This Trait Shows Variable Expressivity with Respect to the Extent of Blotching

BIOLOGY 321 WINTER 2011 ANSWERS TO ASSIGNMENT SET #3

Answers to Questions in Text

Chapter 2 NOTE ON 38: substitute dominant for mutant in part a of the question. 38 a. This would be an example of a haploinsufficient gene since one copy of the wild-type allele does NOT produce enough protein for normal function. You do not expect the mutation to be recessive.

Note on 48: This trait shows variable expressivity with respect to the extent of blotching. When analyzing the data, a plant should be considered blotched even if there is limited amount of blotching

Problem 73

Chapter 6 12.

22.

2 25.

26. NOTE on allele designations in answer to 26. An uppercase A is given to the mutant allele even though it is only incompletely dominant to the wildtype allele. Although this is sort of confusing it is not an unusual designation

3 28

4 Answers to Additional Study Problems Ì Problem 1 ABO blood group: A and B are codominant alleles, O is recessive to both. MN blood group: M and N are codominant alleles; therefore a blood group of N means the individual is homozygous for that allele. Rh+ is dominant to Rh -.

Child 1 belongs to the husband; child 2 belongs to the lover; child 3 could be the progeny of either the husband or the lover. For child 1 and child 2, which genotypes are critical in excluding the husband or lover as the parent? Which genotypes are consistent with either being the parent?

Ì Problem 2 pleiotropic (multiple traits affected) sex-influenced, variably expressed

Ì Problem 3 Answer is d: Note transmission through unaffected female in IVth generation. (a) is eliminated by transmission of the trait via female IV.4 and by the III.1 who did not inherited the trait from his father; (b) extremely unlikely, (c) very unlikely and also female IV.4 is not consistent with this notion

Ì Problem 4 Random inactivation during embryogenesis of one X chromosome in female mammals. See class notes on calico (tortiseshelll) cats and Carbon Copy, the first clone cat.

Ì Problem 5 Statement is true. See discussion of variable expressivity, incomplete penetrance and phenocopy in your text and the handout distributed in class. The term genetic heterogeneity is used to describe situations where mutations in different genes can result in similar phenotypes.

Ì Problem 6 a. pleiotropic

Ì Problem 7 cch = chestnut c= cremello a. Chestnut = bbcchcch Cremello = bbcc Palomino = bbcchc Note: that the palominos don’t breed true b. bbcchc X bbcchc → 1/4 bbcchcch 1/4 bbcc 1/2 bbcchc bbcchc X bbcchcch → 1/2 bbcchc 1/2 bbcchcch bbcc X bbcchc → 1/2 bbcc 1/2 bbcchc c. Mate chestnest with cremellos d. 0, 0, 1

Ì Problem 8 Most obvious feature of this pedigree is that only females are affected. Two possible explanations: (1) autosomal dominant inheritance with expression of the trait limited to females. (2) X-linked dominant trait with homozygotes and hemizygotes inviable. This would explain the other striking feature of this pedigree which is that the trait is never passed via the male line. This explanation is probably the best.

Ì Problem 9 No. See discussion of variable expressivity in textbook and lecture notes. In the case of monozygotic twins, genetic background cannot explain the variable expressivity.

Ì Problem 10 First assess genotype of girls: Based on information in the problem, the twins should both be heterozygous for the hemophilia allele -- whether or not they are mono or dizygotic twins. Then assess doctor’s statement: The different clotting times does not provide evidence for or against the twins being monozygotic --because of the element of chance in the expression of X- linked alleles in heterozygous females. The female with the normal clotting time shows a “recessive” expression typically seen with the allele. Enough of her cells are expressing the wild-type allele for her clotting time to be normal. In contrast, the delayed clotting time of her sister suggests that, by chance, during X-inactivation most of the cells (that end up producing clotting factors) inactivated the normal allele. Since X-inactivation occurs independently (cell by cell in each embryo), the different phenotypic display in the two sisters is not surprising. State Conclusion: They could be mono or dizygotic.

Ì Problem 11A 0.55 = 0.5 (genotypically wildtype offspring) + 0.05 (0.5 x0.1 offspring het for trait but doesn’t exhibit phenotype)

Ì Problem 11B correct answers: a and d

Ì Problem 12 _C__incomplete dominance C haploinsufficient _A__variable expressivity I incomplete penetrance I or C polymorphic A autosomal recessive C X-linked dominant I* sex-limited/sex-influenced A pleiotropic A loss-of-function

* Since the parents are related, this could be a sex-limited, X-linked recessive trait, but this is not the simplest explanation. It could also be a sex-limited autosomal trait, but, again, there is no direct evidence one way or the other.

