Need some help with 3, 5, 6 and 7 3. After graduation, you and 19 friends build

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Need some help with 3, 5, 6 and 7

3. After graduation, you and 19 friends build a raft, sail to a deserted island, and start a new population, totally isolated from the world. Two of your friends carry (that is, are heterozygous for) the recessive Cr allele, which in homozygotes causes cystic fibrosis. A. Assuming that the frequency of this allele does not change as the population grows, what will be the instance of cystic fibrosis on your island? B. Cystic fibrosis births on the island is how many times greater than the original mainland? The frequency of births on the mainland is .059%. 4. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:

Explanation / Answer

3A) Since cystic fibrosis occurs in recessive condition

Both of them are heterozygous , then the frequency of CFTR will be

C Cf * C C f

CC(normal) , 2CCf(carrier), CfCf( diseased),

FREQUENCY OF GETTING DISEASE WILL BE 1/4= 0.25

B) Frequency of CFTR births on mainland is .059%, whereas on the island it is 0.25%. It is near about 5 times more than the mainland.

4) B(brown) is dominant over b(white)

40% of them are white means they have bb genotype.

Incomplete Question (Kindly elaborate)

5) From the three given genotypes AA, BB, AB it can be calculated

Total number pf persons= 200+25+75= 300

Persons with type A= 200 = 200/300= 0.66+ 1/2(0.25)

= 0.66+ 0.125 = 0.785

Persons with type B= 25 = 25/300 = 0.083+1/2(0.25)

= 0.208

Pesons with type AB(it conatins both the alleles A and B) = 75 = 0.25/2= 0.125

6) Red sided individuals= 396

tan sided individuals= 557

total individuals= 396+557= 953

Given- red is totally recessive

a) allelic frequency of each=

SQUARE ROOT OF GENOTYPIC FREQUENCY= sq root of p2  and   q2  

sq root of p2= 0.64

p+q=1

q= 1-0.64 = 0.36

b) genotypic frequency= the genotypes possible would be RR, rr, Rr

p2 = Red (RR) = 393/953 = 0.415

q2   = Tan(rr) = 557/953 = 0.584

c) number of heterozygotes 2pq = 2* 0.64*0.36= 0.46

0.46% of total population

d) Phenotypic frequencies will be

Phenotypic frequency of dominant (red ) will be RR+Rr= 0.415+0.46= 0.87

Phenotypic frequency of recessive will be rr>

RR= 0.415

rr= 0.584

7) total = 1000

blood type INDIVIDUALS

M 490

N 90

MN 420

frequency of M allele = 450/1000+ 0.5(420/1000)

=0.45+ 0.21= 0.66

frequency of N allele= 0.09+ 0.5(420/1000)

0.09+ 0.21= 0.3

Since heterozygotoes contain both the alleles M and N

B) mating frequency can be calculated by multiplying the respective frequecny of the mating type alleles

If MM MATES NN = 0.49* 0.09= 0.044

MM MATES MN= 0.49* 0.42= 0.20

NN MATES NN= 0.09*0.09= 0.0081

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