Download Genetics 314 – Spring 2004

Name: _____________________________________
Genetics 314 – Spring 2004
Exam 4 – 100 points
For math related questions please show work as much as possible.
1. Mendel determined how genes were inherited and proposed two laws to describe
the inheritance of genes.
a) What are Mendel’s two laws and how do they relate to the inheritance of
alleles/genes?
Law of segregation stated that the pair of alleles for a given gene (trait)
separate or segregate in the gametes equally.
Law of independent assortment states that the allelic gene pairs for two traits
will behave independently of each other during gamete formation (unless the
genes for the two traits are on the same chromosome).
b) What in meiosis explains the behavior of alleles/genes as observed by Mendel?
Law of segregation - Separation of homologous chromosome pairs at
Anaphase I in meiosis
Law of independent assortment – Relates to the orientation of the nonhomologous chromosomes on the metaphase plate in
Metaphase I of meiosis.
2. Once Mendel’s laws were rediscovered and accepted it became possible to predict
the probability of recovering specific genotypes from different matings/crosses.
If you cross the following genotypes:
AaBbCcDdEeff x
AaBbccDdEeFF
how may progeny out of a total of 4,000 would you expect with the following
genotypes?
a) aabbccddeeFf
.25 x .25 x .5 x .25 x .25 x 1 = .002 x 4000 = 7.8 or 8
b) AaBbCcDdEeFf
.5 x .5 x .5 x .5 x .5 x 1 = .0313 x 4000 = 125
c) AabbCcddEeff
.5 x .25 x .5 x .25 x .5 x 0 = 0
d) AABBCcDDEeFf .25 x .25 x .5 x .25 x .5 x 1 = .0039 x 4000 = 15.6 or 16
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3. Mendel’s laws can also be used to predict the probability of a specific
combination of events occurring such as the number of boys and girls in a family.
a) What would be the probabilities of the following families occurring
1) 5 boys – 2 girls (7!/(5! 2!)) x .55 x .52 = 21 x .0078 = .164 or 16.4%
2) 3 boys – 4 girls (7!/(3! 4!)) x .53 x .54 = 35 x .0078 = .273 or 27.3%
3) 1 boy – 6 girls
(7!/(1! 6!)) x .51 x .56 = 7 x .0078 = .054 or 5.4%
4) 0 boys – 7 girls (7!/(0! 7!)) x .50 x .57 = 1 x .0078 = .0078 or 0.8%
b) Would the probabilities change for the various combinations if they had to
occur in a specific order? If yes, how would they change?
Yes, they would decrease to the level of (4) or 0.8%. The reason for this is
that if a specific order is required then there are no other possible
combinations making (N!/(W! X!)) = 1 and the probability would be just
0.57 or 0.0078.
4. You are studying a trait in fruit flies and discover that there are multiple alleles
that affect expression of the trait. When you identify them all you find you have
10 alleles that can affect expression.
a) How many different genotypes are possible for your trait?
Number of genotypes = (n(n + 1))/2
Number of genotypes = (10(10 + 1))/2 = 110/2 = 55
b) What type of alleles would you need to have the same number of phenotypes
as genotypes? Briefly explain your answer.
For the number of phenotypes to equal the number of genotypes all the
alleles must be codominant so both alleles will be expressed in the
heterozygote. If any alleles are recessive then it will not be possible to
differentiate between a homozygous dominant individual and a
heterozygous individual carrying the recessive allele.
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5. You discover a white marigold in your garden and wonder if you have discovered
a new mutation that you could possibly market. Looking at the Burpee flower
catalog you see there is already a white flowered marigold for sale.
a) What type of genetic test would you do to determine if your white flowered
plant is due to a mutation in a different gene or is a mutation in the same gene
as the Burpee marigold?
You would do a complementation test to determine if you had a mutation
in a different gene or the same gene as the Burpee marigold. A
complementation test is where you cross the two mutant (white)
marigolds and see the color of the progeny.
b) Diagram how you would do the test and the two possible outcomes you would
expect that would answer whether you have a mutation in the same gene or a
different gene.
2 genes
1 gene
your white x Burpee white
your white x Burpee white
F1 colored
F1 white
Genetically it would look like this:
AAbb x aaBB
F1 AaBb
a1a1BB x a2a2BB
F1 a1a2BB
6. You are asked to determine how much phenotypic expression of a trait is under
genetic control for several different organisms.
a) What two factors control phenotypic expression of a trait?
Genotype of the individual and the environment
b) How could you determine the level of genetic control involved in expression
of a trait in plants?
Either grow genetically identical plants in different environments to
determine the environmental affect on expression of a trait or grow
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genetically different individuals in the same environment to determine
the genetic affect on expression of a trait.
c) Could your system in plants work in humans? If not how can you determine
the level of genetic control of expression of a trait in humans? Give an
example on the type of data you would expect for:
1) under genetic control
2) not under genetic control
No, due to the challenge in manipulating or controlling human subjects to
study the role of the environment or genotype on trait expression. One
way to do such a study in humans is to study concordance data for trait
expression between identical and fraternal twins.
1) If the concordance level is higher in the identical twins compared to
the fraternal twins then the trait is under more genetic control.
2) I the concordance level is at the same level between identical and
fraternal twins then the trait is under more environmental control
than genetic control.