Ì Problem 13 Correct Answers: a and d Scenario a: If a loss-of-function is recessive, then addition of one wild-type gene copy should restore a normal phenotype Scenario b: untreated hets for a recessive allele will show no phenotype so the gene therapy test would not be instructive Scenario c: By definition, two copies of a haploinsufficient gene are required for a normal phenotype. So the addition of a single copy would not restore a wild-type phenotype. Scenario d: The het will show a mutant phenotype. Restoration of two copies of the wild-type gene (one introduced) should result in a wild-type phenotype

Ì Problem 14

6 __T__ [The mutant allele is pleiotropic] __T__ [The mutant allele shows variable expressivity] YES, not everyone shows both traits associated with the mutatnt allele __F__ The mutant allele represents a polymorphism: NO, it is very rare __F__ The transmission pattern is consistent with an X-linked dominant mutant allele showing incomplete penetrance (both traits taken together): NO, see left side of pedigree __F__ The transmission pattern is consistent with an X-linked dominant mutant allele showing complete penetrance (both traits taken together). __T__ The transmission pattern is consistent with an autosomal dominant mutant allele showing incomplete penetrance (both traits taken together). __F_ The transmission pattern is consistent with an autosomal dominant mutant allele showing complete penetrance (both traits taken together).

For the next set of questions assume that this pedigree shows the inheritance of a common set of traits -- such as red hair and freckles in a British population. Assume that these phenotypes result from a loss-of-function mutation in a gene coding for a receptor protein that controls melanin production. Indicate True, False or N (not enough information provided to assess). Indicate false, if any part of the statement is false.

__T__ The transmission pattern is consistent with autosomal recessive inheritance of a completely penetrant mutation (both traits taken together). _ F___ The transmission pattern is consistent with autosomal recessive inheritance only under the condition of incomplete penetrance (both traits taken together). __T__ The mutant allele is pleiotropic. __T__ The mutant allele represents a polymorphism

Ì Problem 15 Part A: See lecture notes and review class discussion of factors influencing the expression of a phenotype. NOTE: The category of chance in this context does not include the chance that a person received the mutant allele. The mutant genotype is a given. You must think of post-inheritance “chance” events that could influence the expression of the mutant allele -- such as variation in the development of the nervous system or other revelant developmental events. Part B: Genetic background and environment are likely to be significant here. But since only a single patient of African descent was examined, chance factors affecting expression could be operating.

Ì Problem 16 Circle True or False or I (not enough info to decide) F This loss-of-function mutation represents a polymorphism for the pax-6 gene. T The pax-6 gene is haploinsufficient. F The pax-6 mutation is pleiotropic (True) and completely dominant to the wild-type allele (False – it is incompletely dominant).

Ì Problem 17. I’ll let you read the article and figure out the answers.

Ì Problem 18#1: answer is b.

7 Ì Problem 18#2:

T F It is known that many genes interact to specify the formation and functioning of structures required for normal hearing (such as the cholea in the inner ear). This means that deafness cannot be inherited as a standard Mendelian single-gene trait. One-two sentence explanation/defense of your answer: If an allelic variation in one of these many genes can cause deafness, then the trait is by definition a single gene trait.

Examine the Table below. For each statement below choose T F or I (not enough info to decide) T F I The gene that codes for the apoB-100 protein is haploinsufficient. T F I The gene that codes for the ARH protein is haploinsufficient. T F I In the general population, the frequency of the mutant LDL receptor allele is 0.1%. This mutant allele would be considered a polymorphism in this population. T F I This data in this table indicate that hypercholesterolemia (high plasma cholesterol levels) is genetically heterogeneous.

Ì Problem 19 UNINFORMATIVE: O X O In either scenario, the trait will breed true. POTENTIALLY INFORMATIVE: AB X O. In the one gene scenario, the kids would have to have either A or B blood types. In the two gene scenario, if the AB parent is AABB, all kids would be AaBb and have the AB blood type. If parent is AaBb, then basic mendel tells you that the kids could be AaBb, or aaBb or aaBb or aabb. I’ll let you sort through the rest of this question on your own.