BIO 111, Fall 2001
Review 2 - Genetics ANSWERS
Provide thorough, but concise, answers to 7 of the following 8 questions.
All questions are worth 10 points.
There is no time limit to this test.
The review is closed book.
Calculators are permitted.
All answers must be typed.
Please staple all pages together and do not put your name on any page
other than this cover page.
Return completed review to Mrs. Hartsell (Watson 162) no later than 4:00PM,
Friday October 26, 2001.
Name (please print):
Pledge and signature:
1. Briefly define the following terms:
Locus - Physical location of a gene or genetic marker in the genome
Chromosome - Single piece of double-stranded DNA
Allele - Specific variant form or a gene
Gene - Specific locus of DNA that is transcribed
Draw a picture to fully demonstrate you understand these terms.
2. What is a restriction endonuclease (AKA restriction enzyme)? Explain how these enzymes relate to RFPLs.
Enzyme that cleaves DNA at a specific sequence. Can be used to identify sequence differences in DNA. RE may cleave one variant of DNA, but not another
3. Describe the two major mechanisms by which of meiosis leads to genetic variation?
Independent assortment of chromosomes
Crossing over
4. For the following cross, AaBb x Aabb, show a Punnet square and provide the expected genotypic and phenotypic ratios of the progeny. Assume the genes are located on different chromosomes.
| AB | Ab | aB | ab | |
| Ab | AABb | AAbb | AbBb | Aabb |
| ab | AaBb | Aabb | aaBb | aabb |
Phenotypes: AB:Ab:aB:ab::3:3:1:1
5. For the following cross, AaBb x aabb, the following numbers of offspring of the various phenotypes are observed:
| Phenotypic Class | Number |
| AB | 88 |
| aB | 415 |
| ab | 92 |
| Ab | 405 |
Explain these results.
Parental Types: Ab and aB
Recombinant Types: AB and ab
% Crossing Over = 180/1000 = 18%
Genes are linked, 18 MU apart
In first parent, allelic arrangement is Ab//aB
6. Some members of the Davidson College community are found to be wildcatase deficient. A sequence analysis of the wildcatase gene in numerous individuals reveals that many different mutations in this gene result in wildcatase deficiency. Describe two different types of mutations that could result in this disease and the effect on transcription, translation, and/or enzymatic activity that you would expect each mutation to have.
Some examples:
Nonsense mutation - single nucleotide change in mRNA, translation stops too soon, truncated protein, probably no enzyme activity
Missense mutation - single nucleotide change in mRNA, possible amino acid change in protein, possible effect on enzyme activity
Insertion or deletion - mRNA contains insertion or deletion, reading frame error in translation, probably no enzyme activity
Promoter mutation - up or down regulation of transcription, too much or too little protein
7. Pretend you used PCR to amplify the D1S80 locus of your DNA, your biological mother's DNA, and your biological father's DNA. The resulting PCR products then were analyzed by gel electrophoresis. Draw a picture representing the resulting gel. Briefly explain the important parts of your picture.
Each allele must be inherited from one of the parents
8. A) To test the effects of cycloheximide, an inhibitor of translation, on chlamy flagellar regeneration, you must make 100mL of a 150mM cycloheximide solution. How would you make this solution?
(281.4)(0.1L)(0.15M) = 4.221g cycloheximide dissolved in a final volume of 100 mL
B) Using this 150mM cycloheximide stock solution, how would you make 1mL of 5mM cycloheximide?
V1C1 = V2C2
(XmL)(150mM) = (1mL)(5mM)
X = 0.033mL or 33uL of 150mM needed
NOTE: The formula weight of cycloheximide is 281.4.