Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Finish writing the complementary sequence: What is as an RNA: What is the probab

ID: 161907 • Letter: F

Question

Finish writing the complementary sequence: What is as an RNA: What is the probability that a random 4 nt sequence is GGGG? What is the probability that a random 4 nt sequence is GATC? Probability that a random 6 nt sequence is GGATCC? Consider a random 10,000 bp circular piece of DNA. How many GGGG sites are expected? (Consider both strands). Consider a random 10,000 bp circular piece of DNA. How many GATC sites are expected? (Tricky) Consider a random 10,000 bp circular piece of DNA. How many GGATCC sites are expected? Consider a random 10,000 bp circular piece of DNA. How many 8nt AGGATCCT sites are expected? How many different 30 nt RNA sequences are possible? On average, how many ways can a 10 amino acid peptide be encoded?

Explanation / Answer

3) Four bases A, T, G and C

Probability of getting any base out of four bases = 1/4

Probability of G at position 1 is 1/4

Probability of G at position 2 is 1/4

Probability of G at position 3 is 1/4

Probability of G at position 4 is 1/4

So 1/4 x 1/4 x 1/4 x 1/4 = 3.9 x 10-3 or 0.39 %

Probability of getting GGGG is 3.9 x 10-3 or 0.39 %

4) Probability of getting GATC is also 3.9 x 10-3 or 0.39 %

5) Probability of getting GGATCC

There are six nucleotides so its probability is (1/4)6

Probability of getting GGATCC = 2.44 x 10-4 or 0.0244 %

6) Number of GGGG sites

In question 3 we have calculated the probability of getting GGGG is 3.9 x 10-3

Circular DNA with 10000 bp and if both strands are considered 20000 bp.

3.9 x 10-3 x 20000 = 78

Number of GGGG sites = 78

7) One strand is considered so 10000 bp

Probability of getting GATC is also 3.9 x 10-3

3.9 x 10-3 x 10000 = 39

Number of GATC sites = 39

8) Number of GGATCC sites

Probability of getting GGATCC = 2.44 x 10-4

2.44 x 10-4 x 10000 = 2.44 or 2 sites

9) Number of 8 nt sites

(1/4) 8 x 10000 = 0.152

10) 20 n where n is the length of peptide

20 10 = 1.024 x 1013 combinations

11) 4 n

4 30 = 1.15 x 10 18 combinations

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at drjack9650@gmail.com
Chat Now And Get Quote