MG 605 Lecture

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• MG 605 Lecture 2
• Did not cover in lecture the following (related
  to 1st lecture)
• You are responsible for this material
• Telomere Structure, telomerase, cancer.
• Enzymology of DNA replication and proof reading
• DNA melting and renaturation

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DNA
RNA
Ptn
Repair Replication
DNA
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One Gene/one gene product (or enz.) concept.
A
B
C
D
E
1
2
3
4
A is substrate B is product 1 is enzyme encoded
by gene 1. If genes encoded by 1,2,3,4 mutated,
then enzymes 1-4 may not work. Ex mutate 3
(deleted gene) no conversion C to D. Also no E
is made. If E is essential (suppose it is an end
product), cells cannot live/grow. This is
strategy used to deduce biosynthetic pathways in
cells.
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• One Gene one enzyme model deduced from  Expts by
  Beadle et al using the haploid organism,
  Neurospora SEE FIG 9-1 for details.
• Mutagenize (UV or X-rays)
• Isolate auxotrophic mutants (cannot grow on
  minimal media)
• Use amino acid supplemented minimal media to
  identify amino acid required for growth

Supplment to medium
Mutants plated on minimal medium growth scored
as or -.
Mutant

• To solve All grow on G. Thus G is at the end
  of path and is reqd end product. NONE grow on
  E thus is early step. Compound A gives growth
  only in 1 mutant (5). Thus it must be after E.
  C gave 2 positives and is next. Using this logic
  E A C B D G
• To identify blockage (mutation) use similar
  logic.
• Mutant 5 grows on A but not E thus E to A is
  catalyzed by enz 5
• 4 cant grow when fed A or E thus 4 mediates A
  to C step.

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Similar logic is applied since you already have
the order of the pathway E A C B
D G For ex Mutant 3 cannot grow when fed
ABCDE and only grows when G is added thus, 3
encodes the enzyme that mediates the D to G
stepand so forth.
Supplment to medium
Mutants plated on minimal medium growth scored
as or -.
Mutant
5
4
2
1
3
E A C B D G
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Amino acid sequence
A repeating structure
H bonds Stabilize shape
Alpha Helix (another is beta pleated sheet, Fig
9-6)
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Native folding Electrostatics H-Bonds Van der
Waals Hydrophobic interactions Hydrophilic
interactions
AA residues far apart can interact in the folded
str.
Myoglobin structure
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Subunits
Small ptn lt100 aa Larger gt1500 aa
Some ptns are monomers, multimers (homomultimers,
heteromult.)
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Linear seq. of nucleotides in a gene determines
the linear seq. of aa in a ptn!
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• Ptn structure (architecture) Key to function of
  that ptn.
• Mutations alter the structure of the ptn
• a new DNA sequence results in a new aa seq. of
  the polypeptide
• Altered aa means altered structure and MAY make a
  gene product unable to perform its normal (wt)
  action in vivo.
• Bottom Line ptn is non-functional and a
  mutation is manifest
• READ about X-ray determ of ptn str p 279 no
  lecture on this.
• READ about how enzyme work (Fig. 9-20) REVIEW WEB
  MOVIES LECTURE 2.

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• Not all mutations lead to inactive gene products
• Temperature sensitive alleles
• WT at non-restrictive (or permissive) temperature
• mutant at restrictive temp.

