bimm110 - CBB Genetics Immo E. Scheffler Fall 2000 FINAL...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CBB Genetics Immo E. Scheffler Fall 2000 FINAL EXAMINATION All answers are to be written into the Blue Book. Leave the first page blank for scoring. The number of points for each question is equivalent to the number of minutes recommended. Please note the question number clearly at the top of each page of the Blue Book. "A fact is a contestable component of a theoretically constituted order of things". M.E. Hawksworth, 1989 (Postmodernist Rhetoric is not acceptable on this examination) Multiple Choice Questions For each of the following questions, select the one best answer. QUESTION 1 (3 minutes) A boy with Duchenne muscular dystrophy was born to a man and woman with no familial history of the disease. The most likely explanation for this occurrence is: A. infidelity. B. a point mutation in the dystrophin gene. C. a recombination event in the dystrophin gene gave rise to a frameshift mutation leading to an untranslatable mRNA. D. a CGG expansion resulted in the disruption of the promoter of the dystrophin gene. E. a translocation resulted in the disruption of the dystrophin gene. Answer: C Duchenne muscular dystrophy (DMD) is a lethal, X-linked recessive disease affecting approximately 1 in 3300 live male births. The disease becomes symptomatic in early childhood. Inability to walk occurs by the end of the first decade, and death usually occurs by the second decade. Nearly all patients show the complete absence of the protein dystrophin, which is abundantly found in skeletal and cardiac muscle. The DMD gene, located at Xp21, codes for this protein. It is approximately 2300 kb in size, making it one of the largest genes known of any species. It contains 24 regions of 109 amino acids, that are similar but not identical repeats of each other. In women, the similarity of these sequences can lead to the misalignment of homologous material at meiotic synapsis. In combination with a recombination event, this misalignment gives rise to frameshift mutations, leading to an untranslatable mRNA. This series of events occurs at an extremely high rate of about 1 in 10,000. In fact, one third of the DMD cases in each generation arise from this mechanism. Infidelity is not a likely explanation for the child because the trait is an X-linked recessive condition. A male child must receive the DMD trait on the maternal X chromosome. The father of the child only contributes a Y chromosome. Many genetic diseases, such as cystic fibrosis, are due to a single point mutation within the coding region of the gene. Diseases such as Fragile X syndrome and Huntington disease, are caused by the expansion of a trinucleotide repeat. Finally, rare cases of DMD in females have been caused by an X chromosome-autosome translocation with the breakpoint on the X chromosome within the DMD gene. After X inactivation occurs in the female zygote, the only cells which survive have the normal X chromosome inactivated. This gives rise to a female who is heterozygous for the DMD, but phenotypically expresses the disease. QUESTION 2 (3 minutes) Klinefelter's syndrome can arise from all of the following events EXCEPT: A. B. C. D. E. nondisjunction in meiosis I of the male nondisjunction in meiosis II of the male nondisjunction in meiosis I of the female nondisjunction in meiosis II of the female nondisjunction in the first mitotic division in the zygote Answer: B Klinefelter's syndrome, XXY, is characterized by poorly developed secondary sexual characteristics, small testes, hyalinization of the seminiferous tubules and sterility. Many affected males are tall and eunuchoid. Klinefelter's syndrome can arise from nondisjunction in any division in a female and from nondisjunction in meiosis I in the male. It could also arise from a failure in an early mitotic division in which XX and Y went to one pole and Y alone went to the other. Cells from the XXY line would be viable, while cells containing only the Y would die. Klinefelter's syndrome cannot arise from nondisjunction in meiosis II of the male because the possible gametes arising from a failure in this division are either XX or YY. Since a female can only contribute an X chromosome, possible children from a meiosis II failure in the male would be either XXX or XYY. The chart below illustrates the possible offspring from failures during different meiotic divisions in the different sexes. QUESTION 3 (3 minutes) The autosomal recessive Tay Sach disease is caused by a deficiency for hexosaminidase A and is lethal in childhood. In a population of Ashkenazi Jews, blood testing showed