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BIOL 411K/511K
EXAM I Spring 2008 Clinical Chemistry II: Molecular Genetics Name:______________________________ (252 Points total)
Definitions and descriptions (187 pts)
- (2 7 pts ) Select the correct pipettor for each volume and fill in the proper settings: a. 127 μl (5 pts) Red Black Black P Red P Black Black Black Black P Black P Black Black Red
b. 320 μl (5pts) Red Black Black P Red P Black Black Black Black P Black P Black Black Red
c. 7.3 μl (5pts) Red Black Black P Red P Black Black Black Black P Black P Black Black Red
OR
d. 0.25 μl (5pts) Red Black Black P Red P Black Black Black Black P Black P Black Black Red
e. 1350 μl (briefly describe the proper procedure here) (7 pts) __pipette 675 μl twice_____________________________________________ Black Black P Red P Black Black Black Black P Black P Black Black Red
Red^0
- ( 5pts ) The Gilson micro-pipettors used in the lab are negative displacement pipettors. Describe the proper use of this type of pipettor. Include in your description what you should specifically do and not do.
- Pipette slowly (especially large volumes (>500 μl) or high viscosity. 2.Discharge air above liquid (first stop)
- Release slowly to take up liquid
- Place to same position (below liquid, to side of tube, whatever so long as it is always done in the same way.
- Discharge by depressing to the second (last stop) under the liquid.
- Remove pipette while still depressed fully (to prevent taking liquid back out)
- ( 5pts ) What are two likely sources of pipetting error?
- not using same technique each time/ dispense in same region-wall or under liquid; Pre- wet tip; routinely observe amount in tip; etc
- Incorrrect used of stops
- Incorrectly set pipette
- Pipette not calibrated
- What is a bacterial restriction enzyme used for in molecular biology? (5 pts) Type II restriction enzymes produce dsDNA fragments by hydrolysis of the phosphoester bonds on boh sides of a palindromic sequence. It is used (in our examples) to produce a sequence specific cleavage specific patter (or DNA fingerprint). Less important for this class: it is primarily the enzyme that makes cloning possible.
- The most fundamental technique used this semester in this lab (in each lab actually) is PCR. Carefully diagram below the products present following the cycles indicated if the reaction tube contained only the two template strands given below. Be careful to reproduce the a,b,c or d or a’,b’,c’ or d’ markers so I will know precisely which strand you are diagramming. Don’t forget to position the target gene in the copies as well. (30 pts) a. Not including the two original template strands, diagram the products in your reaction tube following cycle one of denaturation, annealing and extension, assuming all reaction components are present. (5 pts) b c d F a’ b’ c’
R
a b c d a’ b’ c’ d’ f-primer binding site r-primer binding site Target gene
b. Not including the two original template strands, diagram the products in your reaction tube following cycle two of denaturation, annealing and extension, assuming all reaction components are present. (5 pts) b c d F a’ b’ c’ R b c d F a’ b’ c’ R b’ c’ R b c F c. Not including the two original template strands, diagram the products in your reaction tube following cycle three of denaturation, annealing and extension, assuming all reaction components are present. (5 pts) b c d F a’ b’ c’ R b c d F a’ b’ c’ R b’ c’ R b c F b c d F a’ b’ c’ R b c F b c F b c F b’ c’ R b’ c’ R b’ c’ R d. Not including the two original template strands, diagram the products in your reaction tube following cycle three of denaturation, annealing and extension, if you forgot to add the r-primer, but all other reaction components are present. (5 pts) e. Not including the two original template strands, diagram the products in your reaction tube following cycle two of denaturation, annealing and extension, if you forgot to add the dNTP mix , but all other reaction components are present. (5 pts) NOTHING b c d F b c d F b c d F
- (10 pts) Inhibition of amplification is always a possibility in PCR, given the extreme sensitivity of the thermal cycling conditions and enzymes. Describe a. (5 pts) The controls necessary to diagnose amplification inhibition correctly
- A known positive control to establish that the reactants and condition are appropriate.
