Download Genetics 314 – Spring, 2005

Name: _______________________________
Genetics 314 – Spring, 2005
Exam 4 – 100 points
1. What are Mendel’s two laws and how do they relate to events in meiosis?
Law of Segregtion – relates to the separation of the homologous chromosomes in
Anaphase 1 of meiosis
Law of Independent Assortment – relates to the orientation of the different
homologous chromosome pairs when they align on the metaphase plate during
Metaphase I of meiosis.
2. You are studying six traits in a plant and all six traits are on different chromosomes.
You mate the following genotypes:
Aa Cc Ee HH Jj ll x Aa Cc ee Hh Jj Ll
a) Out of 2000 progeny how many of the following genotypes would you expect to
recover?
1)
2)
3)
4)
5)
AA CC Ee HH JJ Ll
Aa Cc Ee Hh Jj Ll
aa cc ee hh jj ll
Aa cc Ee Hh jj ll
Aa Cc ee HH JJ Ll
= (.25 x .25 x .5 x .5 x .25 x .5) x 2000 = 3.9 or 4
= (.5 x .5 x .5 x. .5 x .5 x .5) x 2000 = 31.25 or 31
= (.25 x .25 x .5 x 0) x 2000 = 0
= (.5 x .25 x .5 x .5 x .25 x .5) x 2000 = 7.8 or 8
= (.5 x .5 x .5 x .5 x .25 x .5) x 2000 = 15.6 or 16
b) Were any of your answers to 2a equal to zero? If yes briefly explain or diagram
why you could not recover that genotype from this cross.
Yes, the answer to ‘3’ was zero because it is not possible to get the genotype ‘hh’
from a cross between a homozygous dominant (HH) individual and a heterozygous
individual (Hh). In the progeny, the only two genotypes possible are HH and Hh
and they will occur in a 1 : 1 ratio.
3. Since you recently finished a degree at the University of Idaho that required a genetics
class, you are asked by some recently married members of your extended family if
you could help them determine the odds of having certain combinations of kids. If
the couples want to have 6 children, calculate the probabilities for the following
combinations:
a)
b)
c)
d)
5 boys – 1 girl
4 boys – 2 girls
3 boys – 3 girls
0 boys – 6 girls
(6!/(5! x 1!)) x .55 x .5 = 6 x .56 = .094 or 9.4 %
(6!/(4! x 2!)) x .54 x .52 = 15 x .56 = .234 or 23.4%
(6!/(3! x 3!)) x .53 x .53 = 20 x .56 = .312 or 31.2%
(6!/(0! x 6!)) x .50 x .56 = 1 x .56 = .0156 or 1.5 %
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Name: _______________________________
4. You are studying a dominant trait and discover an odd inheritance pattern. It appears
that whenever a male that shows the trait is mated to a recessive female all the
progeny show the dominant trait but when a female that shows the trait is mated to a
recessive male none of the progeny show the trait. Further study shows the gene is
not on a sex chromosome so is not sex-linked. Based on this data what type of
genetic control is at work and how would it explained the observed results?
This is a case of epigenetics, where imprinting has occurred and whether an allele is
expressed or not depends on which parent is the source of the allele. This will
change how the allele appears to be inherited and whether the allele is dominant or
recessive. In this case, if the male donates the allele it is expressed in the progeny
but if the source of the dominant allele is the female the dominant allele is silenced
and only the allele donated from the male parent is expressed. Such imprinting can
also occur favoring an allele donated from the female over the same allele being
donated from the male.
5. You are studying blood types and are told that there are five different alleles for blood
type instead of three.
a) With five alleles how many genotypes would you have if four of the alleles were
codominant and one was recessive?
The formula for this (n(n +1))/2 = number of possible genotypes
In this case n = 5 so the number of genotypes = (5( 5 + 1))/2 = 30/2 = 15
b) Would you expect the same number of phenotypes as genotypes? Briefly explain
your answer.
