Download Exam #2 KEY

Name _________________ KEY ___________________
Biology 201 (Genetics)
Exam #2 KEY
31 October 2003

Read the question carefully before answering. Think before you write. Be concise.

You will have up to one hour to take this exam. After that, you MUST stop no matter where you
are in the exam.

If I can not read your handwriting, I will count the question wrong.

Sign the honor pledge if applicable.

Good luck!
I pledge that I have neither given nor received unauthorized assistance during the completion of this
work.
Signature: _________________________________________________________
10pts
.
1. Consider two pairs of genes (A and a) and (B and b). An organism of unknown genotype is crossed
to a tester strain. Progeny are produced with the following genotypes: 10 AaBb, 10 aabb, 40 aaBb,
40 Aabb. This problem came from Problem #1 done at the end of Chapter 8 material in lecture.
a. What is the genotype of the unknown parent? AaBb
b. What is the arrangement of the alleles in the unknown parent? Ab
aB
c. What are the locations of the genes?
Map units = # recombinant progeny/#total progeny *100
= 20/100 *100
= 20 map units
10pts
.
2. Consider the mRNA which encodes a protein:
#1
#2
5’AAGGGCGGAAAAUGCAGUCUAAACCACGUGAUAAGCAUGGGUCGUAAGAACCA
Note 1: Start and stop codons are indicated in red above. The first nucleotide of the mRNA does
NOT represent the first amino acid on the encoded polypeptide. Furthermore, the reading frame is
set by the first AUG which is preceeded by a ribosome binding site/Shine-Delgarno sequence.
a. Imagine that there is an insertion of one nucleotide in the mRNA at arrow #1. What is the effect
of this insertion on the encoded protein?
An insertion at arrow #1 will shift the reading frame by 1 nucleotide so that all the down stream
codons are off by one. This results in the incorporated amino acids being different. Specifically
after the AUG  CAX GUC UAA . Thus, the insertion “uncovers” a stop codon UAA and the
protein will be altered and much shorter than the normal protein.
b. Imagine that there is an insertion of one nucleotide in the mRNA at arrow #2. What is the effect
of this insertion on the encoded protein?
An insertion at arrow #1 will shift the reading frame by 1 nucleotide so that all the down stream
codons are off by one. This results in the incorporated amino acids being different. However, since
this insertion is past the stop codon, there will be no effect on the encoded protein.
3. problem came from Problem #3 done at the end of Chapter 8 material in lecture
10pts In flies: v+ = red (normal) eyes and v= vermillion eyes; n+ = normal wings and n = no wings; and b+ =
.
normal body color and b = black body. Flies with no wings and black body were crossed with
vermillion eyed flies. The F1 progeny were all normal.
v+n b / v+n b X v n+ b+ / v n+ b+  v+n b / v n+ b+
Female F1 progeny were crossed with male flies that had vermillion eyes, no wings, and black
bodies. The progeny are shown in the table below.
v+n b / v n+ b+ X v n b / v n b
phenotype
genotype
#
Red eye, no wing, black body
v+ n b / v n b
592
ANSWER
Part a
NR
Black body
v+ n+ b / v n b
40
SCO
No wing
v+ n b+ / v n b
5
DCO
Vermillion eye, no wing
v n b+ / v n b
45
SCO
normal
v+ n+ b+ / v n b
94
SCO
Vermillion eye
v n+ b+ / v n b
580
DCO
Vermillion eye, no wing, black body
vnb/vnb
89
SCO
Vermillion eye, black body
v n+ b / v n b
3
DCO
total = 1448
a. IN THE BLANK COLUMN WITHIN THE TABLE ABOVE, indicate the non-recombinant
progeny (as NR), the progeny that resulted from single crossovers (as SCO), and the progeny
that resulted from double crossover (as DCO).
10pts
.
10pts
.
b. Provide the order of the three genes.
Possibility 1: v+n b / v n+ b+ double crossover in this F1 female will give gametes of v+n+ b
and v n b+ which should be very rare  If these are fertilized with v n b, get F2 that are
v+n+ b / v n b and v n b+ / v n b. Since these are not the most rare progeny, this is NOT the
right order
Possibility 2: : v+ b n / v b+ n+ double crossover in this F1 female will give gametes of v+b+
n and v b n+ which should be very rare  If these are fertilized with v b n, get F2 that are
v+b+n / v b n and v b n+ / v b n. Since these ARE the most rare progeny, this IS the right
order
From: Chapter 10
notes slides 10-13
15pts
.
4. Name this structure. Nucleotide or deoxyadenosine triphosphate
Clearly box and label the nitrogenous base (adenine).
Clearly circle and label and the deoxyribose.
Label the 5’ and 3’ end of the molecule.
Indicate with a star what part of the molecule allows the two strands of DNA to be held together.
Draw in the next dNTP to be added by DNA polIII.
(Since I did not specify if the dATP shown is the template or new strand, I accepted either a dNTP
