Dr. Ballard
BIO 184
Page 1
Fall 2006
EXAM 1
KEY
1. (10 points)
The anticodon of a tRNA is 5’ – GUU – 3’. What amino acid does it carry? Indicate how you arrived
at your answer.
This anticodon will base-pair with codon 5’-AAC-3’ in the mRNA. This codon codes for asn, so
the tRNA will be carrying asn.
2. (12 points)
Translate the following mRNA into a polypeptide. Label the C and N termini of your polypeptide.
5’ – CCCUUUGACGGCGCAAAGGUGGGGCUUUAGGCCCGACUC – 3’
N – pro-phe-asp-gly-ala-lys-val-gly-leu-C
3. (10 points)
Explain, in clear and precise terms, why (in the most fundamental sense) all living organisms must
consume energy to survive.
In the Universe, entropy (disorder) is always increasing. Life is a highly ordered system and
must constantly fight against entropy. Energy is therefore required to drive life processes;
life processes are not spontaneous.
4. (10 points)
Medical researchers are studying a new human virus that attacks and breaks down red blood cells,
causing life-threatening anemia in its victims. In one of their experiments, they denature and
precipitate the viral proteins so that they can study the virus’ genetic material. Their analysis
reveals the following nucleotide composition: 33.4% C, 16.1% U, 35.0% G, and 15.5% A.
In which form is the virus most likely carrying its genome: ss DNA, dsDNA, ssRNA, or dsRNA?
Explain your reasoning. (Note: ss=single-stranded, ds=double-stranded)
Ds RNA. The presence of “U” indicates that the genome is made of RNA and the fact that
the amounts of C=G and U=A (at least roughly) indicates that the genome is probably doublestranded, obeying Watson-Crick base-pairing rules.
5. (12 points)
The following sequence is found at the 3’ end of a prokaryotic mRNA:
Dr. Ballard
BIO 184
Page 2
Fall 2006
5’ – UAGGGUUUUAGCGAUCUUAGGGAGAAGGACUAAGAUCGCUAAAAGCGGGCCUUUUUUUU-3’
Underline the regions of the sequence that signaled the RNA polymerase to terminate transcription
when this mRNA being synthesized.
6. (10 points)
List and briefly describe two different ways in which small RNAs like tRNA, rRNA, and 7SLRNA
are more similar to proteins than they are to DNA and mRNA.
a.
b.
They fold into complex secondary and tertiary structures that are critical for their
functions, just as proteins do.
They perform important activities in the cell just like proteins. DNA and mRNA are
carry information but are not actively involved in cellular function.
7. (12 points)
Below is a diagram of a eukaryotic cell. Using numbers and arrows, show the movement of the
protein insulin from the time its mRNA first begins to be translated to the point at which the
mature protein is secreted from the cell.
Dr. Ballard
BIO 184
Page 3
Fall 2006
True (T) or False (F)? (24 points)
If the statement is false, cross out the underlined word and replace it with the correct word.
Example:
F
DNA forensics
Dr. Ballard’s research expertise is in 23 Century Romulan cuisine.
rd
_____
Griffiths (1928) discovered the process of bacterial transcription.
_____
In eukaryotic mRNAs, the “5’ cap” consists of a methylated GTP added on backwards.
_____
RNA and DNA both have 3’ OH groups on their ribose sugars.
_____
In the early 1900s, most biochemists thought that the genetic material was a protein.
_____
Translation is catalyzed by RNA polymerase.
_____
At promoters, DNA is unwound at regions rich in A and T nucleotides.
_____
The secondary structure of proteins is stabilized by disulfide bonds.
_____
Rho is a DNA binding protein.
_____
During translation elongation, tRNAs enter the ribosome at the P site.
_____
A missense mutation substitutes one amino acid for another.
_____
A silent mutation does not change the polypeptide sequence of the protein coded by the
gene.
_____
A mutation that changes UGG to UGA is called a reading frame mutation.
_____
Sigma factor is necessary for prokaryotic transcription elongation.
_____
Eukaryotes have four different RNA polymerases.
_____
The base-pairs inside a DNA double helix are stabilized by covalent bonds.
_____
DNA binding proteins usually interact with DNA in the major grooves.
Dr. Ballard
BIO 184
Page 4
THE GENETIC CODE:
5' Base
3' Base
______________________________________________________________
|
U
|
C
|
A
|
G
|
|
_
|
|
_
|
_
|
U
| UUU |=Phe | UCU
| UAU |=Tyr | UGU |=Cys |
U
| UUC_|
| UCC =Ser | UAC_|
| UGC_|
|
C
| UUA |=Leu | UCA
| UAA |=Stop | UGA =Stop |
A
| UUG_|
| UCG
| UAG_|
| UGG =Trp |
G
|
|
|
_
|
|
C
| CUU
| CCU
| CAU |=His | CGU
|
U
| CUC =Leu | CCC =Pro | CAC_|
| CGC =Arg |
C
| CUA
| CCA
| CAA |=Gin | CGA
|
A
| CUG
| CCG
| CAG_|
| CGG
|
G
|
_
|
|
_
|
_
|
A
| AUU |
| ACU
| AAU |=Asn | AGU |=Ser |
U
| AUC |=Ile | ACC =Thr | AAC_|
| AGC_|
|
C
| AUA_|
| ACA
| AAA |=Lys | AGA |=Arg |
A
| AUG =Met | ACG
| AAG_|
| AGG_|
|
G
|
|
|
|
|
G
| GUU
| GCU
| GAU |=Asp | GGU
|
U
| GUC =Val | GCC =Ala | GAC_|
| GGC =Gly |
C
| GUA
| GCA
| GAA =Glu
| GGA
|
A
| GUG
| GCG
| GAG
| GGG
|
G
______________________________________________________________
Fall 2006