Fall 1999 Biology 111 Exam #2 - Genetics, through linkage
There is no time limit on this test, though I have tried to design one that you should be able to complete within 3 hours, except for typing. There are four pages on this test. You are not allowed to use your notes, old tests, any electronic sources, any books, nor are you allowed to discuss the test with anyone until all exams are turned in at 11:30 am on Monday October 18. EXAMS ARE DUE AT CLASS TIME ON MONDAY OCTOBER 18. You may use a calculator and/or ruler. The answers to the questions must be typed on separate sheets of paper unless the question specifically says to write the answer in the space provided. If you do not write your answers on the appropriate pages, I may not find them unless you have indicated where the answers are.
Please do not write or type your name on any page other than this cover page. Staple all your pages (INCLUDING THE TEST PAGES) together when finished with the exam.
Name (please print):
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How long did this exam take you to complete (excluding typing)?
1) What Chlamydomonas structure or structures can you see better with fixed cells than you can with live cells? Give two reasons why.
It is easier to see flagella because they glycoproteins are stained dark with the Lugol's fixative and they are not moving. It is also easier to see the pyrenoid body since it is so rich in glucose as a starch storage oranelle.
2) If we are going to do flagella regeneration again next lab, why do we have to do a control at all? Why can't we just do the experimental cells this week and use last week's data as the control?
Since there may be some differences in the condition of the cells (healthy cells last week but sick cells this week) or the room (e.g. temperature), we need to treat the control cells exactly like the experimental cells with only one variable changed. By repeating the standard regrowth, this will insure identical treatment.
3) What is the mating efficiency in this field of view diagramed below? The formula for this calculation is:
% mating efficiency = 2 (no. of zygotes) ÷ [2 (no. of zygotes) + (no. of gametes)]
7 zygotes and 4 gametes: 14/14 + 4 = 77.8%
|Picture 4A||Picture 4B||Picture 4C|
4) Above are 3 pictures taken from a QT movie of a single cell during one cell cycle. If I tell you that when this cell was in G1 it had 24 chromosomes, answer these questions:
a) what phase is shown in Picture 4A?
b) what is probably happening during this phase?
.recombination or crossing over
c) what phase is shown in Picture 4B?
d) what is happening to sister chromatids during this phase?
Sister chromatids are not separated duirng meiosis I so they are traveling together to one pole or the other.
e) what phase is shown in Picture 4C?
f) what is happening to sister chromatids during this phase?
Sister chromatids were separated during anaphase II and now travel to opposite poles.
5) A man and a woman produced 4 children. The first child is a girl, the next three are boys. The odd numbered children have a rare genetic phenotype of ears-pointing-backwards (very advantageous in cold climates with a lot of wind). Child number one marries a man who also has this phenotype and they have two wild-type boys.
a) Draw this family pedigree on your answer sheet.
b) Is the disease a dominant, recessive or codominant one?
Explain your answer to receive full credit.
.Dominant, because the parents of the F1 generation both have the disease but their children do not. This cannot happen in a recessive disease. There are no partial phenotypes so it is not codominant.
c) What were they odds that any couple would have one girl
and three boys? Show your work to receive partial credit on wrong
GBBB = 1/16 OR
BGBB = 1/16 OR
BBGB = 1/16 OR
BBBG = 1/16
Therefore in any order there is a (1/16 + 1/16 + 1/16 + 1/16) = 4/16 = 1/4 chance of having 1 girl and 3 boys.
d) What were the odds that the F1 couple would produce two
boys that were wild-type? Show your work to receive partial credit
on wrong answers.
If D = disease and
d = wt alleles
F1 parents genotypes are Dd X Dd: (to produce a child that is dd is 1/4 chance and 1/2 that the child will be male) X (1/4 X 1/2) = 1/64
6) Using the pedigree you have drawn for question 5a, provide the genotypes of all six children in a way that is consistent with your answer to question 5b.
See Diagram for genotypes
7) Define the terms: locus; allele; and gene. Draw a picture of two chromosomes to demonstrate you understand these terms.
