column j, and T is the grand total. The table below shows the calculations to obtain the table with expected values: Expected Values Column 1 Column 2 Total Row 1 402X416 = 314.346 130 416 = 101.654 416 Row 2 402 x 116 = 87.654 130 x 116 = 28.346 116 Total 402 130 532

Final Table:

	Observed	Expected
Phenotype	Freq (Oi)	Freq Ei	(Oi-Ei)	(Oi-Ei)^2	(Oi-Ei)^2 / Ei
A-B-	311	314.346	-3.346	11.1957	0.03562
A-bb	105	101.654	3.346	11.1957	0.11014
aaB-	91	87.654	3.346	11.1957	0.12773
aabb	25	28.346	-3.346	11.1957	0.39497
Total:	532	532		44.7829	0.66844

a) H0: AaBa and AaBb are independent
H1: AaBa and AaBb are not independent

b) Degrees of freedom : (2-1)(2-1) = 1

c)
Critical X^2: 3.841456

d) Reject H0 if Chi-square value > Chi-square table/critical value i.e. 3.841456

e) Test Statistic, X^2: 0.6684
since chi-square value < Chi-square critical value so we do not reject H0

f) Thus we conclude that these data approximate the expected phenotypic ratio

g) since P-Value: 0.4136 > alpha 0.05 so we accept h0
Thus we conclude that these data approximate the expected phenotypic ratio and also
AaBa and AaBb are independent