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Additional Complex Problems Involving |
This is a problem that was on the final exam in genetics in both 1998 and 1999.
Only about 20 out of 280 students got this
right. Most students did not even attempt the question. Others tried to answer this
question as an essay question. Still
others used the wrong expected ratios. I will go through the logic of the problem as an illustration of how to
analyze problems of this type.
(1 0 points) In TV Bugs, Cable (A) is dominant to rabbit ears (a) and Color (B) is dominant to black and white (b). Workly Drudge crosses a completely homozygous Cable, b&w bug to completely homozygous rabbit ear, Color, bug. He forgets to look at the F1 progeny. In the F2 he obtained 3408 Cable, Color, 1772 Cable, B&W, 2592 Rabbit, Color and 228 rabbit, b&w progeny for a total of 8000 TV bugs. These results had a Chi-square for goodness-of-fit of 1287.7 (null hypothesis ; 9:3:3:1; 3 df). Workly hypothesized that the results were due to linkage. He immediately did a Contingency Chi-Square test and obtained a Chi-square value of over 8000!! Workly tells his advisor that he has "proved" that the results are entirely due to linkage. The professor, Tadhg McCuil, insists that Workly has not entirely explained the results. Using your knowledge of genetics and statistics, how can the F2 results be explained. You must support your answer with a mathematical proof!
(NOTE: DO NOT REPEAT THE CHI-SQUARE TEST FOR GOODNESS OF FIT TO A
9:3:3:1 RATIO OR THE CONTINGENCY CHI-SQUARE!!!!)
What do we know from the Problem: In the original cross, as clearly stated, a completely homozygous Cable, b&w bug was crossed to a completely homozygous rabbit ear, Color, bug. That is, the original cross was:
AAbb x aaBB
The F1 progeny would have the genotype AaBb.
We presume that all of the F1 progeny would have the dominant Cable, Color phenotype but we do not actually
know this. Workly did not examine the F1 progeny closely!!!
At some point
the experiment would have had to have been redone and the phenotype of the
F1 progeny examined
more closely.
Goodness of Fit and Independent Assortment. We were already told that the observed data do not fit a 9:3:3:1 ratio. Furthermore, we have been told that the loci are not independently assorting (significant contingency Chi-square test). From the wording, the deviation from independent assortment is due to linkage. The question then is what else is going on.
What is being asked:
When there is a deviation from the expected dihybrid ratio (in this case 9:3:3:1). There are four possibilities:
a. Chance
b. Problems with segregation at one of both loci.
c. Problems with independent assortment of the two loci.
d. Problems with segregation at one of both loci and additional problems with independent
assortment of the
two loci.
Each of these possibilities requires a different statistical test. In the case of this problem, the tests for possibility a (chance) has been done and the observed distribution is not due to chance. Workly has already done the test for possibility c (independent assortment). There is a problem with independent assortment which you have told is linkage. This effectively eliminates possibly b. Tadhg McCuil implies that possibility d (deviations from segregation and independent assortment) is the correct answer. This means that we must check the both the Cable locus and the Color locus for problems with segregation. That is, we collapse the data and do the Chi-square test for goodness of fit for the expected monohybrid ratios.
(Test taking skills: On the exam several students told me to check for problems with segregation at each locus but then did not do so! Suggesting a course of action is not mathematical proof. )
What can we infer? Above we listed everything that we knew from the problem. Now we can infer several other things.
1. Expected Ratios for the monohybrid crosses:
If the original dihybrid cross was AAbb x aaBB, then the collapsed F1 crosses are:
AA x aa
and
BB x bb.
With a null hypothesis of complete dominance and no problems with segregation, in the F2 we would expect to see Cable: rabbit in a 3:1 ratio and Color: b&w also in a 3:1 ratio. For 8000 progeny this would be 6000 Cable: 2000 rabbit and 6000 color: 2000 b&w.
(Test taking skills 2: The next most common error on the exam was to have the wrong expected values. Many students told me that I should expect a 1:1 ratio so that there should be 4000 Color to 4000 b&w. If you take the time to write out your monohybrid crosses, this error should not occur.)
What do we need to do next?
(1) collapse the data as we have done
before to obtain observed values
(2) calculate the expected values
(3) check the goodness of fit of observed to expected values using the appropriate Chi-square test.
| Cable, Color | Cable, b&w | Rabbit, Color | Rabbit, b&w |
| 3408 | 1772 | 2592 | 228 |
The c2
value is 47.7. The critical value is 3.84 for 1 df and a level of significance of .05.
We
reject the null hypothesis that there is no problem with segregation at the
Cable locus. Now we only need to test the Color locus. I collapse the data for color and check for the expected 3:1 ratio.