the frequency of heterozygotes to be 0.1. What is the probability that the first child of two individuals from this population with no family history of the disease will be affected with Tay Sach disease? A. B. C. D. E. 0.0025 0.0625 0.1122 0.25 A probability cannot be calculated from the information given. Answer: A Since Tay Sach is a lethal in childhood, the only individuals carrying the gene are heterozygotes. The only mating which produces progeny with the disease will be the mating between two heterozygotes. Using the blood test data, the probability that a random individual is a heterozygote is 0.01. The probability that two heterozygotes will produce a child with Tay Sach disease is 0.25. So the overall probability that two individuals will produce a child with Tay Sach is (0.1)(0.1)(0.25) or 0.0025. What about the other choices? Choice B, 0.0625 is a garbage answer which looks like it would be a reasonable guess for this kind of probability. It is the product of (0.25)(0.25). Aside from that, it is difficult to come up with reasonable probabilities in this problem which multiply out to that number Choice C, incorrectly assumes that the parents had a sibling with Tay Sachs. This meant that the grandparents carried the disease, and their normal children had a 0.67 chance of carrying the allele. This gave the probability that the parent's child would have Tay Sach as (0.67)(0.67)(0.25) or 0.1122. Choice D, 0.25 incorrectly assumed that the parents are carriers, so the probability of a Tay Sach child is equal to the probability that two heterozygotes will produce a child with the disease or 0.25. Choice E, incorrectly assumes that because q2, the frequency of the disease cannot be calculated from the problem, and hence the allele frequencies, that an estimate of the frequency of heterozygotes cannot be made. Without the probability of a heterozygous individual, the problem cannot be done. QUESTION 4 (3 minutes) A 28 year old woman and a 25 year old man present for genetic counseling. Both are Caucasian and have one sibling affected with cystic fibrosis. The most appropriate method to assess the risk of transmitting cystic fibrosis to a potential child would be: A. B. C. D. E. Biochemical testing Karyotype analysis Fluorescence in situ hybridization (FISH) Polymerase Chain Reaction (PCR) Western blot analysis Answer: D Cystic fibrosis is an autosomal recessive disease caused by point mutations or small deletions in an integral membrane protein which functions as a chlorine transporter. Although many mutations within the gene cause the phenotype, it is not practical or cost effective to sequence the entire region to screen for a carrier. However, in the Caucasian population, the most frequent mutant allele causing cystic fibrosis is due to a small deletion at phenylalanine 508 in exon 10. It accounts for greater than 50% of mutant cystic fibrosis alleles in Caucasian populations. An amplification of this region using the polymerase chain reaction (PCR) can be done, and the PCR products sequenced and compared against the normal sequence for this region. If comparison reveals the deletion, the donor of that template DNA would be classified as a carrier. It the mutation is not present, the probability that the donor of that DNA is a carrier of cystic fibrosis is greatly reduced, but still exists. Most diagnostic laboratories will use this method to screen for between 4 to 10 of the most common mutations. Note, that if a person belongs to a different ethnic group, PCR must be used to amplify the exons which contain the majority of the CF mutations within that particular ethnic group. Answer A, biochemical testing, is usually used to detect a defective enzyme or a reduced amount of the normal enzyme. It is more economical than DNA testing for detecting carriers or affected individuals. It is employed in the detection of carriers of autosomal recessive diseases such as TaySachs disease, sickle cell anemia and the thalassemias. Answer B, karyotype analysis is used to determine whether an individual’s chromosomes are grossly normal in their number and structure. It can be used to detect trisomies, monosomies, and translocations as well as large inversions and deletions. Answer C, Fluorescence in situ hybridization (FISH), employs a DNA probe that has been labeled with a fluorescent compound to visualize locations on the chromosomes that have homology to the probe. It can be used to detect microdeletions in a gene of interest. Answer E, Western blot analysis is a technique that is used to detect the presence of antibody against specific proteins in serum. It is most used in laboratory medicine