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Quiz #1, PHYS 3511 (Biological PHYSICS), RB 3049, Monday 22 October 2018, 4:35 PM to 5:15 PM INSTRUCTOR: Apichart Linhananta Student Name: STUDENT ID: This exam book has 5 pages including an equation sheet on page 5. All works must be done on this exam paper. Only one non-programmable calculator is allowed. PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 4) For each question circle the correct answer (A, B, C, D or E).

1. (2.5 points) The figure below shows how an affinity chromatography works. Circle the only correct statement. ANSWER D

A) The largest molecules will reach the bottom first, due to Stokes Formula, which quantifies the drag fluid force on molecules. B) The column is filled with a matrix with small holes that trapped small molecules allowing the largest molecules to reach the bottom first.

C) The matrix has an affinity for particular molecules, and only allows other molecules to passed through. The molecules that passed through are then analyzed. D) The matrix in the tube has an affinity for particular molecules, in this case, it is the yellow molecules. Hence only the yellow molecules are retained in the column. The yellow molecules are then eluted, then analyzed.

E) None of the above statements are true.

2. (2.5 points) The Figure below shows a plot of the fractional occupancy vs. oxygen partial pressure for myoglobin and hemoglobin. Circle the only correct statement: ANSWER D

A) The plot shows that the binding of oxygen to myoglobin is cooperative, in that only four oxygens are observed to be bonded to myoglobin, never one, two or three. B) Oxygen binding to myoglobin is not cooperative. C) Hemoglobin and myoglobin contain four heme groups.

D) Hemoglobin is cooperative in that the binding of one oxygen will increase its oxygen affinity, so that more oxygens will bind to it.

E) None of the above answers are true.

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3. (2.5 points) The figure below illustrates the Hershey-Chase experiment. Circle the only correct statement. (the bacteria are the green tubes, with the phage viruses on the outside).

A) In Figure A, analysis of the bacteria find that the bacteria is radioactive. B) In Figure A, analysis of the bacteria find that the bacteria is not radioactive. C) In Figure B, analysis of the bacteria find that the bacteria is radioactive. D) In Figure B, analysis of the bacteria find that the bacteria is not radioactive. E) None of the above answers are true. ANSWER A and D

4. (2.5 points) Circle the only true statement on the Central Dogma of molecular biology. A) Introns are the coding part of a gene B) Genes are encoded on the protein part of chromosomes

C) RNA has bases Thymine (T), Cytosine (C), Adenine (A), Guanine (G) D) In transcription an mRNA is read to produce an amino acid chain (protein).

E) In translation an mRNA is read by a ribosome to produce an amino acid chain (protein). ANSWER E PART II: FULL ANSWER QUESTIONS (question 5 to 7) Do all two of three questions on the provided space. Show all work. 5. (10 points) Consider an aerobic Bacterium that consumes 1/3 the O2 at its surface is represented by the DE !

!"#𝑟% !&

!"' = 0, with boundary conditions: 1) 𝑐(∞) = 𝑐., the

concentration of O2 at 𝑟 = ∞, must equal the bulk concentration, c0; 2) 𝑐(𝑅) = %0.1

.

A) Show that the solution is 𝑐(𝑟) = 𝑐. #1 −41"'. Proof is by direct substitutions. (2 points)

B) Find 𝑗 = −𝐷 !&!"

, and 𝛷 = (𝑎𝑟𝑒𝑎𝑜𝑓𝑠𝑝ℎ𝑒re) × 𝑗(𝑟) (2 points)

𝑗 = −𝐷𝑑𝑐. #1 −

𝑅3𝑟'

𝑑𝑟 = −𝐷𝑐.𝑅3𝑟%.

𝛷 = (4𝜋𝑟%) × 𝑗(𝑟) =4𝜋𝐷𝑐.𝑅

3 , where I neglected the negative sign since the rate of O2 deposition is an absolute value. C) The activity, Α, is equal to the oxygen consumed by the one kg of bacteria per second

Α =𝛷

𝑀KL&MN"OL,

BACTERIOPHAGESANDMOLECULARBIOLOGY149

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where the mass of the bacteria is 𝑀KL&MN"OL = 𝜌 Q1𝜋𝑅1, with the density of the bacteria assumed

to be that of water 𝜌 = 1000𝑘𝑔 ∙ 𝑚V1, and the volume of the bacteria is Q1𝜋𝑅1(assumed to be a

sphere of radius R). Use the above relation to find the maximum radius, Rmax, of a bacterium. Assume that the mass of the spherical bacteria is 𝑀KL&MN"OL = 𝜌 Q

1𝜋𝑅1, with the density of the

bacteria assumed to be that of water 𝜌 = 1000𝑘𝑔 ∙ 𝑚V1, and the volume of the bacteria is Q1𝜋𝑅1(assumed to be a sphere of radius R):

Α =4𝜋𝑅𝐷𝑐.

