Download 1 Chapter 2 41. Chapter 6 14

BIOLOGY 321 WINTER 2011 ANSWERS TO ASSIGNMENT SET #4
10TH EDITION
Chapter 2
41.
Chapter 6
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Comment on Problem 34: avoid using upper and lower case allele
symbols in situations of incomplete dominance where neither allele is
dominant over the other. It would be better to use a (+) to designate the
+
wild-type allele -- such as f = wild-type and f= wooly:
f+f+ = wildtype f+ f = frizzled and ff = wooly.
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Answers to Additional Study Problems
Ì Problem 1
a. normal, albino, albino, albino, normal
b. complementation; the two albino individuals must have mutations in two
different genes: AAbb X aaBB most likely genotypes
Ì Problem 2
a. The experiment is essentially an in vitro complementation test. Lines H and C
complement each other. In these two strains the mutations that affect anthocyanin
biosynthesis are in different genes.
b. 9/16 of the F2 will be pigmented and 7/16 will have no pigment.
c. Lines H and W fail to complement each other. The mutations in these strains
are allelic -- that is, in the same gene.
Ì Problem 3
(i) c (ii) True, one X-linked (strain #3) and four autosomal
(iii) b Mutant 5 does not show X-linked inheritance. Mutant 3 is X-linked
Ì Problem 4A Answer is 1/2. The aa genotype is epistatic to genotypes at the B
locus. NOTE: this is a test cross, not a self cross.
Ì Problem 4B Be sure to review your lecture notes on the complicated
relationship between genotype and phenotype. Since the penetrance is different in
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different populations, differences in genetic background (suppressor or modifier
mutations) may influence the expression of the trait. For example, the gene pool of
the outbreed population (penetrance is 40%) may contain a suppressor mutation
that in combination with the glaucoma mutation results in a normal phenotype.
The gene pool of the inbred population may not contain that suppressor allele. It is
also possible that there are environmental factor that influence the expression of
the trait that are different for the two populations.
Ì Problem 5 NOTE: a cat must have a dominant allele of the agouti locus in
order to show tabby stripes. If a cat is non-agouti, it will be completely black.
a. tBtB PP AA male X tMtM pp aa female
F1 tBtM Pp Aa
F2 black kitten t-t- P- aa
aa is epistatic to the tabby locus
b. 9/64 c. 1/4
Ì Problem 6 A:
Gene B: presence or absence of scales B+ = scales form (dominant) b = no scales
Gene A: controls arrangement of scales
a+ = wildtype (scattered) A = linear arrangement (dominant)
B: The bb genotype is epistatic to genotypes at the A gene.
Parental
F1
F1 Cross A
F2
Aa+ bb
naked
a + a+ B + B +
X
wild-type scales
ê
(truebreeding)
1/2 linear scales A a+ B+b
1/2 wild-type scales a+a+ B+b
F1 linear X F1 linear
A a+ B + b X A a+ B + b
ê
9/16 linear A - B+ 3/16 wild-type a+a+ B+4/16 naked -- bb (any genotype at gene A)
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F1 Cross B
F2
F1 Cross C
F2
F1 linear X
F1 wild-type
+ +
Aa B b
a+ a+ B + b
ê
3/8 linear A a+ B+ - (1/2)(3/4)
3/8 wild-type a+a+ B+ - (1/2)(3/4)
1/4 naked Aa+ bb or a+a+ bb 2[(1/2)(1/4)]
F1 wild-type X F1 wild-type
a+a+ B+b X a+a+ B+b (only one gene segregating)
ê
3/4 wild-type a+a+ B+1/4 naked
a+a+ bb
Ì Problem 7 This is an example of additive gene action (red + green = brown)
N= chlorophyll is lost from mature pepper (gene might code for an enzyme that is
directly or indirectly involved in its degradation?)
n= chlorophyll not lost (loss-of-function in enzyme required for degradation?)
