Download 1 Total out of 100

Biol 321 Spring 2012 Midterm Exam [100 points]
NAME________________________________________
•
•
•
REMEMBER: carefully inspect problems and pay attention to detail!
YOU ARE NOT ALLOWED TO USE CALCULATORS OR CELL PHONES DURING
THIS EXAM or to send or receive signals of any kind
READ EACH QUESTION CAREFULLY BEFORE ANSWERING.
Problem (pts.)
1-3
(18 pts.)
4
(8 pts.)
5
(10 pts.)
6
(18 pts.)
7
( 18 pts.)
8
( 8 pts.)
9
(20 pts.)
Score
Total out of 100
1. (8 pts.) By each statement circle True/False/Not addressed in the paper. Answer false if any
part of the statement is false. No explanations required but if you want to defend your answer,
I’ll read it.
Recall the assigned reading The Interpretations of Genes.
F The common model organisms most favored by geneticists tend to share certain traits, such
as rapid development, that tend to maximize effects of the environment.
F Researchers have shown that a species that shows phenotypic plasticity with respect to one
trait will almost inevitably show plasticity in all traits.
T Lysenko was an influential biologist in the former Soviet Union who believed that the
environment determined all aspects of an organism’s phenotype.
Recall the assigned reading Beauty is in the nose of the beholder.
T This paper explores how humans respond to a compound called androstenone (and its close
chemical relatives) that is known to be a key mating pheromone in pigs
T This paper speculated on this compound’s commercial potential ………
1
Recall the assigned reading: Zebrafish researchers hook gene for human skin color.
F Zebrafish researchers and human geneticists use the same name for the gene described in this
paper.
T When injected into mutant “golden” zebrafish, a wild-type allele of the human version of this
gene “rescued” (that is, corrected) the loss-of-function (golden) phenotype.
F Allelic variation in this single gene accounts for over 99% of the skin color variation between
individuals of European and African descent.
2. (2 pts.) Consider a diploid organism where 2n=16. The probability that during meiosis all
the paternal chromosomes will segregate together to the same pole of the cell is:
a. (1/2)16
b. (1/2)8
c. 28
d. (1/4)8
e. (1/4)16
f. 0
3A. (5 pts) The Christmas poinsettia produces modified leaves called bracts that may be white,
pink or red. The color of the bract is determined by the final product of the pathway shown
below. Genes A and B assort independently
A= dominant wild-type allele a = loss-of-function
Pathway
1 wild-type allele b = loss-of-function
B=
dominant
white
A
red
pink
B
IfPathway
a plant of
2 genotype AaBb is test-crossed, what fraction will be red?
Write out the cross and show the logic of your answer.
red
white
pink
AaBb X aabb Cà ¼ Aabb D
= red
[1/2 white aa--, 1/4 pink AaBb]
Pathway 3
white
E
pink
F
red
G sheet.
inhibits
thisAstep
3B (3 pts.) Look at pg 2 of the data/info
Genes
& B are known to interact to specify a
certain phenotype. Examine the “phenotype” value assigned to each genotype and choose the
best interpretation based only on the genotypes given.
a. Loss of function mutations in Gene A are epistatic to gene B
b. A loss of function mutation in Gene A acts as an incompletely dominant modifier of the
phenotypic effect of gene B
c. Gene A and B show additive action in producing a phenotype
d. A loss of function mutations in Gene A acts as a incompletely dominant suppressor of gene B
e. I don’t think any of these descriptions are valid.
2
4. (8 pts.) This pedigree shows the inheritance of an X-linked recessive trait.
a. Examine the pedigree carefully and calculate the probability that the female (indicated by the
arrow) is heterozygous. Circle your answer and clearly show your work on the pedigree.
The recessive mutant allele that the affected male received from his mom can be
traced back to the maternal great-grandmother that he shares with his mate. There
is a [½]3 chance that this female is heterozygous.
b. Calculate the probability that the first child of the couple will be show the recessive
phenotype. Show your work, circle your answer and briefly explain calculation. No credit if
you explanation
There is a [1/2 x 1/8 = 1/16] probability that the first child will show the recessive
phenotype – which is the chance that mom passes on the mutant allele. Dad will
either contribute a Y chromosome or a mutant allele on the X. Either way – an
affected kid if mom has contributed the mutant allele.
3
5. (10 pts.) Examine this pedigree (extra copy below). Ignore the numbers under the symbols.
A slash means the individual is deceased. [Assume phenotypes for all deceased individuals were
known].
For each mode of inheritance listed below indicate:
E = this mode of inheritance is excluded by the data (see also instructions below)
C = this mode of inheritance is consistent with the data
Assume only the complications that are stipulated.
