What is the probability that two parents who are heterozygous for the recessive trait of albinism?
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Answer key to practice problems--1999
500 out of 20 million individuals are homozygous dd (where D = wildtype allele and d = disaccharide intolerance).
Let B = allele for beach-loving; b = bridge-loving
So in the next generation, the frequency of bridge-loving iguanas = q2 =0.08. At this point, the alleles should be at Hardy-Weinberg frequencies, so the subsequent generation will not show a change. This one can be solved only if we make the assumption that everyubody gets to mate, and that all crosses produce equal numbers of progeny. While bridge-loving iguanas are homozygous and will give rise to bridge-loving iguanas only, beach-loving iguanas consist of homozygotes as well as heterozygotes. So we can set up a table as before, but this time only for frequencies of alleles B and b within the pool of beach-loving iguanas:
Beach-loving iguanas from these crosses = (0.48)+(0.192)+(0.096) = 0.768. Bridge-loving iguanas = 1 - 0.768 = 0.232. (If we didn't make the assumption stated at the beginning, we'd just be able to make the general conclusion that homozygosity is expected increase while heterozygosity would decrease.)
Among females, the distribution of genotype frequencies is the usual Hardy-Weinberg frequencies -- homozygous dominant = p2, heterozygotes = 2pq, homozygous recessive = q2 (where p = frequency of dominant allele, q = frequency of recessive allele). But in males, there are no heterozygotes for X-linked traits -- males are hemizygous for such traits. Therefore, among males, p = frequency of the dominant phenotype; q = frequency of recessive
phenotype.
1-1998 The phenotype of a (recessive) maternal effect mutation is that females homozygous for the mutation have offspring that fail to develop normally regardless of their genotypes. If m is the mutant allele, mm (female) x any genotype (male) should give abnormal progeny that fail to develop correctly. In contrast, the mm genotype in males does not affect the progeny: mm(male) x M_ females will give normal, viable progeny. [Since there is no directly observable phenotype of mm females (other than their failure to produce normal offspring), one will have to use other markers to follow the mutagenized chromosomes. For instance, one can mutagenize a stock that is heterozygous for one (or more) known recessive markers on each chromosome, mate these with non-mutagenized strains not carrying the recessive marker alleles, and cross the F1 progeny with each other. The F2 progeny of interest will be those displaying the recessive marker traits--since the only source of the recessive allele is the lone homolog (for each phenotype) that was in the mutagenized animals, we will know that we have two copies of a chromosome that had undergone mutagenesis--and therefore potentially homozygous for a new mutation. In real life, one would also use balancer chromosomes to prevent crossovers in the mutagenized chromosomes.]
4-1998 For the sake of simplicity, I shall designate the allele frequencies as: p(Si) = ap(Sy) = bp(Sg) = cThe distribution of genotypes then is: (a+b+c)2 = a2 + 2ab + 2ac + b2 + 2bc + c2 = 1 icky yucky grossThe frequencies of icky and yucky slugs are compound terms and cannot be calculated directly. However, the frequency of gross slugs = 0.2; c2 = 0.2, therefore c = 0.45 But b2 + 2bc = 0.3 (= phenotype frequency of yucky slugs)Substituting for the value of c in this equation, we getb2 + 0.9b - 0.3 = 0Solving for b, we get b = 0.26. a = 1 - (b + c) = 0.29. p(Si) = 0.29p(Sy) = 0.26p(Sg) = 0.45
The math Case 1: probability of a chance match = (0.01)(0.02)(0.003)(0.01)(0.07)(0.04)(0.13)(0.08)(0.04)(0.05)32 = 1.1 x 10-14 -- i.e., the probability of getting this combination of alleles just by chance is about 1 in 100 trillion. Case 2: probability of a chance match = (0.2)(0.4)(0.15)(0.35)(0.4)(0.3)(0.3)(0.6)(0.2)(0.3)32 = 1.7 x 10-4 -- i.e., a little over 1 in 6000. (The factor of 32 comes in because there are two ways
of getting each heterozygous combination, and 5 loci we're looking at = 25 = 32.) We'll see more of this factoring-in-2 business when we get to population genetics.)