R= red pigment produced (enzyme converts yellow pigment to a red pigment?)
r = yellow pigment produced (loss-of-function mutation in enzyme required to
convert yellow pigment to red pigment)
Cross #1 Red NN RR X brown nn RR (green chlorophyll + red pigment)
F1 = NnRR
F2 3/4 red (N-RR) and 1/4 brown (nnRR)
F2 ratios in quarters since parental genotypes differ by only one gene
Cross#2 Parental strains must differ in two genes since F2 ratios in 16ths
Parental: Yellow NNrr X nnRR brown
F1 Red NnRr F2 N-R- red N-rr yellow nnR- brown nnrr yellow-green
Ì Problem 8 d. The D gene modifies but does not mask (and therefore is not
epistatic to) the M gene
Ì Problem 9
a. Since the strains were known to carry mutations in different strains, they should
show complementation.
b. The friend overlooked the fact that, by definition, a loss-of-function mutation in
a haploinsufficient gene will not be recessive to the wild-type allele. Therefore a
complementation test will be invalid
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Ì Problem 10
GENOTYPES
Heavy
(true-breeding)
breeding)
Hypothesis #1
One gene 2 alleles called a1 and a2
Incomplete dominance
Hypothesis #2
One gene, 3 alleles
a1 = heavy
a2 = medium a3 = no-spots
> = Complete dominance
a1 > a 2 > a 3
Hypothesis #3
Gene A involved in anthocyanin
biosynthesis; a is a recessive allele that
produces a non-functional enzyme
Gene S determines the extent of
pigmentation in leaf: S- heavy ss
medium
Medium
No-spots
(true-
a1a1
a1a2
a2a2
a1a1
a2a2
a2a3
a3a3
ssaa
Ssaa*
SSaa
SSAA
ssAA
ssAa
*operationally true
breeding since
genotype hidden
(ii) Cross heavy and no-spot true-breeding strains and self F1
If hypothesis #1 is correct:
F1 will be all medium and F2 1/4 heavy, 1/2 medium and 1/4 no-spot
If hypothesis #2 is correct:
F1 will be all heavy and F2 3/4 heavy and 1/4 no-spot
(iii) a. not possible to generate a true-breeding medium strain if hypothesis #1 is
correct
b. If hypothesis #2 is correct, you can’t generate medium plants from your
original true-breeding heavy and no-spot strains. Go back to the field and collect
medium spotted plants:
(iv) SSAA X ssaa à F1 SsAa heavy
à F2 9/16 heavy S-A- 3/16 medium ssA- 4/16 no-spot ssaa, S-aa
(v) Yes, this is consistent with your hypothesis. Your second no-spot strain was
SSaa in genotype:
SSAA x SSaa à F1 SSAa heavy à F2 3/4 SSA- heavy 1/4 SSaa no-spot
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(vi) Accidentally omitted from original problem: Assuming hypothesis #1 is
correct, explain the variation in spotting:
Variable expressivity due to one or more of the following
• variation in genetic background: a modifier gene segregating in the population
• environmental effects on expression of the trait
inherent randomness in the formation and positioning of the spots during leaf
development
Ì Problem 11 (i) answer is d
(ii) Mutant #1 females aaBB X Mutant #3 males Ab
â
F1 females AaBb
â
haploid male progeny:
1/4 AB purple
1/4 Ab black
1/4 aB black
1/4 ab black
(ii) Genes A and B are both required for biosynthesis of scarlet pigment.
a and b are recessive, loss-of-function mutations which affect scarlet biosynthesis.
The synthesis of the black pigment is unaffected by these mutations.
Ì
Problem
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Based on this information, next to each term indicate:
A if the term applies to the m allele and associated phenotypes
C if the information given outright contradicts the term
I if there is insufficient information to determine whether the
term applies
_C___complete dominance
_A__ haploinsufficient
_C___mm complements R+R__I__variable expressivity
_A___mm supresses R+R__A__ polymorphic
__C*__ mm is epistatic to R+R- (*wrong term here)
What genotype would have 100% therapeutic effect? R+R+mm
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Ì Problem 13
The following terms apply:
_C_ pleiotropic
_I_ multifactorial
_C_ sex-linked
_*_haploinsufficient
_A_ autosomal
_C/I_ polymorphic (not A)
_I_variably expressed
_A_ genetic heterogeneity
_A_ incomplete penetrance probably due to a dominant suppressor allele
_C_ incomplete penetrance probably due to a recessive suppressor allele
_I_ incomplete penetrance due to environmental variation
_C_ incomplete penetrance due to complementation
*Answer is either I or A for modifier and C for DFNB-26
Ì Problem 14
a. complementation test
b. This test is used to determine whether mutants with the same (recessive)
phenotype have defects in the same gene or different genes
c. The mutant Strains C and B fail to complement and have mutations in the same
gene. Strain C complements both D and O, which also complement each other. In
summary, these 4 strains carry mutations in three different genes. Mutations in any
one of these genes can cause a Rex phenotype.