as a confirmatory test for HIV infection. QUESTION 5 (3 minutes) The Rh blood group has two alleles, the Rh-positive allele [R] and the Rh-negative allele [r]. Erythroblastosis fetalis (EF) is a hemolytic anemia of newborn infant caused by the transplacental transmission of maternally antibody against Rh-positive blood. Rh-positive babies born to Rhnegative mothers (rr) are at risk for the condition. If the frequency of the Rh-negative allele is 0.3, the frequency of matings at risk for producing a child with EF is: A. B. C. D. E. 0.063 0.082 0.126 0.164 The frequency cannot be calculated from the information given. Answer: B P(mating at risk for erythroblastosis ) =P(Rh+ male x rh- female) = P(RR male x rr female) + P(Rr male x rr female) = p2q2 + (2pq3) This table shows the possible matings relative to the Rh blood groups. The highlighted matings are between Rh positive males and Rh negative females. ♂ p2 (RR) 2pq (Rr) q2 (rr) ♀ p2 (RR) p4 2p3q p2q2 2pq (Rr) 2p3q 4p2q2 2pq3 q2 (rr) p2q2 2pq3 q4 In this case, the frequency of Rh- = r= 0.3; The frequency of Rh+ = R = 0.7 Plugging these numbers into the first formula gives us: (0.7)2(0.3)2 + 2(0.7)(0.3)3 = 0.082 What about the other answers?? Answer A, 0.063, is the probability of a child having erythroblastosis or P(erythroblastosis ) =P(Rh+ male x rh- female) P(Rh+ child) = P(RR male x rr female)P(Rh+ child) + P(Rr male x rr female)P(Rh+ child) = p2q2 + (2pq3)1/2 = p2q2 + pq3 Answers C and D, 0.126 and 0.164, result when the answer to either A or B is multiplied by two. Remember,if the sex of the parent didn’t matter, there were often 2 ways of fulfilling the question. (See problem set XVII, numbers 1 and 5 for more practice.) QUESTION 6 (3 minutes) An RFLP located next to the centromere of chromosome 21 has two haplotypes, A and B. An AA woman and an AB man produce a child with Trisomy 21 that is AAB. The child’s trisomy arose from a failure in which of the following: A. B. C. D. E. nondisjunction in meiosis I of the male nondisjunction in meiosis II of the male nondisjunction in meiosis I of the female nondisjunction in meiosis II of the female The failure cannot be determined from the information provided. Answer: E The RFLP detects a region near the centromere of chromosome 21. The region around the centromere exhibits a phenomenon called crossover suppression. Since genetic exchange cannot happen in this area, the probe is a reliable marker for the individual chromosomes. The male is AB and the woman is AA. The child is AAB. If chromosomes failed to disjoin in Meiosis I of the mother, eggs containing AA would be produced. If a B sperm fertilized this egg, an AAB child would be produced. If sister chromatids failed to disjoin in Meiosis II of the mother, an AA egg would be produced. If a B sperm fertilized this egg, an AAB child would also be produced. If chromosomes failed to disjoin in Meiosis I of the father, sperm containing AB would be produced. If that sperm fertilized any egg from the mother, an AAB child would be produced. In fact, the only meiotic failure that would be unable to produce an AAB child would be a failure in Meiosis II in the male. If sister chromatids failed to disjoin, AA or BB sperm would be produced. Once one of these sperm fertilized one of the mother’s eggs, either an AAA or an ABB child would be produced. QUESTION 7 (3 minutes) A 26 year old woman presents to your clinic during a spontaneous abortion from which she recovers uneventfully. She and her 29 year old husband have been trying to have a child for the last 3 years, during which time the woman has had 5 spontaneous abortions. The karyotype of the aborted fetus contained 46 chromosomes, and all pairs were normal in except for the pairs shown below. Came from the normal parent 2 p 1 1 2 q 5 4 3 2 1 4 3 2 1 1 2 1 2 3 4 1 3 2 3 4 5 Chromosome 7 Mother's Karyotype Father's Karyotype Karyotype of Spontaneously Aborted Fetus Came from the parent with the translocation. Note both of the centromeres are from the 7 chromosome, so Adjacent II disjunction occurred. The event which lead to the child's abnormal karyotype was: A. B. C. D. E. 5 4 p 1 1 q 2 3 2 1 1 2 3 4 1 2 3 4 5 Chromosome 12 Note-In the adjacent I segregation pattern, the cell correctly places homologous centromeres in different cells. In the adjacent II segregation pattern, a breakpoint of the translocation is so close to a centromere that the cell can’t tell which centromere it is looking at. To our eyes, it looks like homologous centromeres went to the same pole at the end of Meiosis I. By looking at the centromeres, you can quickly determine which type of segregation occurred. adjacent I segregation adjacent II segregation alternate segregation a single recombination event within a pericentric inversion a single recombination event within a paracentric inversion Answer B The couple is experiencing infertility and a high rate of spontaneous abortions because the father has a reciprocal translocation between chromosome 7 and chromosome 12. His karyotype is 46, XY, t(7;12) (12qter→12q23::7p13→7qter; 12pter→12q23::7p13→7pter). During meiosis I, the four chromosomes involved in the translocation will pair together in a structure called a quadrivalent, which is pictured below. 