Question

In a typical Mendelian cross (i.e. independent assortment of the genes is expected) between two heterozygous plants (Rr x Rr)

We obtain the following results. RR=360, Rr=480 and rr= 160

Draw your own table, calculate your allele frequencies, genotype frequencies (observed and expected) and use your observed and expected counts to find X².

What will be the degree of freedom?

Do we accept or reject the null hypothesis of HWE?

Answer 1

Expected:

Rr x Rr

	R	r
R	RR	Rr
r	Rr	rr

RR = 250

Rr = 500

rr = 250

Observed:

RR = 360

Rr = 480

rr = 160

Phenotype	Observed(O)	Expected (E)	O-E	(O-E)2	(O-E)2/E
RR	360	250	110	12100.0000	48.4000
Rr	480	500	-20	400.0000	0.8000
rr	160	250	-90	8100.0000	32.4000
	1000	1000			81.6000

Chi-square value = 81.6

Degrees of freedom = Number of phenotypes – 1

Df = 3-1=2

Critical value = 7.81

The chi-square value of 81.6 is greater than the critical value of 7.81. We can reject null hypothesis