No, because there is one recessive allele the number of phenotypes will be less than
the number of genotypes. This is because heterozygous individuals for the recessive
allele will have the same phenotype as the homozygous dominant for that particular
dominant allele. The only way to have the number of phenotypes equal the number
of genotypes is to have no recessive alleles.
6. You are observing a trait in mammals and notice that you do not always observe
phenotypically what you would expect based on the genotype.
a) What two factors control phenotypic expression?
Genotype of the individual and the environment.
b) How would you determine which factor was the major controlling factor for
phenotypic expression in:
1) Animals?
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Name: _______________________________
With animals you would need to hold one or the factors constant so you could
measure the effect of the other factor. With animals you could develop genetically
identical animals through breeding or possibly cloning and place them in different
environments. Any differences observed for a trait would then be due to the
environment. You could also place genetically different individuals in a controlled
environment and then any differences observed for the trait would be due to
genotypic differences.
2) Humans?
Breeding humans is frowned upon so determining the level of genetic and
environmental control of expression of a trait is usually done by studying twins.
Studying the concordance of expression of a trait in identical and fraternal twins
can be used to determine the amount of genetic re environmental control there is for
expression of a specific trait. If the level of concordance is higher for identical twins
then there is more genetic control and if the level of concordance is the same
between the two types of twins then the expression of a trait is under more
environmental control.
7. You mate true breeding chinchillas always mating a black chinchilla with a spotted
chinchilla. All the F1 chinchillas have a black coat. You mate the F1 chinchillas and
recover the following F2 progeny:
phenotype
black
spotted
white
total
number
224
62
24
310
a) You suspect you are dealing with epistasis. What in the F2 progeny supports this
hypothesis? Briefly explain your answer.
If a new phenotype is observed in either the F1 or F2 generation then epistasis
becomes a possible answer for variation in the genetic ratios observed for a trait. In
this case a white phenotype appears in the F2 generation even though a white
phenotype was not observed in the parents of F1 generation.
b) What type of gene action are you observing?
Determine the observed ratio using a 2 gene epistatic model for calculating the ratio
Black – 224/310 x 16 = 11.56
Spotted - 62/310 x 16 = 3.2
White - 24/310 = 1.24
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Name: _______________________________
An observed ratio of 11.56 : 3.2 : 1.24 is close to a 12 : 3 : 1 ratio which is indicative
of single dominant epistasis.
c) Test your hypothesis using Chi square analysis (please show your work).
Obs
224
62
24
exp
(310/16) x 12 = 232.5
(310/16) x 3 = 58.1
(310/16) x 1 = 19.4
ob – exp
(ob – exp)2
-8.5
3.9
4.6
72.25
15.21
21.16
(ob – exp)2/exp
72.25/232.5 = 0.31
15.21/58.1 = 0.26
21.16/19.4 = 1.09
χ2 = 1.66
degree of freedom (df) for this model is determined by the formula n-1 = df so in this
problem n = 3 so the df = 2
With 2 degrees of freedom and at a probability of .05 the Chi square value is 5.99.
Since 1.66 < 5.99 you would accept the hypothesis that this is single dominant
epistasis.
8. While working with the chinchillas you notice that there it is possible to get a sable
chinchilla but looking at the pedigree you start to suspect that the trait is sex-linked.
a) What is the name of the crosses you would make to test your hypothesis?
To determine if sex linkage is occurring you would do a reciprocal cross and observe
the progeny to determine if the expression of the trait in the progeny differs
depending on the parent expressing the trait.
b) Diagram the crosses you would make and the progeny you would expect if the
sable coat color was recessive to the dominant black coat color and was sexlinked.
The two crosses that would be made would be a cross where the female expresses
the sable trait and the male expresses black, and a second cross where the female
expresses black and the male expresses sable.
Crosses:
XsXs x XSY
XSXS x xsY
Progeny
XSXs – females black
XsY – males sable
XSXs – females black
XSY – males black
Since results differ between the reciprocal crosses it appears the trait is sex-linked.