No need to do the Chi-square for goodness of fit. The fit is exact. Segregation at the color locus is
fine. Conclusion: The professor is right! There is a problem with linkage (discovered by Workly with the contingency
Chi-square) and a problem with segregation at the Cable locus. In addition to being required problem, the TV bug question
was used for extra credit both on the exam and as a
special problem on the website. Extra credit (10 points) Referring back to question 8, Workly Drudge completes his
statistical analysis and heads back to the lab. He test crosses
an F1 TV bug (AaBb) to a completely homozygous rabbit
ears, b&w bug (aabb). He obtains the following results (N = 6000): Workly studies the results and says. "It was always obvious that the genes
were linked and in repulsion. I even calculated that they were 20 map units
apart. Now I understand what else is going on. I guess I should have looked
at my F1 progeny more closely". A. What is the expected ratio in the backcross if the genes are in repulsion
and 20 map units apart? B. With all the data from Problem 8 and these additional results, explain the
genetics underlying this problem. C. What would Workly have seen in his F1 ? ANSWERS TO EXTRA CREDIT A. You are told that the genes are in repulsion and
20 map units apart. This means that the parental gametes are Ab and aB and that the
recombinant gametes are AB and ab. If the genes are 20 map units part,
then the recombinant gametes make up 20% of the gametes while
the parental gametes make up 80% of the gametes. Thus, from the heterozygous parent we get four gametes AB, Ab, aB and ab in
the frequencies of .10:.40:.40:.10,
respectively. From the other parent (aabb) we get one
gamete (ab) with a frequency of 1.0. The expected distribution of phenotypes
from the test cross, assuming a null hypothesis of on problems with segregation
at either locus, would be:
(Test taking skills 3:
In answering this question, most common error was to ignore information about linkage and tell me
that the expected ratio was 1:1:1:1. Be sure to use all of the information
that you have).. B. Again, there is problem with segregation at the Cable
locus. If
one collapses for Color you expect C. If Workly had looked at the F1 progeny he would have seen that about 80%
of the progeny D. Individuals which have the phenotype
rabbit, b&w from the test cross
have the genotype aabb
a. If the individual is aabb then all of the progeny will be rabbit, b&w.
Calculating Expected Ratios From a Cross Involving Two or More
Genes. Quick Links F2 Cross Also See
Phenotypes
Cable
Rabbit
Total obs
5180
2820
8000 exp.
6000
2000
8000 |obs-exp-.5| = d
819.5
819.5
d2
671500
671500
d2/exp
111.9
335.8
447.7
c2 = 447.7
Phenotype
Color
b&w
Total obs
6000
2000
8000 exp.
6000
2000
8000
Phenotype
Obs. Number
Exp. Number
Cable, Color
483
Cable, b&w
1917
Rabbit, Color
2519
Rabbit, b&w
1081
(D) Workly decides to do the definitive cross to prove his explanation of
the genetic system.
He crosses individuals which are rabbit-ear, b&w from his
test cross above to individuals
which are rabbit-ear, b&w that are known to
come from a pure-breeding line. What would
Workly see in the progeny of this
cross?
Phenotype
Genotype
Freq.
Expected
Observed Cable, Color
AaBb
.10
600
483 Cable, b&w
Aabb
.40
2400
1917 Rabbit, Color
aaBb
.40
2400
2519 Rabbit, b&w
aabb
.10
600
1081
(the Allele symbols in Bold came from the heterozygous parent).
3000 Color to 3000 b&w
and that is
exactly what you observe. If you collapse over the Cable
locus you expect
3000 Cable to 3000 rabbit but you observe 2400 Cable to 3600
rabbit. That is,
you have 600 too many rabbit and 600 too few
Cable. The best guess from these
data would be
that Cable does not have 100% penetrance in the heterozygote.
In fact, if we observe 2400 Cable
when we expect 3000 Cable
phenotypes, the penetrance would be 2400/3000 or 80%.
were Cable, color while the
other 20% were rabbit, color. In experimental work one cannot
assume that they know the phenotype
of the F1 progeny.
the genotype Aabb where
Cable did not penetrate. If
these individual are crossed to individuals
known to be aabb then we can
expect two different outcomes.
b. If the individual is Aabb then 40% of the progeny will be
Cable, b&w
while 60%
will be rabbit, b&w.
1
Basic F2 cross - autosomal genes
2
Basic F2 cross- X-linked gene
Expected Phenotypes of F2 Cross - Single Autosomal
Traits
Expected Phenotypes of F2
Cross - X-linked gene
4
F2 cross - two unlinked autosomal genes
5
F2 cross - one autosomal - one X-linked
gene
6
Linkage - Coupling and Repulsion
7
F2 cross - two linked autosomal
genes -coupling
8
F2 cross - two linked autosomal
genes - repulsion
9
Measuring the recombination rate from F2
data
Recombination from F2 data - Table of
Values
10
Basic F2 cross- Z-linked gene
(inactive)
12
Expected Phenotypes of F2 Cross -
Z-linked gene (inactive)
13
F2 cross - one autosomal - one Z-linked
gene (inactive)
Detection of Linkage - Basic Info - Two Autosomal
Traits - Complete Dominance
Detection of Linkage - More complex examples
Detection of Linkage - One More Example
Detection of Linkage - Two
Autosomal Traits - Heterozygote has Unique Phenotype
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