Remember that normal diploid cells have two copies of the gene for Enzyme E (Gene E) and two copies of the gene for Enzyme Z (Gene Z). So, assuming that the amount of enzyme in the cell scales linearly with the gene copy number, each copy of Gene E contributes 30 units of enzyme E activity (so a diploid produces 60 units of Enzyme E), and each copy of Gene Z contributes 50 units of Enzyme Z activity. So a cell line that has a duplication of a gene should produce three copies' worth
of enzyme. For Enzyme E, the duplication should result in cells that produce ~90 units, and for Enzyme Z, a duplication of the gene should result in ~150 units. To find the location of Gene E, we look for cell lines that produce ~90 units of Enzyme E, and ask, what's common between these duplications. We see from the table that cell lines 1, 5, and 6 all produce ~90 units of Enzyme E (while the other cell lines produce the normal ~60 units). So these three cell lines must have duplications
of Gene E. The band that is common to these three duplicaitons is band 2 -- that must be the location of Gene E. For Enzyme Z, cell lines 2 through 6 all produce ~150 units instead of the standard 100 units. The band common to the duplications in these lines is band 5; Gene Z must be located there.
With respect to the disease, the boy must be homozygous recessive (because achondroplasia is dominant). If A = achondroplasia and a = unaffected, the boy is aa. With respect to the polymorphic locus, one allele has 12 repeats of CA, while the other has 7 repeats--so his genotype for the polymorphic site is 7,12. (Or 12,7.) Therefore, his overall genotype for these two loci is
aa 7,12.
The primers are : 5'-TGCTCTGGAT-3' and 5'-TCCGAGAGCC-3', which correspond to the yellow, boxed segments (immediately flankng the greyed segment) below: Note added 10/26/99: The way the question is worded, it is actually possible to amplify an even smaller fragment, by choosing primers within the grayed segment as shown below: In this case, only the grayed segment would be amplified, giving a product length of 26 bp.
(from 1998) The Lod score graph tells us that the pedigree data favor a map distance of 5 cM between Gene 1 and PS1; a map distance of 15 cM between Gene 1 and PS2; a map distance of 10 cM between Gene 1 and PS3, etc. A map that is consistent with these interpretations is: Note: The answer to Q. 5 has been corrected 10/19/99.
Putting this information together -- A/a, D/d and F/f are in the same linkage group; B/b and E/e are in a separate linkage group. The linkage relationships can be depicted as:
The parental genotypes are TTFF x ttff to give TtFf. Therefore, the parental genotypes for gametes made by the F1 plants are TF and tf. If the two loci are unlinked, we expect the four progeny phenotypes (TF, Tf, tF, and tf) in equal proportions. Because there are 1000 proteny total, we expect
250 of each phenotype if the loci are unlinked. If the two loci are linked at at map distance of 44 cM, we expect 44% of the gametes to be recombinant -- i.e., 44% of the progeny should show the recombinant (non-parental) phenotype. As shown above, the parental types are TF and tf, so we expect 44% of the progeny to add up to Tf and tF, or 22% each. The parental types
then should be 56% of the progeny = 28% each. So for 1000 progeny, we expect 280 each of TF and tf, and 220 each of Tf and tF. Clearly, the observed progeny numbers don't match either scenario. So let's do a chi-square analysis on the two data sets, for the two sets of expectations, and see if we can find statistical evidence against either model. Scenario 1 -- the loci are unlinked The corresponding P value is just over 0.05 -- just above the standard cutoff for rejecting the null hypothesis (that the deviation from expected is just due to chance).
Scenario 2 -- the loci are linked at 44 cM
Again, the corresponding P value is just over 0.05. What does this mean for deciding between the two modes of inheritance? The statistical analysis tells that the data are consistent (just barely) with either model -- so we cannot decide between the two models based on this statistical test. At least two approaches are possible to settle the question. One is simply to collect more data (repeat the crosses, count a lot more progeny) and repeat the statistical analysis in the hopes that one hypothesis or the other can be rejected with more data. However, if T/t and F/f are linked, it should be possible to find genes in the interval between them that are linked to both. That way, we'd be working at smaller map distances, and thereby have a better shot at establishing linkage.
