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Unformatted text preview: BICD 100, Fall 2008
Examination  2
Dr. J. B. Ghiara The exam consists of 6 pages including this cover page.
Rules: Please refer to your syllabus for details on Exam and Grading policies.
1. The Exam will end promptly at the end of the assigned time.
2. Your pens, #2 pencils and an ID card (student or driver's license) are the only personal
items you may have with you during the exam.
3. Please only use pencils to fill in the circles on the scantron forms. If you use ink, the
scantron machine will mark your answers wrong and your score will be zero. If you change
an answer, erase it very clearly. If the machine detects two answers, it will mark it wrong.
4. Cheating will not be tolerated and will result in a failing grade for this course. When you are done with your exam, bring this exam and your scantron form
to one of the TAs. You will need to show your ID to the TA and get checked
off the roll sheet before you can leave. This is proof that you took the exam.
© J. B. Ghiara, November 2008.
All 6 pages of this exam are copyrighted material and the property of Dr. J.B. Ghiara.
All copying and/or posting of this material is strictly prohibited.
Please fill in your names and Student ID numbers on this page and on page 6.
LAST NAME:__________________________________
FIRST NAME:__________________________________
STUDENT I.D. #:_______________________________
(5 point deduction if the names of students on both sides are not filled in)
Full Name of student sitting to my left is: __________________________
Full Name of student sitting to my right is: __________________________ Now please fill in your name and student ID number on the back of the scantron
form. You need to fill in the corresponding circles otherwise you will not get a score
for this exam.
Original scantron forms will not be returned to you. A copy of pages 5 and 6 will be
handed back in class next week, on Tuesday. Now that we are done with that, GOOD LUCK! STUDENT INITIALS:____________________ Section A: Please answer questions 1 through 11 on the scantron form. Mark only
one answer (the best answer). Numbers in parentheses at the end of the questions
represent the maximum points awarded for those questions.
(13 points)
Answer questions 1 and 2 based on the information given below:
Two gene loci, P and Q, are unlinked (and thus assort independently). Allele P is dominant over allele p but
alleles Q and q show incomplete dominance. Indicate the probabilities of producing the following:
1. A Pq gamete from an PpQq individual?
A) 1/16
B) 1/8
C) 1/4
D) 1/2
E) 3/4 2. How many different phenotypes could occur in children of parents, both of whom have the PPQq
genotype?
(1)
A) 2
B) 3
C) 4
D) 6
E) 9 (1) Answer questions 3, 4, and 5 based on the information given below:
In Mandells, the allele for huge ears is incompletely dominant over the allele for tiny ears, and thus
heterozygous Mandells have normal sized ears. What ratios of ear sizes would you expect to see among
progeny generated from the following crosses of Mandells?
3. Huge ears × Normal ears
A) 100% normal ears
B) 50% huge ears; 50% normal ears
C) 100% tiny ears
D) 25% huge ears; 50% normal ears; 25% tiny ears
E) 100% huge ears (1) 4. Normal ears × Normal ears
A) 100% normal ears
B) 50% normal ears; 50% tiny ears
C) 100% tiny ears
D) 25% huge ears; 50% normal ears; 25% tiny ears
E) 100% huge ears (1) 5. Tiny ears × Normal ears
A) 100% normal ears
B) 50% normal ears; 50% tiny ears
C) 100% tiny ears
D) 25% huge ears; 50% normal ears; 25% tiny ears
E) 100% huge ears (1) Answer questions 6, 7, and 8 based on the information given below:
Fruit color in eggplants is inherited as an incompletely dominant characteristic: Homozygous dominant
plants produce purple fruit, heterozygotes produce violet fruit and homozygous recessive plants produce
white fruit. You cross two eggplants with violet fruit. Among the offspring, 20 bear purple fruit, 47 bear
violet fruit and 33 bear white fruit. You want to perform a chisquare test, to compare the observed 2 STUDENT INITIALS:____________________
numbers with the ratio you expect for incomplete dominance and determine if the difference between
observed and expected values could be a result of chance.
6. What are the expected values you would need to use in the formula to calculate your χ2 statistic? (1)
A) +5, 3 and +8 for purple, violet and white fruit, respectively
B) 53 and 47 for heterozygous and homozygous fruit, respectively
C) 20%, 47% and 33% for purple, violet and white fruit, respectively
D) 20, 47 and 33 for purple, violet and white fruit, respectively
E) 25, 50 and 25 for purple, violet and white fruit, respectively 7. For the above experiment, your calculated χ2 value should be:
A) 187/50
B) 0.98
C) 0.16
D) 16
E) (0.10)2 8. Your calculated χ2 value turns out to be less than 5.991, the 5% critical value for the correct number of
degrees of freedom for this experiment. You could say
(1)
A) the difference between the observed and expected numbers of offspring is due to something
other than chance alone
B) the difference between the observed and expected numbers of offspring could be due to chance
alone (2) Answer questions 9, 10 and 11 based on the information provided in each question:
9. The genotype of a woman who has AB negative (AB ) blood is: (HINT: Assume the alleles d and h
have the same meaning as we used in our lecture on blood types)
(1)
A) ABhh
B) AaBbdd
C) ABdd
D) Both A and C are possible
E) Both B and C are possible 10. A woman with A negative (A) blood is pregnant and the fetus’s father has A positive (A+) blood.
