PLEASE ANSWER ALL PARTS!!!!! Leucine Glutamic acdPhenylalanine Glycine Serine Gl

Question

PLEASE ANSWER ALL PARTS!!!!!

Leucine Glutamic acdPhenylalanine Glycine Serine Gly Phe Aspartic acid Tyrosine Ala Cysteine Trp Leu Arg Lys Pro Tryptophan Lysine His Thr Gln Met leArg Proline Asparagine Histidine Clutamine Arginine In this problem, we will compute the probabilities of finding specific DNA sequences in a perfectly random genome, for which we assume that the four different nucleotides appear randomly and with equal probability (a) From the genetic code shown in Figure 1.4, compute the probability that a randomly chosen sequence of three nucleotides will correspond to a stop codon. Similarly, what is the probability of a randomly chosen sequence corresponding to a start codon? ts ch. (b) A reading frame refers to one of three possible ways that a sequence of DNA can be divided into consecutive triplets of nucleotides. An open reading frame (ORF) is a reading frame that contains a start codon and does not contain a stop codon for at least some minimal length of codons. Derive a formula for the probability of an ORF having a length of N codons (not including the stop codon). (c) The genome of E. coli is approximately 5 x 106 bp long and is circular. Again assuming a that the genome is a random configuration of base pairs, how many ORFs of n length 1000 bp (a typical protein size) would be expected by chance? Note that there are six possible reading frames

Explanation / Answer

a) total number of combination of 3 nucleotides from 4 nucleotides =43 = 64

UAA, UAG andUGA and stop codon, so total no of the stop codon combination = 3

Therefore probability of getting a start codon upon choosing three nucleotides randomly= 3/64

b) total number of codon = 64

there are total 61 codon which encodes amino acid so the probability of having any one of the codons that encodes protein =61/64

Therefore, probability of finding codon of N length = 61N/64

3) Here, Given size of genome = 5x 106 bp and length of ORF is given =1000bp

expected number of orf in one reading frame= (5x 106)/1000 = 5000 ORF

As there are total number of possibility of reading frame is 6

so one can expect total number of ORF in given E.coli genome = 5000x6 = 30,000ORF

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