The haploid human genome contains about 3 times 10^9 nucleotides. On average, ho
ID: 87901 • Letter: T
Question
The haploid human genome contains about 3 times 10^9 nucleotides. On average, how many DNA fragments would be produced if this DNA was digested with restriction enzyme (a 6- base cutter)? (a 4-base cutter)? How often would an 8-base cutter cleave? A population on a desert island has only green and blue eyed inhabitants. The green allele is dominant to the blue allele. If there are 13, 482 inhabitants and 1, 512 of the island's inhabitants have green eyes, then what is the frequency of the green eye allele in this population?Explanation / Answer
19) The frequency of a 6 base cutter is 4^6= 4096. The probability of having a restriction site for PstI will be one after every 4096 bp. Hence if the DNA is 3 X10^9 nt long, then the fragments produced will be 3 x 10^9/4096= 732,421.88 fragments. The frequency of a 4 base cutter is 4^4= 256. Thus, the probability of Rsa I restriction site to occur will be one after every 256 bp. Hence if the DNA is 3 X10^9 nt long, then the fragments produced will be 3 x 10^9/256= 11,718,750 fragments. The frequency of 8 base cutter is 4^8= 65,536. This enzyme will have one restriction site after every 65,536 bp.
20) 1,512 people have green eyes which is dominating thus remaining population will have blue eye which is recessive i.e. 13,482- 1,512 = 11,970. Let the recessive allele be q and dominant allele be p.
Frequency of individuals= individuals/ total population
q2= 11,970/13,482 = 0.8878
q= 0.942
According to hardy-weinberg equation, p+q=1
P=1-q= 1- 0.942 = 0.058 = 5.8%
Thus the frequency of green eye allele in the population is 5.8%.
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