Which of the following is true?
(1)
A) The father is Rh+ so the fetus has to be Rh+
B) Unless the pregnant mother is given Rhogam, she will develop antiRh antibodies (usually after
childbirth but this could happen earlier if there is bleeding at the placenta)
C) The fetus could be Rh – and if so, there would be no risks due to the Rhesus factor associated
with this pregnancy.
D) Both B and C are possible.
E) The father only contributes sperm. The blood is made in the fetus so it will have A blood and
all will be well.
11. Ptosis (droopy eyelid) may be inherited as a dominant human trait. Among a group of people who are
heterozygous for the ptosis allele, 10 have ptosis and 30 have normal eyelids. What is the penetrance
for ptosis?
(2)
A) 100%
B) 75%
C) 50%
D) 33.3%
E) 25% 3 STUDENT INITIALS:____________________
Section B: Please answer questions 12 through 21. Mark the correct answer for
each of these True/False questions on your scantron forms.
Bubble in A for true, B for false.
(11)
12) In Mendel’s dihybrid cross, if the probability of getting wrinkled green seeds was 1/16 and the
probability of getting wrinkled yellow seeds was 3/16, we can say that the probability of getting
wrinkled seeds regardless of color would be 1/4. T
13) In diploid individuals, the phenotype encoded by a recessive, hemizygous allele will be observed. T
14) A reciprocal cross can be used to determine whether an individual is homozygous or heterozygous for
autosomal alleles. F
15) While the XO genotype results in a woman with Turner’s syndrome, in Drosophila, the XO genotype
gives rise to a male fly. T
16) A person with the genotype XYY will have no Barr bodies in their cells. T
17) The coefficient of the second term in the expansion of the binomial (a+b)9 will be 9. T
18) Mullerian ducts die out in male embryos because the testes produce MIS. T
19) Malelimited precocious puberty is an example of a sexlimited characteristic. T
20) Human X and Y chromosomes do not have any homologous regions. F
21) Women (XX) that are heterozygous for an Xlinked gene have patches of cells that express one allele
and patches of cells that express the other. T
22) In mice, the allele Y is dominant for coat color (Yellow) but recessive for lethality so heterozygous
parents will produce viable (living) offspring in the ratio 2:1 (2 yellow:1 nonyellow) [Assume a
very large number of progeny in a series of crosses] T 4 STUDENT INITIALS:____________________
Section C: Please answer the following short answer question. (6) 23) Mating between a whiteeyed fruit fly and a heterozygous, redeyed (wild type) female results in 101
offspring. What is the probability that the F1 generation will have 100 redeyed males? HINT: Do not
use the binomial expansion to solve this problem. Use the formula with the factorials! When answering
the following parts of question 23, show your work explicitly writing out any Punnett squares/
formulas/ calculations.
a. Write the genotypes of the parents (the whiteeyed and redeyed flies that mate, as stated
above. Assume the male has an X and a Y chromosome and the female has two X
chromosomes, as normally expected) (1) XXw
b. Draw a Punnet Square, clearly showing the possible genotypes of the progeny. (2) Xw
X XXw XY Xw c. Y XwXw XwY Write the formula (the one with the factorials) that you would use to calculate the
probability of the F1 generation having 100 redeyed males (remember the total number of
offspring is 101), with all the correct numbers included (i.e. replace n, a, b, s and t with the
numbers you would need to actually calculate the probability). [As long as all the correct
numbers are in place, you do not need to calculate the final answer.] (3) a=redeyed male = ¼
b = not redeyed male = 3/4 n! asbt
s!t! = 101! x (1/4)100(3/4)1 100!1! 5 STUDENT INITIALS:____________________ Sections
A and B
C
Total Points
24
6
30 Your Score Grader
Scantron machine LAST NAME:______________________ FIRST NAME:____________________
STUDENT I.D. #:_______________________________
Original scantron forms will not be returned to you. A copy of Pages 5 and 6 will be
handed back in class next Tuesday. If you would like to pick up pages 5 and 6 of the
exam some time after it has been distributed in class, please sign below to give
permission for it to be left for you on the shelves outside my office (3080A York Hall).
_________________ Student Signature NAME OF MY TA:____________________ 6 ...
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 Fall '08
 Nehring
 Genetics

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