The flaw is that the F1 progeny, although heterozygous for sneezy and jumpy, are homozygous for itchy. So while recombination between sneezy and jumpy can be detected, there's no way to detect recombination involving itchy. ... no change in genotype (ijs and i++ giving i++ and ijs) Note: i = itchy, j = jumpy, s = scratchy; only one chromatid of each homolog is shown He should be using a fully
heterozygous (ijs/+++ in any cis/trans configuration) and a homozygous recessive (ijs/ijs) for his mapping cross. Assuming that we are starting with the dominant alleles in cis in the heterozygote (i.e., +++/ijs), then parental, double-crossover, and single-crossover products can be predicted as follows:
Finding the correct gene order H = Hairy, P = purple, T = Thorny; and lower case denotes the recessive phenotypes. The parental non-crossover (NCO) allele combinations are Hpt and hPT (these being the most abundant progeny phenotypes), while the double-crossover (DCO) classes are
HPT and hpt. To find the correct gene order, we start with the known NCO types, and see if a double crossover yields the known DCO types. If it doesn't, the order must be wrong; we try a different gene order (the critical information is the gene in the middle). Trial and error (trying each of the three genes in the middle) establishes that H must be the middle gene: Now we can start calculating map distances:
The important thing to remember is that in order to map the genes, we need to be able to detect recombination, and that in order to detect recombination, one of the parents has to be fully heterozygous. Here, the genes are on the X chromosome -- so that parent by default has to be the female (the male only has one X -- no recombination there). There are a couple of ways of setting this up. One option is to make the female heterozygous, and to have recessive alleles on the
male's X chromosome. Then the males and females would consist. To generate heterozygous females, we could cross homozygous dominant females (+++/+++) with recessive males (abc/Y); the females in the resulting progeny would be heterozygotes. When these females are crossed with abc/Y males, the progeny (males and females) would show the non-recombinant phenotypes (abc and +++) as well as the 6 recombinant types: a++ and +bc,
ab+ and ++c, a+c and +b+. A different option is to cross the heterozygous females with males showing the dominant phenotypes (+++/Y). Then the female progeny would all show the dominant phenotypes, and would be ignored; the male progeny would get the single X from the female, and show the same parental and recombinant phenotypes listed above. For an example -- see
Question 1998-2 in Questions from yesteryear.
1998-1
1998-2
This problem is easily solved. Human chromosomes present in cell lines that do not have the insulin sequence can be eliminated from our list of possible candidates. Therefore, any chromosome that is found in cell lines D, E, or F can simply be crossed out from the list of possibilities (eliminated candidates shown below as colored-out boxes).
Of the remaining candidate chromosomes, the only one that is present in cell lines A, B, and C is chromosome 11. Therefore, the insulin gene must be located on chromosome 11. 1997-4
We know that the final genotype has to be gaY. Therefore, at anaphase I, the Y chromosome has to segregate with the homologs carrying the recessive alleles, as diagrammed: Note: In the interests of simplicity, crossing over has been ignored here. Also, the relative sizes of chromosomes and locations of genes is fictitious.
The cross here is AABbDdee x AaBbddEe. We are asked to calculate what fraction of the progeny will have the phenotype ABde. Because these are independently assorting traits, we can calculate the fraction of progeny that will have phenotype A, then the fraction that will have phenotype B, etc., then multiply these fractions to get the fraction that has all the desired phenotypes. Thus -- Therefore, the fraction of progeny expected to be phenotype ABde is
(1)(3/4)(1/2)(1/2) = 3/16. Here, we have to find the fraction of progeny that will have the genotype AabbddEe. Using the same logic as above-- Therefore, the fraction of progeny expected to be genotype AabbddEe is (1/2)(1/4)(1/2)(1/2) = 1/32.
This one is a little tricky. A common mistake is to misinterpret the question to think that the first two children are boys -- when in fact, all we know is that at least two children (in any order) are boys. If the sex of the children is written in order of birth as B (for boy) or G (girl), the possible 3-children families with at least 2 boys are: Only one of the four possible sets has all three children being boys -- so if you know that two of the children are boys, the probability that all three are boys is 1/4.