12qter 12qter 12q23 7qter 7p13 12pter 12q23 12pter 7p13 7qter 7pter 7pter There are three ways that the quadrivalent can separate during meiosis I. They are called adjacent I, adjacent II and alternate segregation. Adjacent II segregation only occurs when the breakpoint of the translocation is so close to the centromere of a chromosome that the cell cannot distinguish between the centromeres. In this specific case, the fetus received a normal chromosome 7 and a normal chromosome 12 from his mother. From his father, the fetus received a normal chromosome 7, and a chromosome composed of 12qter→12q23:: 7p13→7qter. This gives the fetus a total of three copies of the material from 7p13→7qter, and only one copy of material from 12pter→12q23. Note that the zygote has three copies of the centromere of chromosome 7, the hallmark of adjacent II segregation. Since the zygote is unbalanced, it aborts. What about the other choices? Choice A, Adjacent I segregation produces cells which contain different centromeres but duplication and deletions which lead to spontaneous abortion. In this case, the products of Adjacent I segregation would be 12qter→12q23::7p13→7qter and the normal chromosome 12, or 12pter → 12q23::7p13→7pter and the normal chromosome 7. Choice C, Alternate segregation gives the only outcomes that produce viable progeny. The two products are produced. One contains a normal chromosome 7 and a normal chromosome 12, and the second product the balanced translocation. This cell has all the information, it is just in a unique order. In this case, 12qter→12q23::7p13→7qter and 12pter→12q23::7p13→7pter. Choice D, a recombination event within a pericentric inversion leads to the formation of an acentric fragment and a dicentric bridge. This cell will not complete meiosis. Choice E, a recombination event within a paracentric inversion leads to the chromatids that are duplicated for material on one of the arms and deleted for material on the other arm. QUESTION 8 (3 minutes) Lesch Nyhan syndrome is caused by a lack of the enzyme hypoxanthine-guanine phosphoribosyl transferase. In a certain population, if the frequency of affected males is 0.01, what is the probability that a woman selected at random will carry the trait? A. B. C. D. E. Answer 0.0099 0.0198 0.09 0.18 The probability cannot be determined from the information given. B Lesch Nyhan syndrome is an X-linked recessive trait. A male is hemizygous for the X chromosome, and thus has only one copy of each trait. The frequency of an X-linked recessive in males is thus equal to the frequency of the allele in the population. From this, we know that q= 0.01 and p = .99. A woman has two copies of each gene on the X chromosome, so the equation for Hardy Weinberg equilibrium is the same as for the autosomal traits. In this case, a heterozygous female would occur at a frequency of 2pq or 0.0198. What about the other choices? Choice A, 0.0099, is obtained if the incorrect formula, pq, is used for the frequency of female heterozygotes.. Choice C, 0.09, makes the assumption that the trait is autosomal, and so the frequency of affected males and affected females are equal, and also uses the incorrect formula, pq. Choice D, 0.18, makes the assumption that the trait is autosomal, and so the frequency of affected males and affected females are equal and uses the correct formula, 2pq. Choice E, The probability cannot be determined from the information given. If a person is guessing, s/he might find this attractive, however, if you are given the frequency of one of the alleles, you should be able to calculate the frequency of any of the genotypes. QUESTION 9 (3 minutes) In a certain population, the frequency of the alleles of the ABO blood group are A= 0.1, B = 0.3 and O = 0.6. In the following pedigree from that population, the ABO phenotypes of the family are listed beneath their symbols. What is the probability that III-1 will have type O blood? A O A. B. C. D. E. A A B 0.045 0.060 0.112 0.129 0.133 Answer E P(OO child/B male from the population x A female with pedigree data mating)= This expression reads What is the probability of an OO child given a mating between a male from the population that is phenotype B and a female who has pedigree data that is phenotype A? P(AO/A male) P(AO female/A phenotype) P(OO child)= 2pr (p2 + 2pr ) (2/3)(1/4)= 0.36 (0.09 + 0.36 ) (2/3)(1/4)= (0.8) (0.67) (0.25)= =0.133 Matching Questions QUESTIONS 10-12 (3 minutes each) A. B. C. D. For each of the following patients described, choose the most likely mode of inheritance for the patient’s illness. 