adding to the chain or a dTTP H-bonding with the shown dATP)
dTTP
5’
3’ Next nucleotide added here
15pts
.
5. Below is a replication fork of a DNA molecule, where the arrow indicate the direction the
replication fork is moving. Note that this is from slides 15-17 from Ch. 11 and Fig 11-11 (except that 5’ and 3’
ends were inverted to make sure that you truly understood the concepts and did not just memorize the pattern) and Fig
11-6/7).
3’
leading
5’
3’
3’
5’
lagging
5’
Carefully, examine the DNA molecule shown. Then:
a.
Draw in the newly replicated strands.
b.
Indicate the polarity of the new strands.
c.
Clearly indicate which strands are the leading and lagging strands.
d.
Indicate the location of the RNA primers with a box.
e.
If this were a eukaryotic molecule, which DNA strand would need telomerase? lagging
10pts
.
6. The DNA sequence below contains a gene encoding a polypeptide.
3’XXXCTATATTAACCCCCTTCCTCCGCTACGGGTTTATTCCCGGGXXX 5’
5’XXXGATATAATTGGGGGAAGGAGGCGATGCCCAAATAAGGGCCCXXX 3’
Info for this question in many places including:
Chapter 12 (Transcription) Slide 6, 9,
Chapter 13 (Translation) Slide 10
Extra Problem set on Web for these chapters #2 and 8
Yes, I know that there are no 5’ and 3’ ends drawn on the molecule above!
a. What is the sequence of the mRNA? (Hint: Think about what features need to be on a
mRNA molecule that produces a polypeptide/protein.)
5’GGGGGAAGGAGGCGAUGCCCAAAUAAGGGCCCXXX 3’
b. Indicate the 5’ and 3’ ends of the DNA given above and on the mRNA from part a.
c. Show the polypeptide that will be produced.
met-pro-lys
d. Show the codon: tRNA interaction for just the the start codon, labeling 5’ and 3’ ends of
the anticodon.
5’AUG3’ codon
3’UAC5’ anticodon
e. Box the approximate promoter region.
Promoter is on DNA before the start of transcription.
f. Circle the approximate area for the ribosome binding site.
Ribosome binding site (Shine-Delgarno) is on RNA before the AUG..
Multiple choice section: (30 points total – 5 points per question)
Write your answer in the blank provided to the left. If you want to explain your answer, you can do so
next to the question
B
A or B
or C or
D
(D most
likely)
E
A
B
B
7. A virus contains 5% adenine, 25% thymine, 29% guanine, and 41% cytosine. The genetic material
in this virus is:
Correct answer is B; however,
a. double-stranded DNA
since I never explicitly told you
b. single-stranded DNA
that DNA can be single stranded
c. double-stranded RNA
in some cases, I will accept any
d. single-stranded RNA
answer.
e. either a or b
8. You have a DNA molecule that after replication has many changes in the nucleotide sequence of the
new DNA, relative to the old strand. This is most likely due to problems with
a. DNA polymerase I 5’  3’ polymerase activity
b. DNA polymerase III 5’  3’ polymerase activity
Page 219
c. DNA polymerase I 3’  5’ exonuclease activity
d. DNA polymerase III 3’  5’ exonuclease activity
e. DNA polymerase III 5’  3’ exonuclease activity
9. Gene Z encoded Enzyme Z. If you delete the promoter of gene Z,
a. Enzyme Z would have a frameshift mutation
b. The mRNA encoding Enzyme Z would be shorter than normal
c. There would be no Enzyme Z protein
d. There would be no Enzyme Z mRNA
e. C and D
10. Select the posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes.
a. 5'-capping, 3'-poly(A) tail addition, splicing
b. 3'-capping, 5'-poly(A) tail addition, splicing
Extra Problem set on Web for
c. removal of exons, insertion of introns, capping
Ch12-13 #7
d. 5-poly(A) tail addition, insertion of introns, capping
e. A and C
11. A new inhibitor of prokaryotic protein synthesis, Vikocyde, has been discovered in the skin of the
Atlantic salmon. In the presence of Vikocyde, protein synthesis in E. coli initiates, but only
dipeptides (two amino acids linked together) are formed, and these remain bound to the ribosomes.
Vikocyde most likely inhibits protein synthesis by blocking:
a. binding of formylmethionyl-tRNA
b. translocation
Ch 13 Fig 13-7
c. activation of amino acids
d. binding of the 50S ribosomal subunit to the initiating complex
e. formation of peptide bonds (peptidyl transferase activity)
12. In order for the genetic code to be used, each aminoacyl tRNA synthetase must properly recognize
what two things?
a. codons and anticodons
Extra Problem set
b. one specific amino acid and multiple different tRNAs carrying that amino acid
on Web for Ch12c. one specific amino acid and only one type of tRNA carrying that amino acid
13 #12 and
d. multiple amino acids and multiple different tRNAs carrying those amino acids
slide #14 from
e. multiple amino acids and only one type of tRNA carrying those amino acids
Chapter 13
(translation)