We use the binomial distribution to solve this one. Because this is a recessive disorder, and both parents are heterozygous, the probability of an affected child is 1/4. Therefore, let a = probability of unaffected child = 3/4, and b = probability of affected child = 1/4. The equation then is For a family with exactly 2 affected and 4 unaffected children, we use the term 15a4b2 (the exponents indicating the number of a=unaffected and b=affected children). Substituting the probabilities of unaffected and affected children, we get: p(2 affected
children) = 15a4b2 = 15(3/4)4(1/4)2 = 1215/4096 = 0.297. But an easier way is to find the probability of less than
two affected children, then subtract that value from 1 -- (Try it. The longer expression 15a4b2 +
20a3b3 + 15a2b4 + 6ab5 + b6 will give the same result.)
This being a dihybrid cross, we expect a 9:3:3:1 ratio of tall purple : tall white : short purple: short white. For 3200 progeny, the expected numbers are: For df = 3 (i.e,. three degrees of freedom) the chi-square = 2.332, the P value is just over 0.5, which is well above the standard cut-off of 0.05 for rejection of the null hypothesis. Therefore, the null hypothesis (that the deviation from expected values is just due to chance) cannot be rejected.
What are the possibilities here? Possibility # 1: the cross was homozygous purple x homozygous purple; there should be no white-flower progeny Possibility #2: the cross was heterozygote x heterozygote; 1/4 of the progeny should make white flowers. If the seed merchant picks just one seed at random and grows it up, and it makes white flowers -- she knows it must have been a heterzygote x heterozygote
cross. However, if she picks one seed, and it makes a purple-flower plant -- can she then say that it must have been a homozygote x homozygote cross? No, because even in a heterozygote x heterozygote cross, 3/4 of the progeny will be purple, so she has a 3/4 chance of picking a purple progeny even if white progeny are present--i.e., she has a 1/4 (=0.25) probability of missing a white progeny. Suppose she picks two seeds? Then the probability that both will be purple (if it was indeed a
dihybrid cross) = (3/4)(3/4) = 9/16; the probability that she has missed a white progeny plant has dropped to 7/16 = 0.4375. So that's the question -- how many seeds should she sample if she wants the probability of accidentally missing a white-flower seed to drop below 2%. In other words, she needs to sample n seeds such that (3/4)n = 0.02 or, n(log(0.75)) = log(0.02) n = 13.6 So if she samples 14 seeds and they all grow up to make purple flowers, there is < 2% probability that white flower seeds are present but missed just due to chance.
Answers to selections from 1998 1998-1 (i) The disease is probably not autosomal recessive--there are several instances where people marrying into the family have affected children; the people marrying in would all have to be heterozygotes, an improbably scenario. (ii) The pedigree is fully consistent with autosomal dominant where I-1 is heterozygous and 1-2 is homozygous normal, as is everyone marrying into the family. (iii) X-linked recessive can be ruled out, because affected females have unaffected fathers (e.g., II-1, IV-3). (iv) X-linked dominant can be ruled out also, because affected men have unaffected daughters (who would inherit the X chromosome carrying the dominant disease allele from the father).--e.g., II-5. (v, vii) Males and females are affected, so the disease is not Y-linked or sex-limited. (vi) With sex-influenced inheritance, there are two possibilities--dominant in males and recessive in females, or dominant in females and recessive in males. Affected women have unaffected sons (e.g., I-1 and II-3), so it cannot be recessive in women and dominant in men. Likewise, affected men have unaffected daughters (e.g., II-5 and III-6) so it cannot be dominant in women and recessive in men. Thus, the mode of inheritance that best explains the observed pedigree is autosomal dominant. 