d. The simplest explanation for the W data is that this Rex strain is due to a
dominant mutant allele. In this case, the complementation test gives us no
information about allelism with the recessive mutations in the Rex genes defined in
part c. (NOTE: A formal, but very, very, very unlikely possibility is that Rex W is
homozygous for recessive mutations in all three of the Rex genes. This is not a
good answer to the question.)
Ì First Problem 15 From text of problem:
colorless compound à red à
purple
gene A
gene B
Since F1 is purple, parents must be AAbb (red) X aaBB (white)
F2 = 9/16 A-B- 3/16 A-bb 3/16 aaB- 1/16 aabb
The aa genotype is epistatic to allele of the B gene
Answer is d
Ì Second Problem 15 (!) ANSWER is d
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Ì Third Problem 15 :~)
Define allele symbols in this box. Indicate dominance and which trait(s) the
gene(s) controls. BELOW: Indicate all genotypes in each box.
W = prevents formation of pigment (dominantly epistatic to pigment color gene)
w= allows pigment formation
M= dark magenta m = light magenta
Parental: dark magenta (truebreeding) X white
ww MM
X Ww mm
à F1 1/2 dark magenta
1/2 white
ww Mm
Ww Mm
Self F1 dark magenta plants: ww Mm
3/4 dark magenta ww M1/4 light magenta ww mm
Self F1 white plants: Ww Mm self
12/16 white W- - - = W- M- & W- mm
3/16 dark magenta ww M1/16 light magenta ww mm
Predict outcome (phenotypes & ratios) of the following cross:
F1 dark magenta plants X F1 white plants
½ white 3/8 dark magenta 1/8 light magenta
ww Mm
X
Ww Mm
WwM- = white = ½ X ¾ =3/8 white
Ww mm = white = ½ x ¼ = 1/8 white
wwM- = ½ X ¾ = 3/8 dark magenta
ww mm = ½ X ¼ = 1/8 light magenta
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Ì Problem 16
Part A: Propose a concise, title (one phrase or sentence) for this article. The title should
summarize the gene interaction described in the text. Be sure to be explicit as possible and use
appropriate terminology.
STUDENT ANSWER:
SUPRESSION OF THE ALS DISEASE STATE IN NOX1& NOX2 KNOCKOUT MICE
Part B Indicate True, False or N (not enough information provided to assess). 3 pt each. If
you choose N, indicate what additional information you would need.
TRUE Assuming all cases of ALS are genetically determined, the article implies that this
disease state is genetically heterogeneous (or shows genetic heterogeneity).
One sentence explanation/defense of your answer:
The article states that mutations in the SOD gene account for 1-2% of ALS cases indicating that
mutations in other genes are responsible for the majority of cases
TRUE The studies described in the article suggest that there are other genes (in addition to Nox1 and Nox-2) that can influence expression of the ALS mutation.
One sentence explanation/defense of your answer:
See last paragraph. The effect of the Nox-2 knockout on ALS differed in different strains of
mice –probably due to differences in the genetic background.
Student answer: Nox1 and Nox2 are described as two of “several” proteins that generate
reactive oxygen species
TRUE Although not stated explicitly in the article, the results of the studies suggest that
polymorphisms affecting the activity of the Nox-2 gene could result in variable expressivity or
incomplete penetrance of SOD1 mutations in humans.
One sentence explanation/defense of your answer:
Assuming that the observations made using the mouse model extend to humans, then
poymorphisms in the human Nox-2 gene that influence the level of activity of the NOx-2 protein
are likely to affect the phenotypic outcome of a SOD1 mutation by paritally suppressing its
pathogenic effect.
