Molecular Biology Through Discovery
Problem Set 5: The Coding Problem

1. Using a convenient genetic code table, complete the
following:

2. Consider the RNA sequence below. Suppose that the fourth
base, C, were mutated to a U. GAGCGUGCGAACC
  2a. How many amino acids might be affected if the code
were nonoverlapping triplet?

2b. How many if the code were overlapping triplet?

2c. Partially overlapping triplet?

2d. How would your answers be affected if the mutation were a
deletion of the C?

3. Why were no suppressors of FC0 found in segments B1b2 and B2
(see Fig. 2 of Crick et al, 1961)? Let’s find out. In doing so,
I’ll make use of two pieces of special knowledge. ViroBIKE calls
the T4 rIIB gene gene0288 and calls its protein p-gene0288.

3a. Go to ViroBIKE (not CyanoBIKE) through the BioBIKE portal or
the course Resources & Links page, and from the All menu, bring
down the SEQUENCE-OF function. Display the amino acid sequence of
the rIIB protein by executing the function below (doubleclick
SEQUENCE-OF):
Does the protein sequence begin with a methionine? How big is the
protein? From the map of rIIB found in Crick (1962)1 estimate the
amino acid coordinates of the FC0 mutation and of the beginning of
the B1b2 segment.

3b. From the All menu, bring down the READING-FRAMES-OF
function. Display the reading frames of the rIIB gene by
double-clicking READING-FRAMES-OF in: What is the length of the DNA
sequence shown? How does that compare to the length of the amino
acid sequence of the rIIB protein you found in 3a?

3c. How many reading (translation) frames are displayed? Why
that many? Which one gives an amino acid sequence that matches the
amino acid sequence of the rIIB protein?

3d. What is the significance of the second reading frame? Hint:
Note that the M of the first reading frame lies directly under the
A of the first triplet. Where does the C of the second reading
frame lie?

3e. Make an insertion mutant of rIIB, placing an extra A at
position 15 of the gene’s nucleotide sequence. Here’s how. From the
All menu, bring down the INSERT function. Drag the SEQUENCE-OF
function into the target box and edit the entity within SEQUENCE-OF
so that it’s the gene gene0288 (not the protein p-gene0288). Into
the insert-stuff box, type “A” and press the Enter key. Finally,
mouse over the Options icon, select AFTER, and then type 15. Close
the box, producing:

3e1. Execute the function. How does the resulting DNA sequence
(in the result pane) compare with the sequence of gene0288 you
obtained from READING-FRAMESOF? Now bring down from the All menu
the TRANSLATION-OF function. Drag the INSERT function (and all
that’s in it) into the entity box of TRANSLATION-OF.

3e2. Execute the function. How does the resulting amino acid
sequence (in the result pane) compare with the amino acid sequences
in the different reading frames you found in 3b-3d?

3e3. What is the significance of reading frame #2 relative to
reading frame #1? Of reading frame #3 relative to reading frame
#1?

3f. What is the significance of the asterisks (*) that
occasionally appear in the amino acid sequences of some reading
frames? (check to make sure your guess is correct) Do any appear in
the first reading frame? Why not?

3g. The positions of the FC0 mutation and the FC9 mutation are
very close together, according to Fig. 2 of Crick et al (1961), but
the mutations that can suppress FC0 extend into B1b1 but no
further, while the mutations that can suppress FC9 extend into B2.
Considering your answers to the above questions (particularly 3a,
3e2, and 3f), why were Problem Set 5 (Coding Problem) – 3 no
suppressors of FC0 found in segments B1b2 and B2? (and is FC0 an
insertion or a deletion)?

4. Use the simulation of Crick et al (1961) to recreate their
experiment

4a. Create mutants that suppress FC0. List the steps you took,
including the number of plaques you got at each step, the map
positions of each mutant, and the phenotypes of each.