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Name: _______________________________
9. You are asked to evaluate a set of flanking molecular markers that dog breeders would
like to use to identify dogs carrying the allele for a lethal joint disorder. They have
identified three sets of flanking markers shown below. The centa-Morgan (cM)
values indicate the distance the markers are away from the gene.
1) 2 cM--- gene ---4 cM
2) 1 cM--- gene ---3 cM
3) 3 cM--- gene ---2 cM
a) Which pair of flanking markers would be best for selection against the allele for
the lethal joint disorder?
To determine which set of markers would be best to select against individuals
carrying the lethal allele you need to calculate the probability that you could have
both of the flanking markers and not have the lethal allele. This is done by
calculating the probability of a double cross-over.
1) .02 x .04 = .0008
2) .01 x .03 = .0003
3) .03 x .02 = .0006
So with this set of flanking markers set (2) has the lowest probability (.0003 or
.03%) so it would be the best for selection of individuals carrying the lethal allele.
b) Why is it better to have flanking markers to identify a gene instead of a single
marker on one side of the gene?
Using flanking markers instead of a single marker reduces the probability that an
individual positive for the marker would actually not be carrying the lethal allele.
This would reduce the potential for a misdiagnosis.
10. You decide to determine the distance between three genes that control height (tall –
T or short –t) flower color (Red – R or White – r), and seed color (Brown – B or
yellow – b) in a new crop plant. You cross two true breeding parents to produce a F1
that was completely heterozygous for the traits. You cross the F1 plants with a
homozygous recessive plant (tt rr bb) and recover the following F2 progeny:
genotype
number
T R b
t r B
1,200
1,275
T r b
t R B
280
310
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Name: _______________________________
t R b
T r B
155
165
T R B
t r b
15
10
total
3,410
a) What were the genotypes of the true breeding parents?
For genotype you must show all the alleles. In this case the parental genotypes will
be homozygotes for the two largest classes in the testcross population.
TT RR bb and tt rr BB
b) What is the gene order?
Gene order can be determined by comparing the parental allelic combinations to the
allelic combination of the double cross-overs (dco) which are the smallest two classes in
the testcross population
Parental
Double cross-overs
TRb
trB
TRB
trb
In this case the B gene had changed comparing the parental genotypes to the double
cross-over genotypes so the B gene is in the middle and the gene order is either
T B R or R B T
c) Draw a gene map of the three genes showing the genetic distance between the
genes.
To make a map you need to identify which recombinants relate to cross-overs
between T – B and which recombinants relate to cross-overs between B – R.
T–B
Parentals
T b
t B
recombinants
T B
t b
recombinants showing either t – B or t – b allelic combinations include:
155
sco
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Name: _______________________________
165
15
10
sco
dco
dco
Map distances are calculated by the formula (( sco’s + dco’s)/total) x 100 so for T –
B ((155 + 165 + 15 + 10)/3,410) x 100 = 345/3410 x 100 = .101 x100 = 10. 1 map units
B-R
Parentals
R b
r B
recombinants
R B
r b
recombinants showing either R – B or r – b allelic combinations include:
280
310
15
10
sco
sco
dco
dco
Using the formula (( sco’s + dco’s)/total) x 100 the map distance between B – R is:
((280 + 310 +15 + 10)/3410) x 100 = 615/3410 x 100 = .180 x 100 = 18 map units
So the map for these genes is T
10.1 mu
B
18.0 mu
R
d) What is the interference value for these genes?
Interference is the percentage of double cross-overs that do not occur. This is
calculated using the formula: Interference = 1 – (obs. dco/ep. dco)
Expected double cross-overs is calculated using the formula
(freq. sco1 x freq sco2) x total = (.101 x .18) x 3,410 = .0182 x 3,410 = 62
Interference = 1 – (25/62) = 1- .403 = .597 or 59.7% interference
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