7.03 Genetics: Answers to Problem Set #1 (1993)

a) In order to do a complementation mutations strain are dominant to a wild-type test, or recessive. haploid strain one must first determine This is done by crossing to produce a diploid, whether each haploid heterozygous cannot grow without arginine, can see that all the mutations carry a dominant strain are meaningful. mutation, mutation. then the mutation mutant strain. resulting diploid can grow without arginine, then the mutation is recessive. diploid the is dominant. If the If the From the table, are recessive, with the exceDtion of strain 5, which must Therefore, One cannot none of the complementation conduct a complementation tests with this test with a dominant since a diploid resulting from a cross between the dominant strain and any other recessiy_ mutant strain will not be able to grow without arginine. b) If two mutations are in the same gene, or complementation group, then the diploid generated by crossing the two mutant haploid strains will not be able to grow without arginine. / This is because the diploid has two mutant copies of the same gene. ._ "_" This situation ' t' is represented If two mutations diploid generated "complement which the mutation as a "-" in the chart. are in different genes, or complementation Will be able to grow without " each other in the diploid in the other we strain arginine. by providing lies. groups, then the Thus, the two mutants a wild-type copy of the gene in "" " '4 ._4" _..- This situation is represented by a "+" It follows that strains 3,6,8,10 in the chart. comprise one complementaion strains 4,7,9 comprise another, separate, complementation group, whereas group. Strain 5 cannot be placed in a group, since it carries a dominant mutation (see part a.) It cannot be determined if strains 1 and 2 are in the same, or different compiementation groups (see part c.) However, neither are in the same group as either of the two other complementation groups. c) From the last result, we can now determine in the same complementation Therefore, there 3.6.8.10 1 group group, which are 3 complementation 4.7.9 1 group that the mutations is distinct groups, with in strains from the other strain 1 and 2 are two groups. 5 not determined. 1.2 5 1 group undetermined 2 Helpful hint: When a mutant it to the wild type strain. is first identified If the progeny usually the first thing to do is to cross are all wild type then the mutation is recessive and the genotype of the first isolate must have been m/m. If some of the progeny are mutant then the mutation is dominant. It is reasonable to assume that the first animal with a dominant A mutation which is identified is a heterozygote First note that since half the F1 progeny is dominant. The cross is +/M (mice with Six digits). (M/M). x +/+ exhibit male (+/M) F2. 1/4 of the F2 progeny and a six-toed of the F2 mouse mice which M/M. if +/M. Conversely mice are crossed M2/M2 mice with normal are co-dominant of progeny that the F2 mouse were M1/M1 four phenotypic normal should mice should to generate a genes then the genotype should 9/16 will have six toes on all paws (M1/any, paws M2/any). mice are the most important of co-dominant for the genotypes When two F1 1/4 hind paws (see of the F1 mice is +/M1, +/+ and +/+, M2/M2. classes of progeny of be recovered! (+/+, In this case when two F1 be observed (see table below). +/+), will have six toes on 3/16 will have six toes on the hind paws See table below in this test mice with six front paws and normal the front paws (M1/any, rules out the hypothesis are be observed: hind paws, 1/2 M1/M2 will have normal Thus, the normal F2 mice and select male and and M2/M2. Note that 1/16 of the progeny cross. six-toed from this cross then of these test crosses mice with six-toed were M1/M1, +/+), between alleles of the same gene then the genotype front paws and six-toed are in separate while their parents mice are crossed Take several mice are recovered the 1/2 of the F2 will be To distinguish to each other three classes of F2 progeny No completely If the mutations paws, and cross them to each other while their parents digits on aI1 paws, and 1/4 M1/M1 top table below). mutant strain. be M1/M2, M1/M2 are M/M mice) and two +/M from the F1 to generate if a large number Based on the results mouse If the two mutations the F1 mice should +/M2, a test cross. If any normal you have deduced six-toed (+/M) with six toes. which all have six toes then it is highly probable true breeding B to perform to wild type mice (+/+). cross has the genotype female female the mutation or homozygous from this cross will be +/+ with normal these last two classes it is necessary generated phenotype and it gives two +/+ (normal +/M with six toes, and 1/4 of the F2 will be M/M the genotype the mutant We need to get a strain that is true breeding, Mate a six-toed cross them +/M. Normal 3/16 (+/+, M2/any), mice are recovered from this class of F2 mice as their existence alleles of the same gene. that could result from this cross. and Ifthe mutations are co-dominant allelesof the same gene: gametes M1 M1 M1/M1 (6 toes on front paws) M1/M2 (6 toes on all paws) M2 M1/M2 (6 toes on all paws) M2/M2 (6 toes on rear paws) If mutations M2 are in two different gametes genes then the possible +, + progeny +,M2 are as follows: M1, + M1, M2 +,+ +/+, +/+ +/+, +/M2 +/M1, +/+ +/M1, +/M2 +,M2 +/+, M2/+ +/+, M2/M2 +/M1, M2/+ +/M1, M2/M2 MI,+ M1/+, +/+ M1/+, +/M2 M1/M1, +/+ M1/M1, +/M2 M1,M2 M1/+, M2/+ M1/+, M2/M2 M1/M1, M2/+ M1/M1, M2/M2 genotype + / +, + / + +/+, +/M2 +/M1, +/+ all others C +/+, M2/+ M1/+, +/+ Incomplete exhibit have a normal +/+, M2/M2 M1/M1, +/+ penetrance the phenotype means associated number will be recovered from the F1 cross in either case. (.05)x(3/4) If the mutations for the incomplete penetrance) way is to test the genotype mice back to wild type mice. progeny mutations from these test crosses are in different all wild type progeny. mice through the case[!! 