- A positive spike control to a second test with unknown sample present to establish that if the template is present, that amplification does occur. b. (5 pts) What steps you could take to alleviate the problem
- Clean sample a. Precipitate and wash DNA with Ethanol b. Wash DNA on surface of ultra-filter or other support matrix (purification columns are manufactured for this purpose).
- Simply dilute or use less sample
- (20 pts) The focus (so far) of this molecular laboratory has been pathogen identification. What methods (molecular and otherwise) exist for such determinations and what are the relative advantages and disadvantages of each?
- Culture based Adv Low cost, lots of existing data and statistics Disadv Slow, must be culturable, can usually only differentiate a limited range
- Immunological Adv fast, can determine previous exposure (Ab targeted) Disadv significant background and cross-reactivity, cannot establish if organism is still present
- DNA sequence Adv Very specific, moderately fast Disadv Must know a great deal about the target organism, fragile methadology
- PCR Adv Very fast, very specific Disadv Oragnism must be present, easily inhibited
Calculations (65 pts)
Useful conversions and values: The MW of 1 bp is ~ 660 g/mole Avogadro’s number = 6.023 e213 molecules/mole 1 A 260 unit = 50 ng/μl dsDNA
- (15 pts) You purchase a primer that arrives as a dried material with a mass of 2.74 e-3 g. In a 5 ml screw top plastic tube. The molecular weight is 6600 mg/mmol). You need a 200 mole/Liter stock in nuclease-free water. How many l of water should you add? (2.74 e-3 g)/ (6600 g/mol)= 4.15e-7 moles 4.15e-7 moles/ X liters = 2e-4 moles/liter X liters =4.15e-7 moles/2e-4 moles/liter = 2.08 e-3 liters = 2.08 ml = 2080 μl
- (15 pts) Using the 200 Mole/Liter Stock Solution you made up in Question 1, how would you make a 20 mole/Liter Working Solution using 7 l of the 200 Mole/Liter Stock Solution? 7 μl/(13 μl + x μl H2O ) = 20 Mole/Liter/ 200 Mole/Liter = 0.1 dilution factor 0.1 = 7 μl/(7 μl + x μl H2O ) (0.1)7 μl + 0.1x μl H2O = 7 μl 0.7 μl + 0.1x μl H2O = 7 μl 0.1x μl H2O = 7 μl – 0.7 μl = 6.3 μl x μl H2O = 6.3 μl/0.1 = 63 μl check result: [7/ (7 +63)]x 200 = [7/70]x 200 = 0.1 x 200 = 20
- a. (5 pts) How many nmoles of DNA are there in 103 μl of a 10.2 pM solution? _______ (1e-11 moles/l)(1.03e-4 l) = 1.03e-15 moles x 1e9 nmoles/mole = 1.03e-6 nmols b. (5 pts) If you added this 103 μl to 06.73 ml of water what will the μM concentration be? (1.02e-11 moles/l)[103 μl/(103 μl + 6.73 e3 μl)] = (1.02e-11 moles/l)[0.015)] = 1.53e-13 moles/l (1.53e-13 M)(1e6 μM/M) **= 1.53e-7 μM
- (25 pts)** Calculate the number of stx 1 gene copies (one per E. coli EDL 933 [genome size = 4,639,221bp) present in a 25 μl PCR reaction containing 1μl of material taken from a 10e- dilution (900 μl total volume) (undiluted stock contained EDL 933 genomic DNA with a 260 nm absorbance of 4.70). (4.70 A260)(50 ng/μl/A260) = 235 ng/μl MW EDL 933 chromosome = (4,639,221bp)(660 g/mole/bp) = 3.06e9 g/mole (2.35e2-5 ng/μl)(1 μl)(1e-9 g/ng) = 2.35 e-12 g/3.06e9 g/mole = 7.19 e-21 moles ( 7.19 e-21 moles)(6.023e23 molecules/mole) = 4.33e-21+23+1 = 4.33e3 molecules/rxn