4b. [Not easy!] Use the mutants created in 4a to make double
mutants and finally a wildtype triple mutant. Again, list the steps
you took, including the number of plaques you got at each step, the
map positions of each mutant, and the phenotypes of each.

5. We live in a world in which genes determine the linear
sequence of amino acids that comprise a protein. There are only 20
possible amino acids that may be encoded (putting aside some
specialized cases), and there are no restrictions as to what amino
acid sequences are possible to encode.

5a. How many possible dipeptides are there? In other words, if
you chop up all possible proteins (every conceivable sequence) into
two amino acid-segments, how many different kinds of amino acid
pairs would you get? The remaining questions concern an alternate
universe in which the genetic code consists of overlapping
triplets, each codon overlapping the next by two nucleotides.

5b. Consider the triplet codon CAG. How many pairs of adjacent
codons are possible in which the first codon of the pair is CAG?
What is the maximum number of dipeptides that can be encoded by all
of those pairs?

5c. How many possible triplet codons are there?

5d. How many possible pairs of adjacent triplet codons are
there? What is the maximum number of dipeptides that can be encoded
by all of those pairs?

5e. Suppose that the overlapping triplet genetic code we’re
considering is degenerate, that is more than one triplet may encode
the same amino acid. If the dipeptides shown below are found in
nature, how many triplets, at minimum, must encode histidine (His)?
His-Lys, His-Ser, His-Leu, His-Thr, His-Phe, His-Pro Lys-His,
Ser-His, Cys-His, Arg-His, Val-His, Phe-His, Glu-His, Gln-His,
Ile-His

5f. It is 1957. There are many partial amino acid sequences of
proteins known, but DNA sequencing is 20 years in the future. Can
you think of a way to use known protein sequences to test the
proposition that the genetic code consists of overlapping
triplets?

5g. You might enjoy reading the following article: Brenner S
(1957). On the impossibility of all overlapping triplet codes in
information transfer from nucleic acid to proteins. Proc Natl Acad
Sci USA 43:687-694

6. In 1954, George Gamow published the first attempt to conceive
a genetic code. You can read about it in this very short
article:
Gamow G (1954). Possible relation between deoxyribonucleic acid and
protein structures. Nature 173:318.
  Although Gamow was a fine artist (he illustrated his
own popular science books), it may be difficult for you to
interpret his rendition of the double helix. Each diamond in the
article’s figure lies within four nucleotides, defined by a
basepair, one nucleotide above it, and one nucleotide below it.
I’ve tried to clarify its message in the figure to the right. Each
diamond is an amino acid binding site, surrounded by four
nucleotides (highlighted in red for one diamond and blue for the
other).

6a. Make up a genetic code that satisfies Gamow’s criteria. The
box shown below should be helpful. I suggest that you first make
some arbitrary assignment (i.e. one of the 64 codons codes for some
arbitrary amino acid), determine what other codons must code for
the same amino acid, and then proceed along the same vein. Note
that in his scheme, there is no concept of 5′ and 3′ and so the
following diamonds are all equivalent:

6b. Based on the genetic code you just made up, what is the DNA
sequence that would encode Gly-Ala-Gly? Phe-Ala-Phe? T G T A T A G
T T G A T A T G T Problem Set 5 (Coding Problem) – 5

6c. Show that no code that follows Gamow’s criteria could ever
encode both of these tripeptides.

7. Suppose that every Virginia resident is to be assigned an ID
number, except that it will be in the form of a DNA sequence. How
long would the DNA sequence need to be to allow for a unique
sequence for every resident? Provide details of your calculation
plus any assumptions you made. Extra credit: Choose the sequence
that would be your own ID.

ADDITIONAL LINKS:
Class page –
http://www.people.vcu.edu/~elhaij/bnfo300/17F/index.html
Links –
http://www.people.vcu.edu/~elhaij/bnfo300/17F/Links/index.html

BioBike – http://biobike.csbc.vcu.edu/