E X-linked dominant with sex-limited expression; dominant allele is rare in the general
population XDR two generation II females unrelated to main pedigree would have to be het
E if rare C if common Y-linked showing incomplete penetrance YIP if rare: unrelated male in
GII would have to have allele)
E X-linked recessive: recessive allele is very rare in the general population XRR
two generation II females unrelated to main pedigree would have to be het
C X-linked recessive: recessive allele is common in the general population XRC
E Autosomal dominant AD affected individuals have unaffected parents
C Autosomal dominant with incomplete penetrance & sex-limited expression ADIP
•
•
•
•
•
For each mode of inheritance that you excluded, circle the portion of the pedigree that
excludes this inheritance pattern and label with the appropriate acronym (indicated by letters
in bold).
You may use the same region of the pedigree to exclude two different modes of inheritance
Note, you do not need to explain why the region of the pedigree excludes the mode of
inheritance-- just circle and label it.
If more than one region of the pedigree excludes a particular mode of inheritance, just
indicate one region.
Be as precise as possible when indicating the portion of the pedigree that is relevant.
4
6. (18 pts) This is a revised version of your study problem.
You like to collect spiders from the wild and take them into your lab to observe their webmaking and prey-catching behaviors. You come across a spider that is very sluggish and a spider
that is hyperkinetic (moving all over in an undirected fashion). Read through entire problem
before working it and consult extra info on pg 1 of DATA/INFO sheet
Parental
sluggish
h+ h+ r r
X
ê
½ normal activity
F1
hyper
H h+ R + R +
h+ h+ R + r
½ hyper H h+ R+ r
NOTE: this cross is consistent with Hyper being recessive or dominant to wildtype
F1 hyper X
H h+ R + r
X
F1 normal Ò F2
h+ h+ R+ r à
1/4 sluggish
h+ h+ r r
H h+ r r
NOTE: this cross is consistent with Hyper being recessive or dominant to wildtype
F1 normal X F1 normal
Ò
3/8 hyper
H h+ R + -
F2
¾
3/8 normal
h+ h+ R + -
normal
¼ sluggish
h+ h+ R + r
X h+ h+ R+ r à
h+ h+ R + h+ h+ r r
NOTE: Since there is only one gene segregating here, this cross only works if
normal activity is recessive to hyper
Part A. (4 pts) Consult the information on pg 1 of the extra sheet. Define allele symbols
(respecting allele symbol conventions), indicate dominance and specify the trait that each gene
controls H= Hyper (excess ligand) h+ = normal ligand
R+ = normal receptor r= loss-of-receptor
rr = sluggish independent of the genotype at the ligand locus
H and R+ are dominant
Part B (2 pts.) What type of gene interaction is operating here? Use proper terminology (no
explanation required) epistasis: rr genotype is epistatic to the H gene
Part C (8 pts.) Indicate genotypes for all crosses above.
Part D ( 4 pts.) Predict the outcome of a cross between F1 hyper X F1 hyper
(phenotypic ratios and corresponding genotypes)
F1 hyper X F1 hyper Ò
H h+ R + r
X H h+ R + r
F2 9/16 hyper
3/16 wild-type
+
H- R h+ h+ R + -
5
4/16 sluggish
-- r r
7. ( 18 pts.) PART A Examine Nature News blurb on pg 2 of the Info/Data sheet. The
researchers isolated 7 different true-breeding mutant strains showing increased alcohol tolerance
compared to the wildtype strain. Examine this first series of crosses and clearly state all
conclusions below each set of observations
For mutant lines #1-5, 2pts reciprocal crosses between true-breeding wild-type (intolerant) and
mutant (tolerant) lines produced male and female progeny with the wildtype phenotype.
Mutant alleles are recessive to the wildtype alleles. Gene or genes defined by the mutant allele
is(are) located on an autosome.
Be careful with your wording: Alleles or traits show dominance. Genes are not dominant or
recessive and neither are mutant lines.
For mutant line #6 2pts reciprocal crosses between true-breeding wild-type and mutant lines
produced male and female progeny that were alcohol tolerant.
Mutant allele is dominant to the wildtype allele. Gene defined by the mutant allele is located on
an autosome.
For mutant line #7 2pts when mutant females were crossed with wildtype males, all progeny
were mutant (alcohol tolerant). When the reciprocal cross was performed, all males were
wildtype and all females mutant.
Mutant allele is dominant to the wildtype allele. Gene defined by the mutant allele is located on
the X chromosome.
PART B A second series of crosses is performed (below). This time the mutant strains are
crossed with each other, and the following results are obtained.
+ = wildtype - = mutant
Mutant 1
Mutant 2
Mutant 1
_
Mutant 2
+
Mutant 3
_
_
Mutant 3
Mutant 4
Mutant 5
Mutant 4
_
Mutant 5
+
+
+
_
_
_
+
_
+
_
What kind of experiment is this? Use correct terminology. 2pts
This is a complementation test to determine which mutations represent alleles of the same gene.
