Summary
Summary Example of all possible calculations in the exams
- Course
- Genomics (GENE3714)
- Institution
- University Of The Freestate (UFS)
Example of all possible calculations in the exams
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Assume that eukaryotes have approximately 24,000 protein-codinggenes and that an
average eukaryotic protein is 375 amino acids long. What would be the
total length (in
Mbp) occupied by protein-coding genes in an average eukaryotic
genome?
24,000 genes x 375 amino acids x 3 bases/amino acid = 27,000,000 bp = 27
Mbp
Consider the formulae below and answer the questions that follow.
Coverage, c = [(number of reads, N) x (length of a read, L)]/(genome length, G)
Coverage, c = NL/G
Probability that a base is not sequenced, P = e -c, where e is the base of natural
logarithms with a
constant value of 2.718
Total expected gap length = G x e-c
Total number of gaps = Ne-c
A genome has the size of 4,459 Mbps. The genome was sequenced
through random 300 bp
fragments to yield 92.49 million reads. 1 Mbp = 106 bp.
What coverage does the sequences generated above represent? (4) (2
decimals)
Coverage, c = NL/G
= (92.49 x 106)(300)/(4,459 x 106)
= 6.22
3. What is the probability that a specific base was not sequenced? (2) (3
decimals)
Probability that a base is not sequenced, P = e -c
= 2.718-6.22
= 0.002
4. How many gaps would you expect in the assembled sequences? (2)
What total gap length
would you expect in the assembled sequences? (2) (2 decimals and
Mbp)
Total number of gaps = Ne-c
= (92.49 x 106)(2.718-6.22)
= 184 104
Total expected gap length = G x e-c
= (4,459 x 106)(2.718-6.22)
= 8 875 766.70 bp
= 8.88 Mbp
5. If the gap length is to be limited to 2 Mbp following sequence
assembly, what coverage
should you aim for during sequencing? (4) (2 decimals)
Total expected gap length = G x e-c
2 x 106 = (4,459 x 106) x e-c
e-c = (2 x 106)/(4,459 x 106)
ln e-c = ln (2 x 106)/(4,459 x 106)
c = 7.71
How many reads will you need to produce if you intend to have no more
than 1000 gaps at twelve-fold coverage? (3)
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