3 3 a M1/M1 most other mice of this genotype of the progeny of the phenotypically If the mutations One way to distinguish whether wild type these two genes then 1/4 or 25% (plus will be phenotypically normal. wild type F2 mice by crossing are co-dominant Note that you will need to put a number A these then all but 5% of the genes then some of the mice put through between might then only 5% of the will have six toes on one set of paws or the other. a test cross to distinguish will will have six genes because If the alleles are co-dominant are in different mouse in part b to determine alleles of the same gene or in separate be wild type. ¢'_o that carry a mutation For example, the test described is to cross the F1 to the wild type. mice should second with that mutation. This confounds M1 and M2 are co-dominant models that not all of the animals of toes even though toes on their front paws. progeny phenotype normal paws 6 toes on hind paws 6 toes on front paws 6 toes on all paws the test cross will give of phenotypically these two models. If the Think about normal why this is Answers to 7.03 problem set #1 3a. The hypothesis we will test is that the inheritence pattern of children II-1 to II-5 is that expected if the disorder is inherited in an autosomal recessive manner. We will test this idea using the chi squared test. All that this test can tell us is how likely it is the observed data fit the expected pattern by chance. If the observed data are very unlikely to fit the expected pattern by chance, then we can reject our hypothesis, the expected pattern, as being improbable. Thus we can never prove anything with this test, but we can show that one possibility is highly unlikely. What is the expected pattern? If the disorder is autosomal recessive and some of the II generation have the disorder, then both the mother (I-2) and father (I-1) must be carriers of the disorder allele. In this case we would expect a 3:1 ratio of heMthy children to affected children, both for male and female offspring. Since we can not classify our observed healthy children into homozygous, DD, or heterozygous, Dd, we will lump them all together. Calculating healthy disordered chi squared: Expected ratio .750 .250 1.000 Expected ratio for five children 3.75 1.25 5.00 LooMng up the p value: Because we have divided our expected the degrees of freedom= 2-1= 1. Observed ratio 3 2 5 frequencies (O-E) 2 .5625 ,5625 (O-E)2/E .15 .45 X2=0.600 into two classes For one degree of freedom a chi squared value of 0.600 gives a p value of between 0.5 and 0.1. There is between a 50% and 10% chance that the observed data fit the expected pattern by chance. Therefore we can not reject the hypothesis as being improbable. We can not draw a conclusion about the mode of inheritence based only on this family. Answers to 7.03 problem set #1 3b. To solve this problem let us assume that the observed data are consistanr with X-linked in.heritence and that the expected data is for autosomal recessive inheritence. I am going to assume that because the disorder is rare the case of disordered parents having children will nor happen frequendy and we can ignore those cases in our calculations. male, healthy male, disordered female, healthy female, disordered Expected ratio .375 .125 .375 .125 1.000 Observed ratio .250 .250 .500 .000 1.000 The number of observed healthy males is equal to N, the total number of observed individuals times the frequency of the group, N(.250). By this method we can calculate all the O and E values from all the groups. Ca.lculaflng chi squared: In order for the results to be significant at the p=.05 level with our three degrees of freedom, then chi squared must be equal to or greater than 7.815. Xa_ ",.ttr< .)1%%cN} ,) ,v_) 2).H 2"( h;_ So we need to examine at least 24 children before we can get enough data to reject r.he autosomal recessive hypothesis, The problem can also be solved with an alternative _{a._'ol _". E_(._ __ form of chi squared. _ t_(._._ _ r_(._)_3 _ _/ : 6n) N ) z3.q . .4-I_-_- _.'1 h;._. Answers to 7.03problem set #1 3c. The first insight you need to make is that you are going to apply bayes theorum to the mother of the granddaughter (II-5) and not to the granddaughter herseif. The question is really- what is the probablility that 11-5is a carrier given that she had two normal sons. The probability that the granddaughter, III-2, is a carrier is then 1/2 of the chance that her mother is a carrier. This is true because the granddaughter