A locus is a position on a chormosome where a gene is located. The gene is the section of DNA that encodes for a particular triat, such as hair color. Alleles are variations of genes that both encode for hair color but might encode different colors, such as blue and black.
8) Here is a piece of coding DNA:
3' TATACCCCGGGAAACTGCTACAACACATTCCCGGG 5'
a) Write down the RNA strand that would be transcribed from this piece of DNA. Do not worry about post-transcriptional processing at all. Be sure to indicate the 5' and 3' ends on your RNA.
5' AU AUG GGG CCC UUU GAC GAU GUU GUG UAA GGG CCC 3'
b) Use the code provided on the last page of the test to translate
this RNA. Try to be as smart as a ribosome and select the correct
reading frame and do not start translating randomly. Indicate
which amino acids are on the amino and carboxyl ends.
9) a) What role do transcription factors play in the production of proteins?
Transcription factors determine which genes can be transcribed or not. Therefore, they turn on and off genes. Without transcription factors for a particular gene, none of that encoded protein can be made.
b) What type of genetic disease would you develop if a rogue
transcription factor were produced and could not be inactivated?
.This sounds like the molecular cause for a dominant disease - making too much of a functional protein.
c) For the woman with the XY karyotype and the SRY nonsense
mutation, who did she inherit this from and will she pass it on
to her children? Explain your answer.
She must have inherited it from her father since it was on the Y chromosome.
She will not pass it on since she is infertile.
10) Explain the role of new mutations in evolution. Give at least one specific example to support your answer.
New mutations provide the genetic diversity that is the raw material for evolution. If the selective pressures change, then the individuals with advantageous traits will survive and reproduce better, thus passing on their "new mutation". Without new mutations, all individuals would be homozygous for the original alleles and there would be no advantage to any individual and changes in the environment could kill the species. For example, some Chlamydomonas gametes were able to mate very quickly while others took a long time. If this mating were taking place in a small puddle on a hot sidewalk, those that took too long would die due to dehydration while the faster maters would survive as zygotes. Thus, the new mutations that allow fast maters to get done quickly provided the diversity to select for the "fast" allele.
11) a) Explain how a mutation in a gene's signal sequence could produce a genetic disease.
A mutant signal sequence would result in the protein not being translated on the rER. Therefore, the protein would wind up in the wrong location and it would not be able to perform its function properly. This type of mutation would result in a recessive mutation - nonfunctional protein produced.
b) List two post-translational modifications and tell me if
each one could lead to a specific genetic disease or not.
.Glycosylation - this is the sugar coating put on glycoproteins. THis could result in a disease if the glycosylation defect were subtle but would probably result in a spontaneous abortion if it were a drastic mutation.
12) Freebie. Skip this one.
13) What was the conclusion drawn from the experiments performed by Sato and Sato using cAMP and measuring the Cl- concentrations? Do not tell me methods or results.
.The conclusion was that the Cl- pump or channel was missing from cf /cf cells.
14) a) Name three species other than humans whose genomes are being sequenced as a part of the Human Genome Project.
mouse, fly, worm, Arabidopsis (plant), yeast - E. coli is OK
b) Why is this a good thing to do?
This is good because many of these species have very similar genes to us and to each other so when the sequences come in, we can compare them. Then we can do experiments on these other species and deduce the consequences for humans without having to conduct human experiments.
15) There are three traits being examined to see if any are linked. (The number of children has broken a new world's record.) This question and your answer must be written using proper Morgan nomenclature.
a) Write the genotypes for the man and woman described below.
.man = cf cf ; l l ; a a
woman = cf cf+ ; l l +; a a +
b) Determine which traits (if any) are linked.
short legs and cf are linked loci, albino is not linked
c) Determine the distance between linked traits if there are
- 5+ 5 + 5 + 5 = 20 = number of recombinant progeny
total number of children = 200
recombination frequency = 20/200 = 10% = 10 map units
A true breeding man with cystic fibrosis, short legs, and albino mated with a woman who was heterozygous for all three traits. They had 200 children!
See Copy of Test with Answers
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