1998-2The disease skips generations, so it is not dominant. The disease being rare, it is unlikely to be autosomal recessive--it would require heterozygotes marrying into the family on at least two occasions. Males and females are affected, so it is not Y-linked or sex-limited. It cannot be sex-influenced, because unaffected parents have affected children. It cannot be X-linked recessive, because an affected daughter has an unaffected father (from whom she got an X). That leaves us with either the rare possibility of heterozygotes marrying in (for autosomal recessive), or some aberrant event, or some mode of inheritance we haven't considered yet. 1998-3 As described in lecture (refer to the part on evidence for random segregation of homologs in meiosis), meiosis in the exceptional females (XXY, homozygous for the X-linked white allele) can give four kinds of gametes because the two X chromosomes can pair up during synapsis, or an X and a Y--in which case the lone X could segregate either with the other X or with the Y. Some of these eggs can give rise to fertile red-eyed males and white-eyed females, the secondary exceptions. NOTE: The grid above shows only the kinds of progeny that can be formed, not the relative numbers. Because synapsis of the two X chromosomes is more probable than synapsis of an X with a Y, the "Y is unpaired" outcome of meiosis I (see the diagram above) is more probable than the "X is unpaired" outcome. Therefore, gamete types 1 and 2 are much more abundant than gamete types 3 and 4, and the progeny numbers are skewed accordingly. 1998-4 Because this is a heterozygote x heterozygote cross (normal = dominant, albino = recessive), we expect to see normal and albino children in 3:1 ratio-- i.e., the probability of a normal child is 3/4, and the probability of an albino child is 1/4. (a) The probability of the outcome described = (3/4)(3/4)(1/4)(1/4)(1/4) = 9/1024 (b) The probability of 2 normal and 3 albino children in any order can be calculated using binomial expansion. Let a = p(albino) = 1/4 and b = p(normal) = 3/4; since there are five children, the equation to use is: The term representing the probability of 3 albino and 2 normal children is 10a3b2. Substituting the values of a and b, we get: p(3 albino, 2 normal) =
(c) The probability that all five will be normal is:
(d) p(at least one albino) = 1 - p(no albino) =
Answers to selections from 1998 The simplest approach is a trial-and-error method: interpret each cross one at a time, and see if your interpretation is consistent with the interpretation of the previous crosses. To begin with, it is clear that there are three phenotypes, so just for simplicity, I am going to assign them 3 allele designations (R, B, W, for Red, Blue, and White) and assume that they are alleles of the same determinant. I may have to revise this initial hypothesis later on--e.g., this may be a case of incomplete dominance between two alleles--but at least for starters, I'm going to assume simple dominant/recessive interactions. Cross (a) -- Red #1 selfed -- yields a 3:1 ratio of red and blue-flowered plants in the progeny. This looks like a typical heterozygous F1 cross, with R being dominant and B recessive. So I'm tentatively assigning Red #1 a genotype of RW. Cross (b) -- Red #2 selfed -- similarly suggests that R is dominant over W; the genotype would be RW. Cross (c) -- Blue selfed -- gives a 3:1 ratio of blue:white; blue must be dominant over white and the genotype of the blue-flowered plant must be BW. At this point, we have a hypothesis for all of the genotypes: Red #1 = RBRed #2 = RWBlue = BWWhite = WW (because it is recessive to both others)We are now in a position to predict the results of the remaining crosses, and seeing if our predictions are met. Cross (d) -- Red #1 x Red #2 = RB x RW:
Cross (e) -- Red #1 x Blue -- should be RB x BW, which should give a 1:1 ratio of red:blue (draw Punnett squares if you're uncertain about this). Again, that's what we see. Cross (f) -- BW x WW should give 1:1 blue and white Cross (g) -- WW x WW gives only white-flowered progeny. So our initial hypothesis appears to be sound as far as we can tell from the data provided. We can predict the results of cross (h): Red #2 x blue = RW x BW:
AO x BO What is the probability that heterozygous parents will have a kid with albinism?II. For two heterozygote parents (Aa), 1/4 of all offspring would be expected to show the recessive trait of albinism.
What is the probability of two heterozygous parents having a recessive offspring?What is the probability of a heterozygous offspring? There is a 50% x 50% = 25% probability that both of the alleles of the offspring are dominant. There is a 50% x 50% = 25% probability that both alleles of the offspring's are recessive.
What is the probability for two parents who are both heterozygous to have a child who exhibits the dominant trait?We know that when two people who are both heterozygous for a simple Mendelian autosomal gene alpha have a child, the probability that the child will show the dominant phenotype is 3/4.
What ratio do you get for two trait two heterozygous parents?Crossing of two heterozygous individuals will result in predictable ratios for both genotype and phenotype in the offspring. The expected phenotypic ratio of crossing heterozygous parents would be 9:3:3:1.
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