7.03 Genetics: Solutions to Problem Set 3 (1997)

*The first mutation borders on an important domain. Any mutation in the domain, although it will change the frame, will destroy the protein function, so the phenotype will still be rII. *The first mutation is very close to either the start or the stop codon so mutations on one side do not change the frame in which the translation occurs. 3, The sequencing of whole genomes is a rapidly developing area in modern genetics. Large-scale sequencing projects have the potential to define new genes without the need for isolating mutations in these genes. However, identifying genes from sequence information has pitfalls and distinguishing genuine coding sequences from fortuitous open reading frames (ORFs) can be more difficult than it might at first seem. If we define ORFs of at least 100 codons to be of sufficient length to be significant, it is interesting to calculate how often an ORF of this length would appear in a random (noncoding) sequence of DNA. (a) Calculate the probability of finding an ORF of 100 codons with no stop codons in random DNA. Now determine how many such chance ORFs would appear on average in a bacterial genome of 10 Mbp. 3 out of 64 codons are stop codons. p(non-stop codon)= 61/64 p(no stop codon in 100 codons)= (61/64) :_*'=8.22* 10-_ average #of ORFs = 8.22* 10-_* 10 Mbp * 2 [multiply by a fator of 2 to account for both directions] = 16.44"104 ORFs. (b) Often bacteria have skewed ratios of G-C and A-T base pairs. For a bacteria whose DNA is 70% G-C calculate the probability of a sequence of 100 codons with no stop codons. (Hint: use the content of A-T base pairs in stop codons to recalculate the average probability of a stop). Again, determine how many such chance ORFs would appear on average in a bacterial genome of 10 Mbp. 35°70G, 35070C, The stop codons p(stop codon)= = 15% A, 15070T are UAA, UAG, AGA. (0.15"0.15"0.15 + 0.15"0.15"0.35 0.019125 + 0.15"0.35"0.15) p(non-stop codon)= 1-0.019125 p(no stop codon for 100 codons)= (1-0.019125) ''_'= 0.145 average # of ORFs = 0.145 * 10 Mbp * 2 = 2.90"10 _ ORFs --_ (c) For a bacteria whose DNA is 70% A-T calculate the probability of finding a sequence of 100 codons with no stop codons. Determine how many such chance ORFs would appear on average in a bacterial genome of 10 Mbp. 35% A, 35% T, 15% C, 15% G. p(stop codon)= (0.35*0.35*0.35 + 0.35"0.35"0.15 = 0.079625 + 0.35"0.15"0.35) p(non-stop codon)= 1- 0.79625 p(no stop codons for 100 codons)= (1- 0.79625) _'x'= 2.49* 10-_ average # of ORFs= 2.49"10 _ * 10 Mbp * 2 = 4.98"103 ORFs