Why weren’t mutants 6 & 7 included in these crosses? 2pts
Mutant 6: Complemention tests aren’t valid for dominant mutant alleles.
The gene mutated in strain #7 is on the X chromosome so it clearly defines a gene different from
the other mutant lines. We cannot conclude that the dominant mutant allele in line #6 is in a
different gene from the recessive mutant alleles in lines #1-5. It is possible to have dominant
and recessive loss-of-function mutant alleles in the same gene.
6
What can you conclude from this second set of crosses? Be very explicit 4pts
The recessive autosomal mutations define 2 different genes:
The mutations in lines 1, 3 &4 are allelic to each other and define one gene: Gene A.
The mutations in lines 2 & 5 are allelic to each other and define a second gene Gene B.
PART C. 4pts If male and female F1 progeny from the cross of mutant 1 X 2 are mated, what
fraction of the offspring will be wild-type with respect to alcohol tolerance (assume independent
assortment). Show genotypes and indicate the logic of your calculation
A = wildtype a = alcohol tolerant
B = wildtype b = alcohol tolerant
mutant 1 aaBB X mutant 2 AAbb à F1 AaBb males X F1 AaBb females à
9/16 A-B- wildtype
8. ( 8 pts.) Rexamine Nature News blurb on pg 2 of the Info/Data sheet.
Your mom emails you and asks you to explain this news article to her. She is an astute
individual and a supporter of government-sponsored research, but is a bit mystified about the
relationship between fly and human alcohol consumption and wonders how the implied
significance of the fly research can be justified. Briefly discuss the rationale for this study. In
your email response :
a. (3 pts.) Briefly state what is meant in general by a “model” . NOTE this question is not
asking what features make a good model organism. One sentence.
need to state explicitly that a model organism is
• the object of intense study with respect to a specific biological process
• and that what we learn from studying this organism can be applied to other organisms or
serve as a basis for understanding the same process in another organism
b. (5 pts.) Provide a justification for focusing on an animal model rather than directly studying
humans. Your justification should have two parts: practical and theoretical. 3-4 well-crafted
sentences.
Need to mention
• practice issues: brood size, generation time, cost of maintaining organisms
• common ancestry of all life on earth, large fraction of genes shared with all
invertebrates, signalling pathways to handle stress and toxic stuff might be shared with
most eukaryotes (including single celled organisms
7
9. ( 20 pts.) See paragraph on pg 3 of INFO/DATA sheet
Part A Indicate True, False or N (not enough information provided to assess). Answer False if
any part of the statement if false. If there are two sentences the first statement is true and you are
to decide if the second statement is true or false.
2 pt if no explanation; 3 pts if explanation required. If you choose N, and an explanation is
required, indicate what additional information you would need.
T F N Loss-of-function mutations in the CFTR gene are not pleiotropic
T F N Loss-of-function mutations in the CFTR gene are variably expressed and incompletely
penetrant. One sentence explanation/defense of your answer
Variable expressivity is clearly indicated, but there is no information regarding penetrance.
A variety of answers were accepted here. BUT recall that penetrance addresses whether there is
any phenotypic effect of the mutant genotype NOT whether a particular effect of a pleiotropic
mutation is seen.
T F N Loss-of-function mutations in the CFTR gene represent polymorphisms in this gene.
One sentence explanation/defense of your answer
The frequency of CFTR mutations in the general population under study is not addressed
T F The information presented here suggests that the cystic fibrosis phenotype is genetically
heterogeneous. One sentence explanation/defense of your answer
The text suggests that all individuals with CF have mutations in the same gene (CFTR)
The existence of modifier genes (like SLC) does NOT mean that the disease is genetically
heterogeneous. There is no reason to think that mutations in modifier gene will cause CF –
in fact just the opposite.
T F N Heterozygotes (like your instructor) for loss-of-function mutations in the CFTR gene
show no symptoms of this disease. This means that the CFTR gene is haplosufficient.
One sentence explanation/defense of your answer
One wildtype allele is sufficient for a normal phenotype at the organismic level
T F The information presented here suggests that relatively common loss-of-function variants
in the SLC26A gene may act as partial suppressors (modifiers) of CFTR mutations
One sentence explanation/defense of your answer
The last few sentences suggests that mutations in the SLC26A gene make the phenotype of
homozygotes for CFTR less mutant – indicating partial suppression. The last sentence
suggests that these mutations may be common enough to qualify as polymorphisms.
Part B 3 pts. Short Answer: A portion of the sentence with the ******* has been deleted.
Based on our class discussion and the context of the paragraph, what do you think the deleted
portion stated? Don’t need specifics here, just general ideas using proper terminology.
need to mention: genetic background, environment and stochastic events
8