Student answer: If polymorphisms in this gene do exist, then the study suggests that lower levels
of Nox2 will lessen the severity of the phenotype produced by SOD1
Ì Problem 17
Part A Indicate True, False or N (not enough information provided to assess). Answer False if
any part of the statement if 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.
FALSE The Tph2 R441H mutant allele represents is a polymorphism.
This allelic variant is very rare
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TRUE The Tph2 gene is haplosufficient in the test of social interaction (Figure 2F)
One sentence defense/explanation: A Tph2 het shows the same phenotype as a wildtype
homozygote indicating that one copy of the wildtype allele is sufficient for a wildtype phenotype
in this phenotypic assay. Note the slight elevation in aggressive attacks seen in the hets is not
statistically significant.
FALSE A mutation in the gene coding for L-aromatic amino acid decarboxylase is likely to be
epistatic to the Tph2 gene.
One sentence defense/explanation: In a biochemical pathway, mutations in a gene catalyzing an
upstream reaction will be epistatic to downstream genes but not necessarily vice versa. With a
loss-of-function genotype in the LAACD gene, the status of the Tph2 gene can still be assessed - for example by measuring production of 5-HT (its product). On the other hand, mutations in
the Tph2 gene are epistatic to the LAADC gene, since even if you have a functional LAADC
gene a Tph2 mutation will block production of its substrate.
èNonetheless, TRUE was also accepted as a correct answer if the explanation focused on the
inability to generate serotonin in the absence of either enzyme.
FALSE In Figure 2, the wildtype alleles of the Tph2 and GSK3 genes show a clear additive
effect on phenotype.
TRUE The loss-of-function mutation in the GSK3 gene acts as a suppressor of the Tph2
R441H phenotype even though the GSK3 heterozygote has no phenotype in the wildtype
background (in the data shown).
One sentence defense/explanation: In Panel 2A & 2B, mice that are either het or homozygous
for the mutant Tph2 allele show a wildtype phenotype when they are also het for the GSK3
mutation.
Part B Short Answer
(i) (3 pts.) The authors of this paper did not put a label on the dominance relationship between
the Tph2 wildtype and R441H mutant alleles. Briefly explain why (1-2 sentences)
The dominance relationship between the wildtype and mutant allele is different in each set of
data: Figure 1A: incomplete dominance Figure 2A mutant allele looks completely dominant
Figure 2B mutant allele looks recessive. Go figure. Student answer: Because the effec ot Tph2
is dom and recessive in different assays. In the immobility test hets are more immobile than WT,
so it would be Dom. In social, hets are the same as WT, (so it would) be recessive.
(ii) ( 3 pts.) The authors of this paper conclude that targeting GSK3 may afford therapeutic
advantages for the management of certain 5-HT (serotonin) related psychiatric conditions.
What observations support this conclusion? Would the “management” involve increasing or
decreasing the level of GSK3 function? 2 sentences
The suppressive effect of a loss-of-function mutation in the GSK3 gene in Tph2 mutant mice is
consistent with the observation that GSK3 activity is elevated in Tph2 mutant animals that are
homozygous for a wildtype GSK3 allele.
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This suggests that lowering GSK3 activity (with a pharmaceutical agent that specifically targets
this enzyme) might relieve symptoms (whatever form they may take) in individuals with Tph2
loss-of-function alleles.
NOTE: there is no evidence that lowering GSK3 activity will cause an increase in serotonin
levels but rather the converse --- lowering serotonin levels results in incrased GSK3 levels
(iii) (6 pts.) Your mom reads about this particular research project on the internet one day and
emails you expressing her indignation at the money and time spent on a mouse project given that
the issue at hand (as proposed in the first line of the paper’s abtract) is the understanding of a
quintessential human trait. Based on your understanding of model organisms, in a 3-5 sentence
email response pursuade your mom that the time and money are likely to have been well spent.
Need to talk about the practical and theoretical justification for using models and indicate that we
share with mice most of our genes and our nervous systems (including signal transduction and
neurotransmitter biochemistry) have much in common with other mammals.
Requests to mom for money and/or care packages OK but not required.
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