7. You have a friend who is a cat breeder. She mates a white cat with a brown cat
and gets all black kittens. When the F1 black kittens are mated to F1 black kittens
from similar matings the following F2 progeny are produced.
phenotype
Black
Brown
White
number
160
42
68
Total 270
observed ratio
160/270 = .5926 x 16 = 9.48
42/270 = .1556 x 16 = 2.49
68/270 = .2519 x 16 = 4.03
a) What type of gene action is responsible for the fur color in these cats?
9 : 3 : 4 – single recessive epistasis
b) Test your hypothesis using Chi square analysis.
observed
160
42
68
expected
270/16 x 9 = 151.9
270/16 x 3 = 50.6
270/16 x 4 = 67.5
ob-exp
160 – 151.9 = 8.1
42 – 50.6 = -8.6
68 – 67.5 = 0.5
(ob-exp)2
(ob-exp)2/exp
8.12 = 65.6
-8.62 = 73.9
0.52 = 0.25
65.6/151.9 = 0.432
73.9/50.6 = 1.463
0.25/67.5 = 0.004
χ2 = 1.899
df = (n – 1) = (3 – 1) = 2, P (.05) 2 df = 5.99 1.899 < 5.99 so accept hypothesis
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Name: _____________________________________
8. Your friend at the cattery is so impressed with you that she comes to you with
another question. Seems she has found a cat from overseas that has a unique
phenotype that she believes has a potential market in the United States. The
problem is she can not produce a true breeding individual for the trait. When she
mates two unique cats together she always gets both unique cats and normals.
She also notices that the litter size is always smaller when she mates the unique
cats than when she mates a unique by normal or a normal by normal.
a) What type of gene action is occurring?
Recessive lethal gene action where an individual homozygous allele does
not exist due to it being lethal in the homozygous condition.
b) Briefly describe genetically what is happening in the matings to get the results
she is seeing.
When the two unique cats are mated there are two types of progeny produced,
unique and normal. This indicates that the unique cats used in the mating
were heterozygous for the trait. The litter size was reduced because instead of
getting a 3:1 segregation ratio a 2:1 ratio was being observed because the
homozygote for unique was being killed.
unique (Uu) x unique (Uu)
1 (UU) dead : 2 (Uu) unique alive : 1 (uu) normal alive
9. After answering her first two questions she now thinks you are the ‘cat’s meow’.
So she brings you one more situation she has occurring in the cattery. She has a
trait, red (orange) fur, which appears to be passed on to progeny differently
depending on the parent. If the female has the trait and is mated to a non-red male
all the kittens are red. If the male has the trait and is mated to a non-red female
only the female kittens are red.
a) What type of gene action is occurring?
Dominant sex linkage with the red (orange) fur allele being dominant and
found on the X or sex chromosome.
b) Describe (can use diagrams) how your answer explains the results she is
seeing.
What the woman produced were reciprocal crosses that demonstrated the
trait was sex linked.
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Female red x Male non-red
(XX)
(xY)
Female non-red x Male red
(xx)
(XY)
Xx female red
XY male red
Xx female red
xY male non-red
10. The friend hears about some new molecular markers that can detect if a cat carries
various genes that predispose them to various genetic disorders that cause early
death. She asks you if you think they would work and you ask to see the linkage
data before answering.
a) What is linkage?
Linkage refers to how close two genes are together on the same chromosome.
b) Why would you want to know the linkage data for the molecular markers for
the various genetic disorders?
Depending on how close the markers are to the gene it is possible for
recombination to occur separating the gene from the markers. Depending on
the frequency of recombination (which is dependent on the distance between
the markers and the gene) it is would be possible to have the markers present
and not have the gene or have the gene but not the markers leading to a
misdiagnosis.
11. You finally get away from the cat lady and decide to study traits in plants. You
find there genes that appear to be all on the same chromosome so you decide to
map them. The three genes are Al, Nd and Va. You mate two true breeding
parents and then mate the F1 progeny to a completely homozygous plant with the
genotype alal ndnd vava. You recover the following progeny:
genotype
number of progeny
Al nd va
al Nd Va
1,326
1,346
Al nd Va
al Nd va
410
418
Al Nd Va
al nd va
250
228
Al Nd va
al nd Va
10
12
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Name: _____________________________________
total
4,000
a) What was the genotype of the two true breeding parents?
Need to show the complete genotype: AlAl ndnd vava and alal NdNd VaVa
b) What is the gene order?
Compare parentals to double crossovers (highest to lowest numbers)
parentals
double crossovers
Al nd va
al Nd Va
Al Nd va
al nd Va
So Nd changed between parentals and double crossovers so Nd is in the
middle and the gene order is Al Nd Va or Va Nd Al
c) Draw a gene map of the three genes showing the genetic distance between the
genes.
Al to Nd
parent
recomb.
sco
dco
Al nd
Al Nd
Al Nd
al nd
250
228
10
12
Map distance = ((sco + dco) / total) x 100 = ((250+228 +10+12)/4000 x 100
= (500/4000) x 100 = 12.5
Nd to Va
parent
recomb.
sco
dco
nd va
Nd Va
nd Va
Nd va
410
418
10
12
Map distance = ((sco + dco) / total) x 100 = ((410+418 +10+12)/4000 x 100
= (850/4000) x 100 = 21.3
Al
12.5 cM
Nd
21.3 cM
Va
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d) What is the interference value for these genes?
First determine the number of expected double cross-overs
(sco1 x sco2) x total = expected dco
(0.125 x 0.213) x 4000 =
0.0266 x 4000 = 106.5 = expected dco
interference = 1 – obs dco/exp dco
= 1 – 22/106.5
= 1 – 0.207
= .793 or 79.3% interference
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