has a 1/2 chance of receiving the disordered X from her carrier morn. Bayes theorum a/low one to calculate the probability an observed effect, probability of a given b, p(a/b) of a cause based on In this case a= probability that II-5 is a carrier given b= she has two healthy sons. p(b/a)= probability that II-5 has two healthy sons given that she is a carrier. This is (1/2)(1/2)= 1/4 p(a)= probability that II-5 is a carrier. Assuming X-linked recessive inheritence, II-5 has a 1/2 chance of being a carrier. She definitely got a wild-type X chromosome from her father and had a 50-50 chance of getting the disorder X chromosome from her mother. p(b/not a)= probability that II-5 has two healthy sons given that she is not a carrier. If she is not a carrier than there is a 100% chance her sons will not have the disorder. p(not a)= the probability that II-5 is not a carrier. This is 1-p(a)=l/2. PLUG IN THE NUMBERS: p(aJb) = p_b/a_ p(a) p(b/a) p(a) +p(b/not a) p(not a) = (1/4_(1/2) (1/4)(1/2)+ = 1/5 (1)(1/2) So if II-5 has a 1/5 chance of being a carrier, given that she had two healthy sons, her daughter, III-2, has a (1/5)(1/2)= 1/10 chance of being a carrier. PROBLEM 4 a) Under the first hypothesis that the disease allele is recessive, the affected individuals are of the genotype ss (homozygote mutants). We know that. 18% of the population are affected, this means that q2= 0.0018 (Remember that ss= q2, Ss=2pq and SS=p2). We can now calculate q which is equal to (0.0018) 1/2, q= 0.042 and p=l-q, p---0.958. 2pq which is the number of heterozygotes the sicklemic should be equal to 2x0.042x0.958= Weinberg equilibrium the heterozygotes population carriers of the disease and in this case 0.08. If the population were in Hardy- then 8% should be heterozygotes for the sickle cell allele. Since have a distinctive phenotype (sicklemic) we can conclude that the population which is present in 8% of the is in equilibrium. b) Under the second hypothesis that the disease allele is dominant and incompletely penetrant we can calculate the frequency of the disease allele by writing the following equation: q = f (A) = f (AA) + 1/2 f (Aa) Since q is a very small number the expression f(AA) = q2 ,which are the homozygotes mutants, is negligible compared to 1/2 f (Aa) and we can write: q = 1/2 f (Aa) = the sum of sicklemic and sickle cell individuals = 1/2 0.0818 = 0.041 In this problem you can't say anything about Hardy-Weinberg equilibrium since you can't distinguish in the sickle cell population which ones are heterozygotes and which ones are homozygotes for the sickle allele. This problem can be solved in a number of ways. As long as you justify your assumptions and your reasoning is correct you are going to get full credit. c) Under the hypothesis that the disease is dominant and incompletely penetrant you want to know what fraction of the individuals with sickle cell disease have one parent sicklemic (Ss) or both parents sicldemic (we are going to assume that the parents of the individuals with sickle cell disease have the genotype Ss, the homozygotes for the disease axe a very small fraction). Ss (2pq)x ss (p2) Y The probability of a mating between a heterozygote (2pq)Ss x (2pq)Ss Y and a homozygote wild type is equal to 2pq x p2 x 2 You multiply times two because there are two possibilities to get this mating -- the mother is heterozygote and the father wildtype or that the mother is wild type and the father heterozygote for the sick allele. You then have to multiply the possibility of the mating times the possibility that the mating gives you sickle cell progeny which is 1/2. The probability of having one parent sicklemic is 2pqxp2x2x 1/2= 2p3q The probability of mating between two heterozygotes is 2pq x 2pq. The probability this mating wiI1 give rise to sickle progeny is 3/4. The probability of having both parents sicklemic is 2pq x 2pq x 3/4 = 3p2q 2 The ratio is 2p3q / 3 p2q2 = 2p/3q = 2 x 0.958 / 3 x 0.042 = 15.2 Approximately sicklemic. 1/16 of the sickle cell individuals are going to have both parents that "4 - :" '= % • ,it" ... " ° . • •' "J.: i" d) Giventha_star cell anemia is X-linked, we need to determine how to represent the affected males in the population. If .04, or 1125 of the total X chromosomes in the population carry the star cell allele, and 213 of the X's are in females whereas 1/3 are in males, we can set up an equation as follows: let n=number of individuals 1/2n + 2(1/2n) Therefore, = number of X chromosomes in population = 3/2n 1/25(3/2n) 1/3(1/25)(3/2n) in the population. = number of X's carrying star cell allele, and = number of X's carrying the star cell allele in males = 1/50n. This number divided by the total number of males in the population (n/2), gives the number of males with the star-cell disease, n/50 = 1/25 = q !!TTT n/2 So, we can represent males with the disease Now, the ratio of star cell to "star[emic" merely as q. individuals is represented by: q + q2/2pq .04 + .0016 / .0768 = 0.54, which is roughly a 1:2 ratio. e) Ratio of star cell males to star cell females: q/q2 = .04/.0016 f) Ratio of star cell males to females q/50 q2/50 roughly a 1:2 ratio. + 2pq/50 = = 25:1. given incomplete penetrance: .0008 .000032 + .001536 = 0.51, which is =