Question

On the planet Susru, there are two types of bears; those who like honey-nut cheerios, and those who like multi-grain. The phenotype is determined in an epistatic way by two loci:

HNNT, with alleles H (dominant) and h (recessive), and
MLTGRN, with alleles M (dominant) and m (recessive).

In a cross of a HHMM bear and an hhmm bear, and the F1s like honey-nut. A cross of two F1 bears produces the following sums two-locus genotype counts:

All F2 bears with at least one H allele but no M alleles: 1001.
All F2 bears with at least one M allele but no H alleles: 754.

What number of F2 bears with the hhmm genotype would produce an F2 data set that is NOT consistent with a duplicate dominant mode of inheritance at the 10% level of significance (Hint: Think Chi-square)?

Answer 1

Dominant mode of inheritance ratio is 12 : 3 : 1

F2 :

All F2 bears with at least one H allele but no M alleles: 1001 (Phenotype : Honey-nut and Genotype : Hhmm)

All F2 bears with at least one M allele but no H alleles: 754 (Phenotype : Multigrain and Genotype : hhMM, hhMm)

12/16 x 100 = 75 %

3/16 x 100 = 18.75 %

1/16 x 100 = 6.25 %

If 75 % is 1001, then 100 % is 1335

75 % = 1001

100 % = ?

(100 x 1001) / 75 = 1355

Therefore, (1/ 16) x 1355 = 84

	Observed	Expected
Honey nut (Hhmm)	1001	1001
Multigrain (hhMM, hhMm)	754	250
Plain (hhmm)	118	84

The given data is not consistent with ratio of dominant mode of inheritance (12 : 3 : 1)

Therefore,

1001 + 754 = 1755

1755 = 93.75 %

118 = 6.25 %

Number of F2 bears with the hhmm genotype that would produce an F2 data set that is NOT consistent with a duplicate dominant mode of inheritance at the 10% level of significance = approximately 118