BIOL 114
Biology Dept
Kenyon College
Solutions -- Bacterial Genetics
Bacterial gene transfer
Plasmids and cloning
lac operon

Solutions--Bacterial gene transfer
Problem 5.
Two processes to create a partial diploid in bacteria are specialized transduction (also called restricted transduction) and F' conjugation.

1.  Specialized (or Restricted) Transduction.  A lysogenic bacteriophage can excise itself so as to carry a piece of host DNA by mistake.  The phage will now carry a second copy of an allele (or linked alleles)  into a host cell.  The new bacterium is a partial diploid for the allele(s).

In biotechnology, a phage chromosome can have a piece of foreign DNA ligated into it in the test tube.  Then the phage DNA is packaged into phage, and it can infect a new host where it either (1) produces many copies of the host gene; or (2) lysogenizes the host, to express the cloned DNA.

2.  F' plasmid.The F plasmid can recombine itself into the host chromosome, then recombine itself out again with some host DNA by mistake.  When it enters the next host cell, it carries a second copy of several genes; again, a partial diploid is created.

In biotechnology, a plasmid can have a piece of foreign DNA ligated into it in the test tube; then the plasmid is transformed into E. coli.  Then the plasmid makes many copies, including the cloned gene.

Problem 1.
A.  The complementation groups are defined by which mutant strains FAIL to complement.
Group A:  1, 3
Group B:  2, 4, 5, 6
There are two different genes so far, A and B.  You would have to screen a lot of other mutants before ruling out yet other groups (genes.)

Group A.  1, 5, 7
Group B.  2
Group C.  3, 4, 6

Solutions, Plasmids and Cloning

Problem on plasmid recombination.  Here are some ways the vector and/or insert DNA could recombine in the test tube:

  • Vector self-ligate in a circle.
  • Vector ligate to another vector.
  • Insert can ligate to itself, or to another insert.
  • Vector can ligate two copies of insert.
Problem on genomic cloning and cDNA cloning:

Genomic Cloning

  • To study how a gene is regulated, you need to include introns which may contain enhancers.
  • To study a gene in its natural host, the whole gene is needed.
  • To find a new gene, you can use any DNA source from any tissue.  Aside from cells of the immune system, all tissues contain the same DNA.
  • The gene sequences are extremely long; will have to be pieced together in several clones.
  • If your aim is to obtain a protein product, the gene cloned contains introns and therefore cannot be expressed in bacteria.
cDNA cloning
  • The entire coding region can be obtained in one vector and transformed into bacteria for expression.
  • You can tell the exact coding region; you do not have to figure out where the introns are.
  • You have lost any regulatory information related to the introns.
  • You need to find a source of mRNA from a tissue that expresses the desired produce.  mRNA is a pain to work with; it readily degrades.
Primers in PCR (Notes 2/20).  You need primers in PCR because the DNA polymerase, as usual, always  needs a 3' OH to attach nucleotides.

Solutions, Mutation problems
Problem 1.  Explain the advantages and disadvantages of different ways to find strains containing interesting mutations.

Observation of large numbers of progeny.
By observing large numbers, we eventually find interesting defective or different phenotypes.
Unexpected phenotypes may appear, that no one would have thought to screen for; such as the famous homeotic mutations, in which limbs appear at the wrong position, in fruit flies.
Defects may be hard to screen for by other methods, because there is no positive selection.
A large number of progeny have to be screened.
Most of the mutations will be point mutations, easily revertable.
Since we don't know what kind of mutation it is, it is hard to find the gene.
Positive selection.  This works for traits which confer survival advantage.
A large number of mutations of the same class can be obtained, and complementation groups determined.  With luck you can define a number of genes contributing to the trait.
It only works for gain-of-function mutations.  Loss of function is much more common.
Negative selection works for traits in which a loss-of-function strain can be made to grow.
Advantage: This approach gives a way to find loss-of-function mutations which could not be found by direct observation (since the strains would be dead) nor by positive selection.
Disadvantage: It only works for conditional mutations, for which we have found a permissive conditions.

Problem 2. Chromosome mutations may occur when homologous recombination makes a mistake.  The Holliday structure can rearrange so as to connect by mistake the same two arms of two homologues, instead of the long end of one with the short end of the other.  Also, when the first strand of DNA cleaves, if it attaches to some other cleaved chromosome by mistake, a translocation will occur.  Translocations and rearrangements are increased by X-rays which break the backbone of DNA.