10. A 20 year old man myoclonic epilepsy with ragged-red muscle fibers (MERRF) Answer: D MERRF exhibits a mitochondrial inheritance pattern. The hallmark of this pattern is matrilineal inheritance. All of the children of an affected woman will be affected since they receive mitochondrial genes only from the female parent. Affected males do not contribute mitochondria to progeny, so their children will not receive the trait. 11. A 12 year old boy with hemophilia Answer: B Hemophilia is an X-linked recessive disorder that affects the blood clotting pathway. The hallmark of X-linked recessive inheritance is an abundance of affected males and an absence of affected females. Males are hemizygous for the X chromosome, so the phenotype is expressed with only one dose of the gene. Females have two copies of the X chromosome, so they appear phenotypically normal although they may carry the recessive allele. Since a male inherits his X chromosome from his mother, if he is affected, she must carry the trait. Other X-linked recessives include glucose-6-phosphate dehydrogenase deficiency, colorblindness, and Duchenne’s muscular dystrophy. 12. A 28 year old man with Marfan’s syndrome. Answer: A Marfan’s syndrome shows an autosomal dominant inheritance pattern. Since the trait is a dominant, the phenotype can be expressed if the allele is present in one dose. Deleterious autosomal dominants occur at very low frequencies, so affected individuals are almost always heterozygous for the trait. Since the gene is located on an autosome, both male and female progeny can be affected. Common autosomal dominants include Ehlers-Danlos syndrome, neurofibromatosis, Niemann-Pick disease and osteogenesis imperfecta. Answer C shows an autosomal recessive pattern. For the phenotype of the autosomal recessive to be expressed, the recessive allele must be present in two doses. Both male and female children can be affected. The hallmark feature is that unaffected parents have affected children of both sexes. Common autosomal recessives include phenylketonuria, sickle cell anemia, and Tay-Sachs disease. Short Answer Questions QUESTION 13 (12 minutes) The polymerase chain reaction (PCR) has become one of the most powerful means of genetic analysis. a. Explain briefly how it is useful in diagnosing fragile X syndrome (a schematic diagram with appropriate labels will be very helpful) With PCR we can use primers flanking the CGG repeat and amplify the corresponding genomic DNA; from a comparison of the PCR product from normal males with that from males with fragile X syndrome one can see the degree of amplification of the CGG repeat (>> 200 repeats in affected males) b. Describe with a schematic diagram what you would expect to see in a nonpenetrant transmitting male and his daughters. A nonpenetrant transmitting male would yield a PCR product of intermediate size, or closer in size, but larger than that from normal males (in the premutation the number of repeats is in the range of 60 – 200); carrier females would yield two diffuse bands, one from their normal X chromosome and the second showing the premutation c. What do we mean when we speak of the phenomenon of anticipation in case of fragile X, Huntington's disease, and similar diseases? Anticipation means that the severity of the disease gets worse in progressive generations of a pedigree, and the age of onset may also become reduced progressively. d. The Dutch-Belgian Fragile X Consortium has published the following paper : "Fmr1 knockout mice: a model to study fragile X mental retardation" Explain why such a mouse knockout model would not be useful to study the Sherman paradox of the genetic of Fragile X syndrome in humans. The mouse with the fmr1 point mutation may show the symptoms of fragile X (i.e the consequences of missing the FMR1 protein, but this would be due to a “classical” mutation (deletion of an exon), and one would not see the stochastic phenomena associated with trinucleotide expansion. QUESTION 14 (12 minutes) a. Define the phenomenon of imprinting in two or three sentences. Imprinting is the phenomenon that describes the differential expression of certain