Craig Venter has constructed 16 distinct RNA strands (think very short books). H
ID: 3301747 • Letter: C
Question
Craig Venter has constructed 16 distinct RNA strands (think very short books). He has prepared three different codon pairs between each of which he plans to attach an ordered sequence of his RNA strands (think books on shelves). He'll use each of 16 strands once, putting from 0 to 16 of them in order between each codon pair. The result is a collection of three new genes. How many distinct sets of three genes can he make? Say one of the genes he creates is denoted AUGCGCCGCCAGUGA. The first AUG and the last UGA are the start-stop codon pair. If we leave the codon pair in place and rearrange only the letters of the middle portion CGCCGCCAG, how many genes can we make? (Think anagrams.) Don't worry about whether the results are real genes. Just for MATH 515 students: We take a set of n distinct strands, order them between three distinct codon pairs as above, and then arrange the three resulting genes in a long ordered row to get a genome. Find a formula for the number of possible genomes.Explanation / Answer
1. If we have n different objects, all of them can be rearranged in n! (n factorial = n(n-1)(n-2)...x1)
Thus, if we have 16 different RNAs, they can be arranged in 16! = 16x15x14x13x12x11x10x9x8x7x6x5x4x3x2 = 20922789888000 ways!
Out of the two codon pairs, one should be the start codon, AUG, common to all the three pairs and the stop codon will not code for any amino acid. Therefore, all the three genes are in fact the same gene.
2. If you have n objects to arrange, of which you have p of one kind, q of a different kind and r of a third kind, the number of combinations of n would be (n!) / (p!xq!xr!)
In CGCCGCCAG, we have 5 Cs, 3 Gs and 1 A, thus the number of genes would be 9! / (5!3!) = 1008 distinct genes.