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)