loci (autosomal) depending on whether the chromosome passed through spermatogenesis or oogenesis. An imprinted gene is expressed only from either the paternal or the maternal (never both) chromosomes. Example: Igf2 is always expressed from the paternal chromosome, never from the maternal chromosome. b. Give a prominent example from the human genetics literature and illustrate diagrammatically how the different symptoms are interpreted. A prominent example from human genetic are the Angelman syndrome and the Prader-Willi syndrome. A small deletion on chromosome 15 can cause Angelman syndrome, if the deletion is on the maternal chromosome, and Prader-Willi syndrome if the deletion is on the paternal chromosome. QUESTION 15 (12 minutes) "Mitochondrial diseases" have become recognized in the past decade, originating from mutations in the mitochondrial genome. They are heritable, but inheritance is not mendelian. a. What is the current explanation (one sentence) for the observation that heteroplasmy in the offspring from a heteroplasmic female is highly variable? There is an apparent bottleneck in the segregation of mtDNA during oogenesis; only a very small number of mtDNAs are selected to be amplified to end up in the oocytes produced b. How would this variability be reflected in the observed symptoms of these children? The symptoms are generally related to the degree of heteroplasmy, that is, the higher the proportion of mutant mtDNAs, the more severe the disease c. How is it possible that an individual suffering from Leber's hereditary optic neuropathy can be homoplasmic for the mutation in the mitochondrial ND4 gene? If the mutation is in a structural gene and causes a missense amino acid substitution, the resulting protein may be partially functional; thus, even when 100% of the protein is mutant, one may have sufficient functional electron transport to support live individuals QUESTION 16 (16 minutes) The Worton team in Toronto identified a DNA fragment from the X chromosome that was also very close to the breakpoint/junction of the X-autosome translocation in the female suffering from Duchenne muscular dystrophy. a. How can we be reasonably certain that this piece of DNA is from the relevant gene (DMD, encoding dystrophin)? The hypothesis is that such females are suffering from DMD because the DMD gene is broken in the translocation; the normal DMD gene on the normal X chromosome is inactive, because the entire normal X chromosome is preferentially inactivated. b. This piece of DNA is used in an analysis of RFLPs in several pedigrees of families in which DMD is observed. With the restriction enzyme Taq1 RFLPs are observed in females with affected children. Draw a pedigree with three generations and several children in each that can be used to illustrate linkage of the locus defined by the RFLP locus and the DMD locus. Also, show a representative schematic Southern blot illustrating the RFLPs in several individuals. c. A lod score analysis between the Taq1 RFLP and the DMD loci shows that θ = 0.0 with a lod score of 3.4. What does that tell us? A recombination frequency of θ = 0.0 indicates that the two loci are very closely linked, i.e. inseparable by recombination; a lod score of 3.4 suggests that the odds are greater than 1000:1 that this value of theta is correct. QUESTION 17 (12 minutes) The following molecules/structures are associated with the phenomenon of X-inactivation in females: XIST RNA, Barr body a. What is their relation to X-inactivation and to each other? What is very special about the XIST RNA? XIST RNA is expressed from the inactive X chromosome and it acts in cis on the chromosome from which it is transcribed to inactivate it; the Barr body represents the inactive X in a highly condensed (inactive) form. It was found that the XIST RNA is tightly associated with the Barr body. XIST RNA is never exported from the nucleus; it has no open reading frame; it is never found on polysomes b. Give a few arguments rationalizing the observation that in females with balanced, reciprocal Xautosome translocations the intact X chromosome is generally inactivated. Such females have a normal X, and an X chromosome carrying an autosomal fragment; if the derivative X chromosome was inactivated, the inactivation may spread into the autosomal region, and hence there would be a deficiency (imbalance) with regard to the autosomal genes. The second derivative chromosome would have a piece of X chromosome on the autosomal fragment; this piece of X chromosome might not be inactivated, since it is removed from the X inactivation center. Hence one would not be achieving dosage compensation with