Genetics (BI 211) Name: Write on back of last page (1 point)
Spring 2013
Multiple choice (2 points each)
1. Spider size is regulated by 4 genes. If two 16 mg spiders mate, while many of the progeny are 16 mg, the range in size goes from 4 mg to 28 mg. What increment in grams does each gene contribute to spider size?
A. 3
B. 4
C. 6
D. 9
E. 12
2. In the ABO blood types, which offspring is not possible?
A. An individual with type O blood from a mother with type A and a father with type AB.
B. An individual with type AB blood from a mother with type B and a father with type AB.
C. An individual with type A blood from a mother with type O and a father with type AB.
D. An individual with type B blood from a mother with type A and a father with type AB.
E. all of the above
3. A female is heterozygous for the recessive X-linked gene for Lesch-Nyhan syndrome. What proportion of her daughters will be carriers for the trait if their father is not affected?
A. 0%
B. 25%
C. 50%
D. 75%
E. 100%
4. What observation did Griffith make in his experiments with Streptococcus pneumoniae?
A. The mouse did not survive when injected with a mixture of live, avirulent (smooth) Streptococcus pneumoniae and heat-killed, virulent Streptococcus pneumoniae.
B. That DNA is the genetic material.
C. The mouse survived injection of live virulent (smooth) Streptococcus pneumoniae.
D. The heat-killed, virulent Streptococcus pneumoniae was lethal to the mouse.
E. All of the above
5. Which activity of E. coli DNA polymerase III is responsible for proofreading the newly synthesized DNA?
A. 5' to 3' polymerase
B. 3' to 5' endonuclease
C. 3' to 5' exonuclease
D. 5' to 3' exonuclease
E. none of the above
1. Why is the upper limit of crossing over 50% recombinant gametes?
Two genes are linked on chromosome 2 of Drosophila melanogaster. After crossing a female parent that is heterozygous at the two loci and a male parent that is homozygous recessive at both loci the following data were obtained. What type of specialized cross is being referred to above? Based on the data what is the arrangement of the female heterozygote’s alleles and how many map units apart are they? Red eyes is dominant to yellow. WT sex combs is dominant to short.
Red eyes, short sex combs 250 flies
Yellow eyes, WT sex combs 250 flies
Yellow eyes, short sex combs 25 flies
Red eyes, WT sex combs 25 flies
2. A Geneticist studying tail length in gerbils noticed that in the wild gerbil tails varied in length in 4 centimeter intervals from short to long with the longest being 28 centimeters. When he crossed a true breeding 12 centimeter tailed gerbil to a true breeding 28 centimeter rose he obtained all 20 centimeter tailed gerbils. When he crossed 2 of the 16 centimeter progeny he obtained a range of sizes from 4-28 centimeters and the 4 centimeter tail occurred in 1/64 of the crosses progeny.
What is the mode(s) of inheritance demonstrated by the above cross? How many genes are involved? How many phenotypes are possible for gerbil tail length from above?
What are possible genotypes of the 12 centimeter gerbil crossed to the 28 centimeter gerbil. What was the genotype of the 20 centimeter progeny?
3. A scientist was studying anteaters and found an unusual population in the wild in which a quarter of the animals had short snouts and 3/16 of the animals had medium snouts. The rest had long snouts. She then performed crosses between these short-snouted animals and anteaters with normal (long) snouts. What she saw in the offspring was either what she expected, which was that all of the offspring had long snouts (cross #1), or something weird, which was that only ¼ of the offspring were normal, half had the abnormal, short snouts, and the last ¼ had medium snouts.
What mode(s) of inheritance is(are) illustrated by this example?
What are the genotypes of the parents of cross 1 and cross 2?
Is it possible to cross a true-breeding medium-snouted animal to a true-breeding
short snouted animal and get a long snouted animal? If not, explain – if so, show
genotypes.
4. Two crosses were performed with eggplants to learn something about the genetics of color (either purple or green) and shape (either plump or lean).
Cross Offspring
purple, plump female X green, lean male 36 purple, lean
40 green, plump
11 purple, plump
12 green, lean
green, lean female X purple, plump male 49 purple, plump females
51 green, lean males
What are the genotypes of the parents of the two crosses. Be sure to be
THOROUGH in how you assign genotypes. In other words, provide all
information that can be derived from these data including potential arrangement of the genes on a chromosome.
5. Draw a biochemical pathway (enzymes and substrates; use your imagination) demonstrating how complementary gene interaction works. Also show it genetically by writing out the normal ratio of genotypes, the altered ratio of genotypes, and the resultant phenotypes based on your biochemical pathway.
6. A white elephant with black tusks was crossed to a grey elephant with white tusks. The F1 generation was all grey elephants with white tusks. The F2 generation had 100 progeny. 72 were grey and 28 were white. 56 had black tusks and rest had white tusks. How many genes are segregating in the original parental cross? What are the genotypes of the parental, F1, and F2 individuals. What mode(s) of inheritance are demonstrated?
7. What would be the effect on chromosome replication in E. coli strains carrying the following mutations?
a. dnaE (polymerase III subunit) ______________________________
b. polA (DNA polymerase I) _______________________________
c. dnaG (primase) ____________________________________
d. lig (ligase) ________________________________________
e. ssb (single stranded binding protein) ______________________________
f. oriC (origin of replication) ____________________________________
g. sigma 70 _________________________________________________
8. Match the chromosomal mutation (1-5) or the disease (6-10) to the correct terms. Multiple terms may be needed for each number. Some terms may not be used at all, others more than once.
Chromosome Term
1. A B C F E_o_D G H a. Monosomy
b. Polyploid
2. A D o E F B C G H c. Chromosome 21
d. XO
3. A B C D o E F E F G H e. Insertion
f. Chromosome 13
4. A B C D o E F F E G H g. Non-reciprocal
h. Pericentric inversion
5. A B D o E F G H i. aneuploidy
j. XXY
6. Downs Syndrome k. autopolyploid
l. Chromosome 5
7. Edwards Syndrome m. tetraploid
n. Reciprocal
8. Turner Syndrome o. Deletion
p. Paracentric inversion
9. Kleinfelter Syndrome q. Trisomy
r. XXX
10. Cri du Chat s. Translocation
9. In human males a small portion of the Y chromosome is called SRY. Characterize the function of SRY as specifically as you can. A non reciprocal translocation of SRY to Chromosome 10 has occurred prior to meiosis, in a male patient accidentally exposed to high doses of radiation at a power plant. Explain briefly how the radiation might have caused the translocation. What percentage of male progeny will potentially be overly aggressive and may end up in prison if this man marries a completely normal woman and has children? Explain.
10. Sometimes during abnormal transcription events small RNA/DNA hybrids are left in the E. coli genome after transcription is over. Normally they are removed by an enzyme called RNAseH. E. coli deleted for OriC are inviable unless RNAseH is also deleted at the same time. These cells grow slowly and are “sick” but still survive. Explain why cells without OriC are inviable and why the cells containing the double deletion of OriC and RNAseH remain viable.
11. In Eukaryotic systems DNA origins of replication are only activated once per cell cycle. However viruses often need to make multiple copies of their genome in one cell cycle, for instance the SV40 animal virus generates 100,000 copies in a single cell cycle. In part, to do this it uses a protein denoted T-antigen. When T-antigen is incubated in vitro with circular viral DNA carrying the SV40 origin of replication, topoisomerase I, and ATP the following structure is observed:
What activity in addition to binding the origin of replication must T-antigen also possess in order to create the above structure? How does this differ from what occurs at the E. coli origin of replication?