Question

HELLO, PLEASE HELP WITH QUESTIONS 8-10 PLEASE PLEASE AND THANK YOU!!

Question Completion Status: BUTTON Questions 5 to 11 are based on the following scenario Question 5: Although the following scenario is hypothetical, it is based on actual facts. African leopards live in a variety of habitats including forests, mountains, grasslands and deserts. Most leopards are light brown with black spots, however, in very rare instances a leopard may be reddish in colour with tan-coloured spots - a leopard with this phenotype is referred to as a strawberry leopard. This phenomenon is called erythrism and it is believed to be a result of a genetic mutation caused by a recessive gene. Erythrism refers to an unusual reddish pigmentation of an animal's fur, hair, or skin. The following image compares the pigmentation. The top image shows a strawberry leopard with the recessive red pigmentation phenotype and the bottom image shows a leopard with the dominant brown pigmentation phenotype.

PLEASE, I DON'T HAVE ANY MORE QUESTIONS, AND I ASKED THIS 2 TIMES ALREADY AND NO RESPONSE. THANK YOU!!! APPRECIATE IT!!

Answer 1

Answer 8:

Given:

There are 112 SS, 98 Ss and 7 ss type individuals.

Thus, total population size

= (112+98+7)

=217

The allelic frequencies of allele S and s be p and q respectively.

Frequency of an allele = (Number of that allele in the population) /(Population size × 2)

Frequency of S allele

= { (112×2) + (98) }/(217×2)

=0.742

Thus, frequency of allele "s"

= (1-0.742)

0.258

Expected number of individuals of each genotype :

Number of SS individuals:

= p^2 × population size

= (0.742×0.742×217)

=120

Number of Ss individuals:

=2pq × population size

= 2×0.742×0.258×217

=83

Number of ss individuals:

= q^2 × population size

=(0.258×0.258×217)

=14

Genotype	Observed Frequency (O)	Expected frequency (E)	(O-E)2/E
SS	112	120	0.533
Ss	98	83	2.710
ss	7	14	3.5

The value of chi square is:

0 - E)

= (0.533 + 2.71 + 3.5)

=6.743

Answer 9:

Here, number of expected genotypes is 3 (SS, Ss, ss) and there are two alleles (S and s).

Thus, degrees of freedom,

=2-1

=1

Answer 10:

Chi square value of 6.733 lies between the between the values 0.01 and 0.005 at one degrees of freedom (from the chi square table).

Hence, the correct choice should be:

0.01>P>0.005