regard to those X-linked genes. QUESTION 18 (8 minutes) Osteogenesis imperfecta type I (OI) is an autosomal dominant that causes a generalized decrease in bone mass (osteopenia) and makes the bones brittle. The disease is often associated with blue sclerae, dental abnormalities and progressive hearing loss. The gene for OI has been cloned and mapped to chromosome 22. However, each family usually has a unique mutation, which would require the sequencing and analysis of more than 5000 base pairs in each of the two genes to identify the exact mutation. For these reasons, the strategy of choice for genetic diagnosis is to trace the inheritance of the disease allele using a RFLP close to the locus. RFLP SC3 is known to map to the q arm of the chromosome 22. It has four haplotypes, A, B, C and D. It is hypothesized that OI is linked to this RFLP. Use the data from this family to answer the following questions. Gen I BD CD II DD Part a B I BC AA AB AC C i III AB AC C i NR NR Part b B i R B I C I R AC C i NR AB B I NR AC C i NR a. What is the linkage phase of II-2? (2 pts) See above b. Write the general Lod score expression for the family of II-2 x II-3. (2pts for determining recombinants versus nonrecombinant progeny. 2pts for writing the general lod expression.) z= log10 (1-θ)5(θ)2 (θ)7 c. The following table is the cumulative Lod values of all the families in the study. State explicitly what the Lod scores indicate about the possibility of linkage between the RFLP and OI. Explain your answer. 0.02 0.05 0.07 0.10 0.12 0.15 0.20 0.30 0.40 θ Lod 3.126 4.000 3.868 3.487 2.673 2.262 1.764 1.566 1.223 The maximum lod score occurs at θ = 0.05; a lod score of 4.0 indicates that there is a 10,000:1 chance of θ = 0.05 is the best estimate for the recombination frequency. By convention genes are considered linked at the highest lod value greater than or equal to 3. QUESTION 19 (12 minutes) Wilson’s disease is a rare autosomal recessive disorder of copper metabolism. The impairment of the normal excretion of copper over time results in toxic accumulations of the metal in liver, brain and other organs. III-1 and III-2 present for genetic counseling. Key: Wilson's Disease Normal a. What is the probability that their first child will have Wilson’s disease? W= normal; w = Wilson’s disease III-1 is not affected with the disease so he received a normal allele from his mother and a w from his father who has the disease. So, III-1 is Ww. III-2 is not affected with the disease so she received a normal allele from her father and a w from her mother who has the disease. So III-2 is Ww. The probability that their first child will be affected is: P(III-is Ww) x P(III-2 is Ww) x P( a ww child is produced) (1) x x (1/4) = (1) ¼ (2 pts) Since many mutant forms of the normal gene have been identified, linkage analysis is the technique of choice for calculating the probability of an affected child. The locus for Wilson’s disease has been mapped to 13q14. An RFLP is located 4 cM from the locus causing Wilson’s disease. The RFLP contains two variable EcoRI sites labeled E1 and E2. Invariant sites (restriction sites that are always present) are labeled E. The following diagram shows the region of genomic DNA with homology to the probe. The results of a Southern blot of the family’s genomic DNA are provided below. Map of the RFLP located 4cM from Wilson's's Disease 10Kb 6 Kb 5 kb probeJ1 E E1 E2 E Haplotype A B C D E1 E2 + + + + - Band Size in KB 10, 6 10, 11 16 21 2pts Note: Highlighted bands are unique to that haplotype. (+ = enzyme cuts; - = enzyme cannot cut) b. Using the restriction map of the RFLP and the data from the Southern blot, determine the RFLP haplotypes for the individuals listed on the Southern blot. (3 points) AA II-1 BC BC DD AC CD CD II-2 II-4 II-5 III-1 III-2 IV-1 D D C C DD C B AA C B C markers 21.0 16.0 11.0 10.0 A 6.0 Use the unique bands to set each person’s haplotype. c. What is the probability that IV-1 will have the disease? Key: Wilson's Disease AA WW BC ww A C W w III-2’s gametes C 0.48 III-1’s gametes C D w CD W D 0.48 W w Normal DD WW BC ww 0.02 C W D 0.02 w A A C A D A C A D W W w W W W W W w C C C C D C C C D w 0.48 w w w W w W w w 0.48 0.02 C C C C D C C C D W 0.02 W w W W W W W w A A C A D A C A D w w w w W w W w w We know that IV-1 is CD, so one of the yellow squares happened. The chance of having Wilson’s Disease and being CD is equal to: P(CD and ww)/ P(CD)= C D w w CD (0.48)(0.02) = 0.0384 (0.25) ...
View Full Document

This note was uploaded on 04/19/2009 for the course BIMM BIMM 110 taught by Professor Mcginnis during the Spring '09 term at UCSD.

Ask a homework question - tutors are online