3 × 𝜌 43𝜋𝑅1,

which gives a maximum radius of

𝑅WLX = Y𝐷𝑐.𝜌Α .

(4 points) D)Find Rmax for Α = 0.006 W[\

]∙^_; D = 2 × 10Vb W

c

]; 𝑐. = 0.2W[\

Wd .

𝑅WLX = Ye%×f.ghi

cj k#..%

ilmid '

(f...^_∙Wgd)#....nilmj∙op'

= 8.2 × 10Vn𝑚 = 8.2𝜇𝑚 (2 points)

6. (10 points) In the diagram below, estimate the length (in bases (bp) and 𝜇𝑚) and mass of the mRNA, and the length (in amino acid (aa) and 𝜇𝑚) and the mass of the protein associate with this mRNA.

.Doing a visual “by the eyes” approximation the length of the mRNA is about ℓ = 2000𝑏𝑝. From table 1.1 on the last page the length of one bp or nt is about 0.3 nm, which gives a length on 600 nm. Using the above data (660uL

Kv→ 330 uL

xM) the mass is about, 𝑀W4yz =

330uLxM× 1 × 10V%{ ^_

uLℓ = 6.6 × 10V%%𝑘𝑔. (5 points)

Since 3 bp or nt corresponds to one amino acid (aa), the associated protein will have an approximate length of ℓ = %...Kv

1|}~~= 666𝑎𝑎. Using data from above ℓ = 666𝑎𝑎 × 0.8 xW

LL=

533𝑛𝑚. Using data from table 1.1, 𝑀v"[MNOx = 100uLLL× 1 × 10V%{ ^_

uL× 666𝑎𝑎 =

6.6 × 10V%1𝑘𝑔. (5 points)

Exercise 3) Problem 4.3 of Chapter 4 Exercise 4) Problem 4.6 of Chapter 4

•4.3Geneticsbythenumbers

(A)

1 dalton = 1Da = 1 × 10V%{𝑘𝑔 or = 1 × 10V%Q𝑔 1 nucleotide pair is about 660 daltons Length of amino acid is about 0.8 nanometer (nm)

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7. Three successive nucleotides along a DNA molecule, called a codon, encode one amino acid. The weight of an E. coli DNA molecule is about 3.1 × 109 daltons. The average weight of a nucleotide pair is 660 daltons, and each nucleotide pair contributes 0.34 nm to the overall DNA length. (a) What fraction of the mass of an E. coli cell corresponds to its genomic DNA?

The mass of the DNA is𝑀uyz = 3.1 ×10b𝐷𝑎 × 1 × 10V%{ ^_uL= 3.1 × 10Vf�𝑘𝑔.

From table 1.1 the mass of an E. Coli is 1.0𝑝𝑔 = 1 × 10Vf�𝑘𝑔. Hence the E. Coli’s DNA to total mass fraction is 3.1 × 10V1 or 0.31%. (2 points)

(b) Calculate the length of the E. coli DNA. Comment on how it compares with the size of a single E. coli cell. ℓ = 3.1 × 10b𝐷𝑎 ÷ 660uL

Kv= 4.7 × 10n𝑏𝑝 (2 points)

and since each nucleotide pair contributes 0.34 nm, ℓ = 4.7 × 10n𝑏𝑝 × 0.34 xWKv=

1.6 × 10n𝑛𝑚 = 1.6𝑚𝑚, which is visible to the naked eyes. (2 points) (c) Assume that the average protein in E. coli is a chain of 300 amino acids. What is the

maximum number of proteins that can be coded by the E. coli DNA? 300 aa corresponds to 900 nt (or bp) of an mRNA (or DNA). Since ℓ = 4.7 × 10n𝑏𝑝, justifiable conversion givesQ.{×f.

�Kv

b.. |}}�l����

= 5222 proteins. (2 points)

(d) On an alien world, the genetic code consists of two base pairs per codon. There are still four different bases. How many different amino acids (i.e. maximum number of amino acids) can be encoded? 4 × 4 = 16, but this is the maximum number, not the actual numbers that will actually be encoded in this alien world. (2 points)

(e) On another alien world, the genetic code consists of four base pairs per codon. There are now three different bases (say A, G, C). How many different amino acids (i.e. maximum number of amino acids) can be encoded? 3 × 3 × 3 × 3 = 81 (2 points)

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