PRELIMS Flashcards
Indicate whether the following peptides are hydrophilic
or hydrophobic:
a. MLWILSS
b. VAIKVLIL
c. CSKEGCPN
d. SSIQKNET
e. YAQKFQGRT
f. AAPLIWWA
g. SLKSSTGGQ
a. MLWILAA — hydrophobic
b. VAIKVLIL — hydrophobic
c. CSKEGCPN — hydrophilic
d. SSIQKNET — hydrophilic
e. YAQKFQGRT — hydrophilic
f. AAPLIWWA — hydrophobic
g. SLKSSTGGQ — hydrophilic
. Is the following peptide positively or negatively
charged at neutral pH?
GWWMNKCHAGHLNGVYYQGGTY
Answer: The peptide is positively charged
Consider an RNA template made from a 2:1 mixture
of C:A. What would be the three amino acids most
frequently incorporated into protein?
Answer: Proline (CCA), histidine (CAC), and
threonine (ACC) would be most frequently
incorporated.
What is the peptide sequence encoded in
AUAUAUAUAUAUAUA . . .?
Answer: The peptide sequence would alternate
isoleucine (AUA) and tyrosine (UAU):
IYIYI. . .
Write the anticodons 5’ to 3’ of the following amino
acids:
a. L
b. T
c. M
d. H
e. R
f. I
Answers:
a. L — UAA, CAA,AAG, GAG, UAG, CAG
b. T — AGU, GGU, UGU, CGU
c. M — CAU
d. H — AUG, GUG
e. R — ACG, GCG, UCG, CCG
f. I — AAU, GAU
A protein contains the sequence LGEKKW
CLRVNPKGLDESKDYLSLKSKYLLL. What is the likely function of this protein? (Note: See
Box A3-4.)
Answer: This protein has a leucine residue at se
quential 7th positions forming a leucine zipper,
found in transcription factors.
A histone-like protein contains the sequence:
PKKGSKKAVTKVQKKDGKKRKRSRK. What
characteristic of this sequence makes it likely to
associate with DNA?
This protein is positively charged, which would facilitate association with negatively charged DNA
A procedure for digestion of DNA with a restriction
enzyme includes a final incubation step of 5 minutes
at 95ºC. What is the likely purpose of this final step?
The 95ºC incubation will inactivate the protein, preventing its activity in subsequent
steps of the assay.
What is a ribozyme?
A ribozyme is an RNA molecule that can metabolize other molecules like an enzyme.
Name the nonprotein prosthetic groups for the
following conjugated proteins:
glycoprotein
lipoprotein
metalloprotein
glycoprotein — sugars
lipoprotein — lipids
metalloprotein — metal atoms
Calculate the DNA concentration in µg/mL from the following information:
a. Absorbance reading at 260 nm from a 1:100
dilution = 0.307
b. Absorbance reading at 260 nm from a 1:50
dilution = 0.307
c. Absorbance reading at 260 nm from a 1:100
dilution = 0.172
d. Absorbance reading at 260 nm from a 1:100
dilution = 0.088
a. 0.307 X 50 ug/mL = 15.35 ug/mL
15.35 ug/mL X 100 = 1535 ug/mL
b. 0.307 X 50 ug/mL = 15.35 ug/mL
15.35 ug/mL X 50 = 767.5 ug/mL
c. 0.172 X 50 ug/mL = 8.60 ug/mL
8.60 ug/mL X 100 = 860 ug/mL
d. 0.088 X 50 ug/mL = 4.40 ug/mL
4.40 ug/mL X 100 = 440 ug/mL
If the volume of the above DNA solutions was
0.5 mL, calculate the yield for (a) – (d)
a. 1535 ug/mL 0.5 mL = 767.5ug
b. 767.5 ug/mL 0.5 mL = 383.8 ug
c. 860 ug/mL 0.5 mL = 430 ug
d. 440 ug/mL 0.5 mL = 220 u
After agarose gel electrophoresis, a 0.5 µg aliquot of DNA isolated from a bacterial culture produced only a faint smear at the bottom of the gel lane. Is this an acceptable DNA sample?
