D. How many times would you expect to find a specific 20 base pair sequence in the human genome?
D. How many times would you expect to find a specific 20 base pair sequence in the human genome?
Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN:9781305389892
Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan
Chapter19: Genomes And Proteomes
Section: Chapter Questions
Problem 1TYK
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![Student Guide
D. How many times would you expect to find a specific 20 base pair sequence in the
human genome?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc62819af-1f81-4d59-bfb3-1ca1702855ed%2F54f16db0-f191-4f38-8242-936305e6dff7%2F062yf0h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Student Guide
D. How many times would you expect to find a specific 20 base pair sequence in the
human genome?
![Part 3. Compare the Specificity of DNA-Cutting Tools
The flexibility and specificity of CRISPR-Cas9 technology offer a large step forward for gene editing.
The first DNA "scissors" were restriction enzymes, which cut DNA at predefined sequences, typically
4-8 base pairs long. For example, EcoRI, a restriction enzyme found in E. coli, will cut double-
stranded DNA at every GAATTC sequence. If EcoRI were added to a sample that contained the entire
human genome, it could cut at every GAATTC sequence.
We can calculate the probability that a particular nucleotide sequence, such as GAATTC, will occur
within a larger sequence. Table 1 below shows the calculated probabilities of finding sequences of
particular lengths. These calculations are based on the assumption that DNA sequences are entirely
random and that every nucleotide position has an equal probability of being A, T, C, or G. Use the
table to answer the following questions.
Table 1. Calculated probabilities of finding a specific sequence.
Sequence
Sequence Probability Calculation
Length
A
AC
GAATTC (EcoRI)
NNNN...
1
2
CO
n
14 = (1/4)¹ = 0.25
1/4*4 = (1/4)² = 0.0625
¼*14*¼*¼*¼*¼ = (1/4)6 = 2.44 x 10-4
14*14*14*. (1/4)
Predicted Occurrence in a
Sequence the Length of the Human
Genome (3,234,830,000 bases)
808,707,500
202,176,875
89,753
(1/4)n* 3,234,830,000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc62819af-1f81-4d59-bfb3-1ca1702855ed%2F54f16db0-f191-4f38-8242-936305e6dff7%2F7slul5p_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part 3. Compare the Specificity of DNA-Cutting Tools
The flexibility and specificity of CRISPR-Cas9 technology offer a large step forward for gene editing.
The first DNA "scissors" were restriction enzymes, which cut DNA at predefined sequences, typically
4-8 base pairs long. For example, EcoRI, a restriction enzyme found in E. coli, will cut double-
stranded DNA at every GAATTC sequence. If EcoRI were added to a sample that contained the entire
human genome, it could cut at every GAATTC sequence.
We can calculate the probability that a particular nucleotide sequence, such as GAATTC, will occur
within a larger sequence. Table 1 below shows the calculated probabilities of finding sequences of
particular lengths. These calculations are based on the assumption that DNA sequences are entirely
random and that every nucleotide position has an equal probability of being A, T, C, or G. Use the
table to answer the following questions.
Table 1. Calculated probabilities of finding a specific sequence.
Sequence
Sequence Probability Calculation
Length
A
AC
GAATTC (EcoRI)
NNNN...
1
2
CO
n
14 = (1/4)¹ = 0.25
1/4*4 = (1/4)² = 0.0625
¼*14*¼*¼*¼*¼ = (1/4)6 = 2.44 x 10-4
14*14*14*. (1/4)
Predicted Occurrence in a
Sequence the Length of the Human
Genome (3,234,830,000 bases)
808,707,500
202,176,875
89,753
(1/4)n* 3,234,830,000
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