Problem 3.  Point mutations, such as transitions and transversions, occur when (a) DNA polymerase inserts an incorrectly pairing base during replication; (b) editing enzymes fail to excise and replace the base.  Transistions (purine for purine, or pyrimidine for pyrimidine) are more common errors than transversions (purine for pyrimidine).  Point insertions or deletions may occur where an incorrect base pair causes a "bulge" that results in the DNA pol III missing or adding a base.

Problem 4.  Transcribe to RNA:

DNA        5' A A T G G G C T A C T  T A G C C A C T A G G C T  T T A G C C 3'
                3' T T A C C C G A T G A  A T C G G T G A T C C G A A A T C G G 5'

RNA (1)  5' A A U G G G C U A C U U A G C C A C U A G G C U U U A G C C 3'

                Now turn the DNA over, to make the next 5' -> 3' transcript:

RNA (2)  5' G G C U A A A G C C U A G U G G C U A A G U A G C C C A U U 3'

Which could make a message?  RNA (1) has an AUG to translate, and a stop codon:
Note the stop codon must be in frame with the AUG.

                AUG  GGC  UAC  UUA  GCC  ACU  AGG  CUU  UAG
Protein        Met -- Gly -- Tyr -- Leu --Ala -- Thr -- Arg -- Leu -- Stop

Mutations Note: many answers are possible.

Frame shift
DNA        5' A A T G G G C T A C T  T A G C C A C T ^ G G C T  T T A G C C 3'
                3' T T A C C C G A T G A  A T C G G T G A ^ C C G A A A T C G G 5'

               AUG  GGC  UAC  UUA  GCC  ACU  GGC  UUU  AG . . .
Protein        Met -- Gly -- Tyr -- Leu --Ala -- Thr -- Gly -- Phe . . .

Base pair substitution (missense, in this case)
DNA        5' A A T G G G C T A C A T A G C C A C T A G G C T  T T A G C C 3'
                3' T T A C C C G A T G T  A T C G G T G A T C C G A A A T C G G 5'

                AUG  GGC  UAC  AUA  GCC  ACU  AGG  CUU  UAG
Protein        Met -- Gly -- Tyr --  Ile --Ala -- Thr -- Arg -- Leu -- Stop

Silent mutation
DNA        5' A A T G G G C T A C T  T G G C C A C T A G G C T  T T A G C C 3'
                3' T T A C C C G A T G A  A C C G G T G A T C C G A A A T C G G 5'

                AUG  GGC  UAC  UUG  GCC  ACU  AGG  CUU  UAG
Protein        Met -- Gly -- Tyr -- Leu --Ala -- Thr -- Arg -- Leu -- Stop

Missense mutation -- see Base Pair substitution above (in this case)

Nonsense mutation (can also be caused by frameshift)
DNA        5' A A T G G G C T A A T  T A G C C A C T A G G C T  T T A G C C 3'
                3' T T A C C C G A T T A  A T C G G T G A T C C G A A A T C G G 5'

                AUG  GGC  UAA  UUA  GCC  ACU  AGG  CUU UAG
Protein        Met -- Gly -- Stop

Solutions for Week 8, Prokaryotic Genetics
Problem 2.   The active components of each operon are shown in red.

p +    lacZ -    lacI +               Expresses B-gal (lacZ), inducible by lactose.
p +    lacZ +    lacI -

p +    lacZ -    lacI -                Never expresses lacZ.
p -  lacZ +    lacI -

p +    lacZ -    lacI +                 Expresses  lacZ, inducible by lactose ( same as first example)
p +    lacZ +    lacI -

p +    o-c         lacZ +    lacI +      Expresses lacZ constitutively.  Though LacI is made, it will
---------------------------------------       never bind the constitutive operator.
p +    o +         lacZ -     lacI +

Problem 4.  Same as problem 5, above.

Problem 5. Will LacZ be expressed in each condition?
Note: For last two rows, other answers are possible.

LacI-   P+  O-c  LacZ+
LacI+  P+  O+   LacZ-
LacI+  P-   O-c  LacZ+
LacI+  P+   O+   LacZ-
       LacI-  P+   O+   LacZ
       LacI   P+   O+      LacZ- 
  LacI-   P-    O-c   LacZ+
      LacI-   P+   O +   LacZ+