Question

1. On the planet Susru, there are three types of bears; those who like honey-nut cheerios, those who like multi-grain, and those who like plain cheerios. The phenotype is determined in an epistatic way by two loci:
a. HNNT, with alleles H (dominant) and h (recessive), and
b. MLTGRN, with alleles M (dominant) and m (recessive)
2. In a cross of a HHMM bear and an hhmm bear, and the Fls like honey-nut. A cross of two F1 bears produces the following sums two-locus genotype counts:
a. All F2 bears with at least one H allele: 801
b. All F2 bears with at least one M allele but no H alleles: 200.
3. What number of F2 bears with the hhmm genotype would produce an F2 data set that is consistent with a dominant mode of inheritance at the 1% level of significance (Hint: Think Chi-square)?
A. 55.
B. 37.
C. 118.
D. 90
E. 15

Answer 1

Answer:

A. 55.

Explanation:

The three types of bears are:

1. Those who like honey-nut cheerios (let that be H)

2. Those who like multigrain (let that be M)

3. Those who like plain-cheerios (let that be P)

We were told that the phenotype is determined in an Epistatis way by two loci: i.e Epistatis Dominance Mode with:

a) HNNT : with alleles Hh

b) MLTGRN : with alleles Mm

In a cross of  HHNT and MLTGRN, all F(1) like honey-nut. i.e (HhMm)

Furthemore, the process continuous with a cross of two F1 bears i.e Interbreeding of  F(1) bears produced bears with:  

Bears who like honey-nut cheerios, Multi-grain and Plain cheerios in F(2) in the ratio of 12:3:1. This can be explained as follows:

Dominant Epistatis for types of  bears in planet Susru.

(the table can be found in the attached file below)

From the table, Let:

H= honey-nut Cheerios  

M= multi-grain

P= plain

From above table,  (H) is dominant to (h) and epistatic to allels (M) and (m). Hence, it will mask the expression of M/m alleles. Therefore in F(2),

• bears with H-M (9/16) and H-mm (3/16) will produce bears who like honey-nut cheerios;

• bears with hhM- (3/16) will produce bears who like multigrain and,

• those with hhmm (1/16) genotype will produce bears who like Plain.

Thus the normal dihybrid ratio 9:3:3:1 is modified to 12:3:1 ratio in F2 Generation.

 

Now that the Epistatis  Dominance Mode of Inheritance ratio is 12:3:1

12/16= 0.75  

=75%  

75% = 801 (i.e Phenotype: Honey-nut and Genotype: HHMM,HhMM,HHMm,HhMm,Hhmm)

3/16= 0.1875

=18.75%

18.75% = 200 (i.e Phenotype: Multigrain and Genotype : hhMM,hhMm).

If 75% is 801, then 100% will be;

(100*801)/75=x

x= 1068

100% = 1068

This implies that, the number of F2 bears with the hhmm genotype (plain) that would produce an F2 data set will be:

(1/16) * 1068 = 67( Expected) but observed will be 55.

(The Chi-Square table can be found in the attached file below)