Name:_____answer key
___________________________
SHOW ALL OF YOUR WORK!
Genetics Exam #2
November 2, 2007
(5) Define and distinguish between silent and nonsense mutations.
silent mutations change the a nucleotide of the DNA but do not change the coded amino acid
nonsense mutations create a stop codon which terminates protein synthesis
(5) Define and distinguish between lytic and lysogenic virus stages.
lytic viruses make hundreds of copies and generally kill the cell by using up its resources
lysogenic viruses use a circular DNA intermediate to integrate into the organism's DNA where it
sits quietly, replicating with the cell and NOT killing the cell
(3) Define gene targeting.
technique using homologous recombination (often in mice) to replace one allele of a particular
gene with anything desired
(3) Define hypomorph.
type of mutation which decreases the amount of protein synthesized
(3) Define pilus.
protein/membrane finger-like projection from bacteria; used by F+ bacteria to bind to a target
(3) Define Holliday junction.
4 stranded cross-shaped intermediate found in DNA recombination
(3) Define restriction endonuclease.
protein which cleaves DNA at particular (usually palindromic) sequences
(3) Define episome.
circular piece of DNA which can exist and replicate independently OR in the host genome
(3) Define lysogen.
bacteria which has a virus integrated into its genome (ie. virus is in the lysogenic stage)
(3) Define Southern blot.
hybridization technique using a labelled probe to identify DNA bands on a blot (after
electrophoresis)
(3) Define ligase.
enzyme which makes a phosphodiester bond between 2 pieces of DNA (ie. joins 2 pieces)
(3) Define artificial chromosome.
vector capable of replicating and dividing with the cell and containing large amounts (up to 1
million bases or so) of cloned DNA
(8) Describe the 2 different types of transposons found in mammals and explain the differences
in their mechanisms of replication.
LTR retrotransposons are transposons which replcate using an RNA intermediate and contain
long terminal repeats that can act as strong promoters. When active, these transposons make
copies of themselves via RNA, make a DNA copy by reverse transcriptase, and the DNA copy
integrates elsewhere in the genome.
DNA based transposons, on the other hand, move around by physically cutting out the
transposon as a circular intermediate and moving elsewhere in the genome. This method does
not make nearly as many copies, as numbers only increase when actively moving around during
S phase of the cell cycle.
(15) You are a scientist trying to develop an improved way of making the antibacterial protein
abac which is normally purified from a severely endangered frog. In order to make abac in
bacteria, you need to clone it into an expression vector. Given the restriction maps below for the
plasmid multicloning site and the abac cDNA (with listed atg start codon and tta stop codon),
how you would go about cloning abac if the promoter must be 5' of the start codon and why?
5'-plasmid promoter-----|----------|--------|----------|------------|-------|------------|--stop codons-3'
EcoN1 SalII AvaI XbaI
NheI EcoRI BalV
abac gene: 5'----|-----|-atg----|-------|----------------|---tta----|-----------|------ 3'
NotI SalII EcoN1 ClaII
XbaI HaeIII EcoR1
The abac gene should be cut before the start codon and after the stop codon to express the
entire protein. SalII and EcoR1 fullfill this condition AND are present in the plasmid
multicloning site. Both the plasmid and abac gene should be cut with SalII and EcoR1 enzyme.
The products should then be gel purified on an agarose gel, and the large plasmid band and abac
DNA cut out of the gel. The cut bands should then be mixed together and joined together with
DNA ligase. Ligated DNA is then transfected into bacteria, grown for antibiotic resistance
(provided by the plasmid), and clones with gene inserted identified by PCR.
(9) Describe the key chemical requirements for a PCR reaction (or any DNA replication) to
proceed, and explain why PCR can obtain such high levels of amplification.
PCR requires several things, particularly a template DNA, oligonucleotide primers (2 of them,
pointed inward toward each other), dNTPs (deoxynucleotide triphosphates), thermostable (ie.
Taq) polymerase, and compatable buffer. Because PCR goes through multiple cycles of
denaturation, annealing, and extension, each cycle doubles the amount of DNA present. This
doubling occurs consistently over 20-30 cycles, allowing a billion or so copies made from a
single template.
(8) Describe the A-P endonucelase repair system, including what types of mutations it is able to
fix.
The A-P endonuclease repair system is used to replace anything that generates a gap in
one strand of the double helix. This system recognizes the empty spot on the DNA and cuts the
broken base out of the damaged strand using an endonuclease. DNA polymerase I adds the
missing nucleotide, and DNA ligase seals the gap to generate a complete double helix. Gaps in
the double helix are produced by depurination reactions (loss of purines by hydrolysis) or DNA
uracil glycosylase, which removes uracil bases created by the deamination of cytosines.
(10) Fill in all the blanks in the complimentation test below.
A
B
C
D
E
F
G
A
-
B
+
-
C
+
-
D
+
+
+
-
E
+
+
+
-
F
+
+
+
+
+
-
G
+
+
+
+
-
(10) Using the interrupted mating technique, a graduate student obtained the following table of
information regarding the listed 5 genes in the bacteria Salmonella. When the graduate student
broke her leg in a car accident, it is up to you to map the genes (position and relative distance) on
the 70 minute circular Salmonella chromosome.
time
0 min
5 min
10 min
15 min
20 min
25 min
30 min
35 min
40 min
45 min
50 min
55 min
60 min
65 min
70 min
Vxa
+
+
+
+
+
+
+
+
+
+
+
+
+
Gal
+
+
+
+
+
+
+
+
+
+
Ipn
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Rep
+
+
+
Ipn- 5-Vxa-15-Gal-10-Pla-25-Rep-15-Ipn
Pla
+
+
+
+
+
+
+
+
(10) PCR is useful for identifying individuals based on samples of DNA, and is often used in
paternity testing. Given the results below for the D1S80 locus, explain who the parents of the
child must be and why.
MW child
900
800
700
600
500
400
females
#1 #2 #3
#4
males
#5 #6 #7
#8
MW
900
800
700
600
500
400
300
300
200
200
100
100