Objective: Get a sense of how genomics, the study of the genome in its entirety, needs to think about how to go about its research. Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment (a “read”) are sequenced. The sequenced reads are assembled together into contiguous sequences (“contigs”) based on sequence similarity. The idea is to sequence enough random fragments so that every nucleotide in the genome is represented on some read. The number of such fragments needed is called the coverage, c. The coverage c can be calculated by the formula RL/G, where R is the number of reads sequenced, L is the average length of a read and G is the total length of the genome. The units of length are bases (b) or base pairs (bp). Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniques are applied to the genome, resulting in 20 reads, with an average length of 50 bp. A very important point is that, even though 20 x 50 = 1000, there is no guarantee that ALL 1000 bp of the genome are represented in the fragments. If the fragments are truly random and uniform, then the probability that a particular nucleotide is not sequenced on some fragment is given by e–c. Therefore, the probability that all of the nucleotides are sequenced is 1 – e–c. Given the values given above, what is the probability that all of the nucleotides are sequenced in that genome?
Objective: Get a sense of how genomics, the study of the genome in its entirety, needs to think about how to go about its research. Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment (a “read”) are sequenced. The sequenced reads are assembled together into contiguous sequences (“contigs”) based on sequence similarity. The idea is to sequence enough random fragments so that every nucleotide in the genome is represented on some read. The number of such fragments needed is called the coverage, c. The coverage c can be calculated by the formula RL/G, where R is the number of reads sequenced, L is the average length of a read and G is the total length of the genome. The units of length are bases (b) or base pairs (bp). Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniques are applied to the genome, resulting in 20 reads, with an average length of 50 bp. A very important point is that, even though 20 x 50 = 1000, there is no guarantee that ALL 1000 bp of the genome are represented in the fragments. If the fragments are truly random and uniform, then the probability that a particular nucleotide is not sequenced on some fragment is given by e–c. Therefore, the probability that all of the nucleotides are sequenced is 1 – e–c. Given the values given above, what is the probability that all of the nucleotides are sequenced in that genome?
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Objective: Get a sense of how genomics, the study of the genome in its entirety,
needs to think about how to go about its research.
needs to think about how to go about its research.
Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment
(a “read”) are sequenced. The sequenced reads are assembled together into
contiguous sequences (“contigs”) based on sequence similarity.
(a “read”) are sequenced. The sequenced reads are assembled together into
contiguous sequences (“contigs”) based on sequence similarity.
The idea is to sequence enough random fragments so that every nucleotide in the
genome is represented on some read. The number of such fragments needed is
called the coverage, c.
genome is represented on some read. The number of such fragments needed is
called the coverage, c.
The coverage c can be calculated by the formula RL/G, where R is the number of
reads sequenced, L is the average length of a read and G is the total length of the
genome. The units of length are bases (b) or base pairs (bp).
reads sequenced, L is the average length of a read and G is the total length of the
genome. The units of length are bases (b) or base pairs (bp).
Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniques
are applied to the genome, resulting in 20 reads, with an average length of 50 bp.
A very important point is that, even though 20 x 50 = 1000, there is no guarantee
that ALL 1000 bp of the genome are represented in the fragments.
are applied to the genome, resulting in 20 reads, with an average length of 50 bp.
A very important point is that, even though 20 x 50 = 1000, there is no guarantee
that ALL 1000 bp of the genome are represented in the fragments.
If the fragments are truly random and uniform, then the probability that a
particular nucleotide is not sequenced on some fragment is given by e–c.
Therefore, the probability that all of the nucleotides are sequenced is 1 – e–c.
particular nucleotide is not sequenced on some fragment is given by e–c.
Therefore, the probability that all of the nucleotides are sequenced is 1 – e–c.
Given the values given above, what is the probability that all of the
nucleotides are sequenced in that genome?
nucleotides are sequenced in that genome?
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