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Homework 4 Solutions

by: Annie Burton

Homework 4 Solutions 220

Marketplace > Chemical Engineering > 220 > Homework 4 Solutions
Annie Burton
GPA 3.45
Biological Systems
Shelly Peyton

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About this Document

Chem Eng 220
Biological Systems
Shelly Peyton
75 ?




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This 2 page Bundle was uploaded by Annie Burton on Friday October 9, 2015. The Bundle belongs to 220 at a university taught by Shelly Peyton in Fall. Since its upload, it has received 66 views.

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Date Created: 10/09/15
HOMEWORK 4 CHEMENG 220 SOLUTIONS 1 Virus M13 DNA content 50 POINTS In all samples of doublestranded DNA the mole percents of A and T are equal as are the mole percents of G and C Results such as this one stood out as odd in the days before the structure of DNA was known Now it is clear that while all cellular DNA is double stranded certain viruses contain sin glestranded DNA The genomic DNA of the M13 virus for example is sin gle stranded In singlestranded DNA A is not paired with 39139 nor G with C and so theA T and C G rules do not apply 2 Resistant bacteria 50 POINTS For either hypothesis you might expect to see about 10 surviving colonies per plate If the bacteriophages induced resistance the surviving colonies would appear in random positions on each of the replica plates If the muta tions preexisted the resistant colonies would appear at the same locations on each of the three replica plates In actual experiments of this kind the surviving colonies appear at the same locations indicating that the muta tions preexist in the population Reference Hartwell LH Hood 1 Goldberg ML Reynolds AE Silver LM 8 Veres RC 2000 Genetics From Genes to Genomes pp 217 218 New York McGraw Hill 3 Titin 50 POINTS A Since an average protein contains about 455 amino acids 50000 dl protein x amino acid110 d it will take a muscle cell about 38 minutes to make it 455 amino acids x sec2 amino acids x min60 sec Since titin is 60 times the size of an average protein the muscle cell will require 38 hours to make it 3000000 dltitin x amino acid110 d x sec2 amino acids x hr3600 secl B It will take a muscle cell about 23 minutes to transcribe an average gene and 23 hours to transcribe titin For the average protein 455 amino acids corresponds to 1365 nucleotides of RNA 3 ntlcodon x 455 codonsl Given that 5 of the initial transcript is converted to mRNA the initial transcript is 27 x 10quot nucleotides 1365 nt x 20 which would require about 23 minutes to transcribe 273 x 10 nt x sec20 nt x min60 secl Because titin is 60 times as big the muscle cell will require about 23 hours to transcribe it 4 Chemical Engineers Bioengineers and the Nobel Prize 50 POINTS 20 points Prize choice and analysis 10 points grammar 10 points plagiarism check 10 points length


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