BIO 184 Lab
Spring 2013
Name _______________________________
LAB PRACTICAL EXAMINATION 4 - KEY (30 pts) 1. (4 pts) DEMO a.
With respect to eye color and wing morphology, what is the phenotype of this fly? Wild eye, miniature wing
b.
What is the sex of this fly? Female How do you know? She doesn’t have sex combs on her cute little forelegs. (This allows me to know for certain.) Another indicator: The tip of her abdomen doesn’t have the broad dark band that is seen in males.
2. (10 pts) Wild-type Cooper’s Hawks have white spots on their wing feathers (S) and curved talons (f). Mutants are sometimes observed with no spots (s) and/or forked talons (F). A dihybrid hawk is test-crossed for with the following results: 27 22 12 11
white spots; curved talons no spots; forked talons white spots; forked talons no spots; curved talons
a.
What are the expected values (assuming the Mendel's Second Law is operating)?
Total = 27+22+12+11 = 72; Mendel’s Second Law states that if a dihybrid is test-crossed there should be a 1:1:1:1 phenotypic ratio in the offspring. 72/4 = 18. Therefore, we expect 18 offspring in each category. 18 18 18 18
white spots; curved talons no spots; forked talons white spots; forked talons no spots; curved talons
b.
Your null hypothesis is that the observed deviation can be explained by sampling error. Perform a Chi-Squared analysis on the data to test this hypothesis. Show all your work.
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BIO 184 Lab
Spring 2013
Chi-Squared = 10.1; df = 3 Use the table to estimate the p value associated with your Chi-Squared analysis. Do you reject or fail to reject the null hypothesis? Briefly explain. Reject. The p value is less than 0.05. Therefore, it is unlikely that the observed deviation from the expected 1:1:1:1 distribution is due to sampling error.
df 1 2 3 4 5
c.
0.95 0.004 0.10 0.35 0.71 1.15
0.90 0.016 0.21 0.58 1.06 1.61
0.70 0.15 0.71 1.42 2.20 3.00
0.50 0.46 1.39 2.37 3.36 4.35
0.30 1.07 2.41 3.67 4.88 6.06
p-value 0.20 1.64 3.22 4.64 5.99 7.29
0.10 2.71 4.61 6.25 7.78 9.24
0.05 3.84 5.99 7.82 9.49 11.07
0.01 6.64 9.21 11.35 13.28 15.09
0.001 10.83 13.82 16.27 18.47 20.52
Is there evidence that the genes are linked? If so, map the distance between them and show all your work. (If not, leave this space blank.) Yes. The S_F_ and ssff are “mirror image” phenotypic classes and have approximately equal numbers of offspring. S_ff and ssF_ are also “mirror images” and have approximately equal numbers of offspring. This is the classic distribution pattern observed when genes are linked; fewer than 50% of the gametes are recombinants and therefore fewer than 50% of the offspring fall into the recombinant phenotypic classes. Map distance = (11+12)/72 x 100 = 32 cM
3. (6 pts) William and Bree are both carriers for Tay Sachs disease, an autosomal recessive disorder that is fatal in early childhood. a.
Bree is pregnant with the couple’s first child. a.
What is the probability that the baby will have Tay Sachs Disease? Draw a Punnett Square to support your answer. T
t
T
TT
Tt
t
Tt
tt
Two carriers have a ¼ chance of having an affected child (tt) in any pregnancy. b.
What is the probability that the child will NOT have Tay Sachs disease? ¾
c.
William and Bree want to have three children altogether. i.
What is the probability that all three children will be normal? Show your calculation.
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BIO 184 Lab
Spring 2013
(3/4)3 = 27/64 ii.
What is the probability that exactly one of the three children will have Tay Sachs disease (any order)? Show your logic and calculation.
Three possible orders for having one affected child: AUU, UAU, UUA Product Rule: Each has a (1/4)(3/4)(3/4) probability of happening = 9/64 They are mutually exclusive (if AUU happens then UAU and UUA won’t happen, etc.) Sum Rule: 9/64+9/64+9/64 = 27/64
5. (6 pts) Examine the pedigree below. I
II
III
IV
A genetic counselor eliminates X-linked dominant as a possible mode of transmission of the disease. Identify two different reasons why the genetic counselor made this elimination. Be specific! 1. Dominant diseases appear in every generation. This disease skipped generations I and IV. 2. If a man has a sex-linked dominant disease, his genotype is XAY. He will pass his XA allele to all his daughters and they will also be affected with the disease. In this pedigree, III-2 has the disease but did not pass it on to his daughters. 6. (4 pts) For each scenario, indicate the most probable mode of inheritance: X-linked dominant, X-linked recessive, autosomal dominant, autosomal recessive, Y-linked. Scenario As far back as anyone can remember, all the men in Mike’s paternal line have been
Most probable mode of inheritance Y-linked
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BIO 184 Lab
Spring 2013
exceptionally tall (over 7 feet). The women in Mike’s family are of normal height. Karen and Sabrina are cousins. Both have sons with the same extremely rare genetic illness. There is no inbreeding in their family. Eva and Paulo have a child with PKU, a genetic illness that leads to mental retardation. They are shocked when they get the diagnosis because neither of their extended families have any history of the disease. In 1785, a small group of Norwegians settled the southern coast of Greenland. Between 1785 and 1900, the community grew to about 240 people despite almost no contact with the outside world. Today, there are about 500 descendants and a large number of them lack nails on their big toes. This is a genetic disorder that is rare elsewhere in the world.
X-linked recessive
Autosomal recessive
Autosomal dominant
7. (4 pts) EXTRA CREDIT You have an ant problem in your home and decide to try a “home remedy” that your mother used when you were a child: equal parts white vinegar and baby powder. The “home remedy” is non-toxic and you’d rather use it than the commercial ant spray you’re currently using. You have two trails of ants of about equal size, one running from your laundry room sink to a crack in the wall and one running from a crack in the flooring up the wall and into the back of the washing machine. Being a good scientist, you decide to compare your home remedy against the commercial ant spray before making the switch. You spray the first trail with your home remedy and the second trail with the commercial ant spray. Then you wait one hour and return to see what happened. The results are below.
# dead ants per square inch of trail a.
Home remedy
Commercial ant spray
31
45
Is your home remedy as good as the commercial ant spray at killing ants? Show how you arrived at your conclusion. 31+45 = 76
Home remedy Commercial ant spray
Observed 31 45
Expected 38 38
Experimental hypothesis: The home remedy works just as well as the commercial ant spray. Therefore, both methods will kill the same number of ants per square inch of trail.
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BIO 184 Lab
Spring 2013
Null hypothesis: Any deviation between the observed numbers and the expected numbers (as predicted by an equal distribution) is due to chance.
The p value is greater than 0.05 and I fail to reject the null. Therefore, there is no evidence that the distribution is not equal. The home remedy works just as well as the commercial ant spray. Yay! b.
Do you feel confident about your finding? If not, what could you do to raise your level of confidence? However…. The p value not very far above 0.05 and my sample size is small. (The calculated p value on Excel is 0.108.) Therefore, I do not feel very confident in my result. It’s possible that my sample size is so small that I cannot detect a difference. I should increase my sample size (do more replicates and count a lot more ants) before reaching a final conclusion. If the two sprays are not equivalent in their effectiveness, the p value will drop as sample size increases and will eventually fall below 0.05. If the two sprays actually do work equally well, the p value will rise as sample size increases.
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