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T4 Replication cycle
Single phage particle attaches and injects
DNA This starts early events expression using
the parental phage genome DNA replication occurs
this defines start of Late events (templating
off progeny genomes). MAJOR amplification of
phage DNA! Lysis and release. Gene control and
Timing is essential
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Benzers Expts. With T4 phage Lets get some
genetic fine tuning. See Ch. 7 (pp 221-224 and
web site)
To quantify phage use plaque assays to
determine PFU/ml. Plaque is a clearing (lysis) of
bacteria on a lawn due to release of progeny
phage.
Benzers system is based upon Crosses within a
single cell between different phage
mutants -infections done at high MOI
(multiplicity of infection). If you have 10
million cells and you add 10 million phage, the
MOI 1)
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Both rII and rII- form plaques (pfu) on B
Only rII forms plaques (pfu) on K
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Here is how the rII system (Benzer) works The
awesome power of genetics
Progeny phage are plated on B this gives total
yield of ALL progeny (rII rII)
Could also count small plaques on B but
resolution would be really low. Counting wt rII
on K MUCH better!
Progeny phage are also plated on K this ONLY
rII or WT recombinants.
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To calculate the recombination frequency pfu on
K (x 2) divided by total pfu on B Or wt phage
divided (x2) divided by total yield of
wtmutant. (x2 since dbl recombinants will form
that are mutants and cant plate on K) With
recombination Frequency deduce gene order (map
mutations) to right or left of each other.
rII1
rII7
rII4
rII5
rII2
rII8
rII3
rII6
Frequency between 14 might be 0.1
Frequency between 18 might be 0.25
Must also check back reversion frequency Plate
parental phage mutants on K, any plaques are due
to reversions and must be accounted for!
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1
Mut1
Plate on K cannot replicate
X
2
Mut2
Plate on K Replicates well
WT

1
Plate on K cannot replicate
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Dbl mutant
Cross two mutants a crossover anywhere between
can result in wt plus double mutant. If
mutations are real close, frequency of cross over
will be LOW. If far apart, it will be much
higher. -This is rcbn WITHIN genes (intragenic
rcbn). Note that rcbn can also occur between
genes (intergenic). -Benzer showed that genes
WERE NOT indivisible but could be subdivided by
recombination (rcbn). This led to idea that
genes are composed of smaller units (now known as
nucleotides) and rcbn takes place between
nucleotides.
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Rcbn frequency expts Allow construction of
detailed maps
8 mutations mapped in two domains
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Deletion Mapping Experiments when chunks go
missing! Benzer found some mutants never gave WT
recombinants These behave like deletns. Also,
he found these NEVER REVERT to WT. Deletions can
be mapped in same manner as with point mutants
and once mapped can greatly speed up further
mapping of other mutations. Lets examine some
basics of deletion mapping If you crossed D1
with D2 you will not get any WT rII progeny since
they are both missing the WT information where
the deletions overlap. Its gone! Also note that
reversion cannot happen with deletions.
Intact WT gene
D1
Deleted regions in green
D2
Another view of a deletion
Intact WT gene
D2 Deletion
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1 2 3 4 5 6
7 8 9 10
WT gene
D-1
3 deletion mutants
D-2
D-3
How deletions help us map point mutants If you
cross D-1 with point mutants 1-5 No wt rII
phage (on K) Cross D-1 with mutants 6-10 wt rII
(pfu on K) will result EXPEDITES locating point
mutants.
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Using deletions to quickly fine mapping point
mutations
1 2 3 4 5 6
7 8 9 10
WT gene
D-1
3 deletion mutants
D-2
D-3
D-4
Three point mutants crossed with each of
above D-1 D-2 D-3 D-4 Mut-20 - - - Mut
ation spans? 1-5 1-8 1-10 Near
1-2 Mut-21 - - - Mutation spans?
5-10 5-8 5-8 Near 5-6 Mut-22 - Mutat
ion spans? 5-10 8-10 8-10 Near 8-10
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Complementation analysis gene function in
pseudodiploids
2 different rII mutants infect non-permissive K
strain. Neither rII can replicate in K
recall Cells infected _at_ hi MOI to make sure that
each host cell has gt1 phage mutant
If the 2 different mutants can C each other,
lysis and progeny phage will result.
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A
B-
Mutant allele 1
Precursor Product A Product B
A-
B
Mutant allele 2
In the same cell these 2 mutants complement
each other and the cells is WT for product
B. -Corresponds to MIXING of gene
products -Recombination does not occur and each
parental genome remains unaltered.
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