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

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)

a. decomposes use carbon from dead organisms as fuel for cellular respiration, returning carbon dioxide back into the atmosphere.

after making a punnet square using bv bv bv bv you will find that the ratio will be 9: 7 or a 128.6% probability.

hope this