This amount of bacterial DNA should produce a bright smear near the top of the gel lane. This DNA is probably degraded and is therefore unacceptable.
Contrast the measurement of DNA concentration by
spectrophotometry with analysis by fluorometry with
regard to staining requirements and accuracy.
Spectrophotometry requires no DNA staining. Fluorometry requires staining of DNA to
generate a fluorescent signal. Fluorometry may
be more accurate than spectrophotometry, since
double-stranded DNA must be intact to stain and
generate a signal, whereas single nucleotides will
absorb light in spectrophotometry.
Calculate the RNA concentration in µg/mL from the
following information:
a. Absorbance reading at 260 nm from a 1:100
dilution = 0.307.
b. Absorbance reading at 260 nm from a 1:50
dilution = 0.307.
c. Absorbance reading at 260 nm from a 1:100
dilution = 0.172.
d. Absorbance reading at 260 nm from a 1:100
dilution = 0.088.
a. 0.307 X 40 ug/mL = 12.28 ug/mL
12.28 ug/mL X 100 = 1228 ug/mL
b. 0.307 X 40 ug/mL = 12.28 ug/mL
12.28 ug/mL X 50 = 614 ug/mL
c. 0.172 X 40 ug/mL = 6.88 ug/mL
6.88 ug/mL X 100 = 688 ug/mL
d. 0.088 X 40 ug/mL = 3.52 ug/mL
3.52 ug/mL X 100 = 352 ug/mL
An RNA preparation has the following absorbance
readings:
A260 = 0.208
A280 = 0.096
Is this RNA preparation satisfactory for use?
The A260/A280 ratio is 0.208/0.096 = 2.17.
This RNA preparation is satisfactory for use.
A blood sample was held at room temperature for
5 days before being processed for RNA isolation. Will this sample likely yield optimal RNA?
This sample will likely not yield optimal RNA due to degradation and changes in gene expression at room temperature.
Name three factors that will affect yield of RNA from a paraffin-embedded tissue sample.
Isolation of RNA from fixed tissue is especially affected by the type of fixative used, the age/length of storage of the tissue, and the preliminary handling of the original specimen
You wish to perform an electrophoretic resolution of
your restriction enzyme–digested DNA. The size of
the expected fragments ranges from 100 to 500 bp.
You discover two agarose gels polymerizing on the
bench. One is 0.5% agarose; the other is 2% agarose.
Which one might you use to resolve your fragments?
The 2% agarose is best for this range of
fragment sizes.
After completion of the electrophoresis of DNA frag
ments along with the proper molecular-weight stan
dard on an agarose gel, suppose (a) or (b) below was
observed. What might be explanations for these?
a. The gel is blank (no bands, no molecular-weight
standard).
b. Only the molecular weight standard is visible.
a. Since the molecular-weight standard is not
visible, something is wrong with the general
electrophoresis process. Most likely, staining
with ethidium bromide was omitted.
b. The presence of the molecular-weight standard
indicates that the electrophoresis and staining
were performed properly. In this case, the
DNA fragments were not loaded, or the
method used to produce the fragments was
unsuccessful
How does PFGE separate larger fragments more efficiently than standard electrophoresis?
PFGE forces large fragments through
the gel matix by repeatedly changing the direction
of the electric current, thus realigning the sample
with spaces in the gel matrix.
A 6% solution of 19:1 acylamide is mixed, deaerated,
and poured between glass plates for gel formation.
After an hour, the solution is still liquid. What might
be one explanation for the gel not polymerizing?
The nucleating agent and/or the polymerization catalyst were not added to the gel solution.
A gel separation of RNA yields aberrantly migrating
bands and smears. Suggest two possible explanations
for this observation
This RNA could be degraded. Alternatively, improper gel conditions were used to separate the RNA.
Why does DNA not resolve well in solution (without
a gel matrix)?
Particles move in solution based on their
charge/mass ratio. As the mass of DNA increases,
slowing migration, its negative charge increases,
counteracting the effect of mass.