Genomics Questions! How many copies of the sequence GGTTATCGAAATCCT would you ex
ID: 147742 • Letter: G
Question
Genomics Questions!
How many copies of the sequence GGTTATCGAAATCCT would you expect to find in the human genome, assuming that a) the human genome is a random sequence with composed of equal proportions of all 4 nucleotides? b) the human genome is a random sequence composed of 40% of GC pairs and 60% AT pairs. c) How many fully complementary PCR primer sites would you expect for an oligonucleotide GGTTATCGAAATCCT in "human genome" as described in (b)? Explain how you got to your conclusion. do you think the "expected number of copies" that you figured out (in a, b, and c) are underestimates or overestimates (compared to what we expect in the real, not random sequence genome assumed above) and why? Note that the answer to this question may depend on the source of this sequence. What would be your answer (underestimates or overestimates?) if: a) This is an entirely random sequence that I made up in my head. b) This sequence is a primer that I designed for my PCR, that is, I used a sequence that I know is present in the genome.Explanation / Answer
a) Number of copies of this sequence can be calculated based on the nucleotide probabilities
the probability of finding any nucletidein a sequence is 1/4
Total probability of the sequence having 15 nucleotides will be= (1/4)15
= 9.3132257e-10
b) Given the percentage of AT 60% AND 40% GC
amount of A= amount of T, Therefore A will be 30% and T will be 30%,
amount of G=amount of C, G will be 20% and C will be 20%
Total probabilbilty od all the four nuclotides will be =1
Probability of A= 30% of 1= 0.3
Probability of T=0.3
Probability of G= 0.2
Probability of C= 0.2
Therefore tota probability for the given sequence GGTTATCGAAATCCT will be
0.2*0.2*0.3*0.3*0.3*0.3*0.2*0.2*0.3*0.3*0.3*0.3*0.2*0.2*0.3= 1.259712E-9
c) Since PCR primers will bind the same site but with having reverse complementarity to that of the given sequence , therfore the probability of the primer is same.
Number of binding sites of the primer pairs will be 2
= 9.3132257e-10/9.3132257e-10 =2
It is an overestimate of the number of copies what we expected in genome.
b) If this sequence is a primer that i designed for my PCR that is, i used a sequence that i know is present in a genome. Any particular Primer can have 2 binding sites in a genome. The same sequence may be present in multiple sites in the genome.
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