PROBLEM SETChapter 6 -- Linkage and Mapping in EukaryotesP. 134, #1. A homozygous groucho fly (gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality). The F1 females are testcrossed, with the following offspring produced: groucho 518 rough 471 groucho, rough 6 wildtype 5 ------- 1,000a) What is the linkage arrangement of these loci?To determine the linkage arrangement of these two loci we need to calculate the map distance between them. The two classes in highest frequency are the non-recombinant classes and the two classes in lowest frequency are the recombinant classes. Thus groucho, rough and wildtype are recombinants. To calculate the map distance, we divide the total number of recombinants by the total number of flies and then multiply by 100. 11 map distance = ------ x 100 = 1.1 m.u. or 1.1 cM 1000b) What offspring would result if the F1 dihybrids were crossed among themselves instead of being testcrossed?This is a little complicated because we need to set-up a Punnett square but the four gametic genotypes will not have the same probability of occurrence of 0.25 each, as would expect if we had independent assortment. The first thing then is to estimate the probability of occurrence for each gametic genotype. To do this we must remember that a map distance of 1.1 m.u. means that a recombinant gamete will occur with a probability of 0.011. Thus the two recombinant gametes (gro,roand+,+) will occur with a combined frequency of 0.011, which when divided evenly between them means that the frequency of occurrence forgro,rois 0.0055 and that the frequency for+,+is also 0.0055. The combined frequency of occurrence for the non-recombinant gametes (gro,+and+,ro) is 0.989 (1 - 0.011), which when evenly divided between them means that the frequency of occurrence forgro,+is 0.4945 and that the frequency of occurrence for+,rois 0.4945. The other thing to remember is that in fruit flies, males do not crossover, thus one of the dihybrids (the male) will not produce any recombinant gametes, and our Punnett square is reduced to a Punnett rectangle.

By summing the frequency of occurrence for each cell in the Punnett Square, we can determine the expected frequency for each phenotype in the F2.groucho gro gro, + + 0.24755 gro gro, ro + 0.00275 ------------------ total 0.2500

rough + +, ro ro 0.24755 + gro, ro ro 0.00275 -------------------- total 0.2500

wildtype gro +, ro + 0.24725 gro +, ro + 0.24725 gro +, + + 0.00275 + +, ro + 0.00275 -------------------- total 0.5000Based upon the expected frequency of occurrence of the flies and assuming 1000 offspring, we would expect to see in the offspring of a dihybrid crosswildtype 500rough 250groucho 250

P. 134, #2. A female fruit fly with abnormal eyes (abe) of a brown color (bis, bistre) is crossed with a wildtype male. Her sons have abnormal, brown eyes; her daughters are of the wildtype. When these flies F1 are crossed among themselves, the following offspring are produced: sons daughters abnormal, brown 219 197 abnormal 43 45 brown 37 35 wildtype 201 223What is the linkage arrangement of these loci?In this case the two loci are on the X-chromosome. We know this because we see a difference in the expression of the traits between the sexes in the F1 generation. However, this does not affect the working of the problem because the ratio of males to females for each phenotypic class is approximately 1:1. We can simply pool the sexes for each phenotype. Thus abnormal brown 416 abnormal 88 brown 72 wildtype 424That two classes in highest frequency are the non-recombinants and the two classes in lowest frequency are the recombinants. To calculate the map distance between the two loci, we use the formula above such that 160map distance = ------ x 100 = 16 cM 1000P. 134, #3. InDrosophila, the loci inflated (if, small, inflated wings) and warty (wa, abnormal eyes) are about 10 map units apart on the X chromosome. Construct a data set that would allow you to determine this linkage arrangement. What differences would be involved if the loci were located on an autosome.To construct the data set, we need to setup the crosses and then calculate the expected occurrence of each phenotype in the F2 generation. We will cross an inflated, warty female with a wildtype male. This will produce in the F1 a heterozygous wildtype female and a hemizygous inflated, warty male. These two individual will then be mated to produce the F2 generation. As the male is hemizygous for recessive traits, this is essentially a testcross as the gamete from the F1 female will determine the phenotype of the F2 offspring. Also, there will be no difference in the frequency of expression of the traits between males and females in the F2 generation. For a map distance of 10, in the F2 generation 10% of the offspring must be recombinants. Thus from 1000 offspring, we expect 100 to be either inflated or warty as these are the recombinant conditions given that the P generation was either inflated and warty or wildtype. Of the 100 recombinant offspring, we will evenly divide them among inflated (50) and warty (50). The nonrecombinant offspring will be divided evenly among warty and inflated (450) and wildtype (450). As these traits are X-linked, we could also evenly divide each of the phenotypic classes evenly between males and females. This would give us a data set that is shown below. sons daughters totalinflated, warty 225 225 450inflated 25 25 50warty 25 25 50wildtype 225 225 450If these traits were on an autosome, the difference would be that to get the F2 generation, the female would have to be backcrossed to a male that was homozygous inflated and warty and could not be crossed to her brother in the F1 generation. In the F2 generation, the data set would look like that shown under the total column above.

P. 134, #4. A geneticist crossed female fruit flies that were heterozygous at three electrophoretic loci, each with fast and slow alleles, with males homozygous for the slow alleles. The three loci were got-1 (glutamate oxaloacetate transaminase-1) amy (alpha-amylase), and sdh (succinate dehydrogenase). The first 1,000 offspring isolated had the following genotypes: class 1) got-s got-s,amy-s amy-s,sdh-s sdh-s 441 class 2) got-f got-s,amy-f amy-s,shd-f sdh-s 421 class 3) got-f got-s,amy-s amy-s,sdh-s sdh-s 11 class 4) got-s got-s,amy-f amy-s,sdh-f sdh-s 14 class 5) got-f got-s,amy-f amy-s,sdh-s sdh-s 58 class 6) got-s got-s,amy-s amy-s,sdh-f sdh-s 53 class 7) got-f got-s,amy-s amy-s,sdh-f sdh-s 1 class 8) got-s got-s,amy-f amy-s,sdh-s sdh-s 1What are the linkage arrangements of these loci, including map units? If the three loci are linked, what is the coefficient of coincidence?In this case only two classes are in highest frequency, which indicates that all three loci are linked. The next step is to determine the gene order. To do this we must compare the parental classes (highest frequency) to the double crossover classes (lowest frequency). As the male parent is homozygous for the slow alleles, we can easily determine the gamete donated to the offspring by the heterozygous female.For the parental classes these are: got-s,amy-s,sdh-s and got-f,amy-f,sdh-fFor the double crossover classes these are: got-f,amy-s,sdh-f and got-s,amy-f,sdh-sA comparison of the gametes from the parental classes with the double crossover classes shows that the locus amy is in the middle.The next step is to calculate the map distance among the loci. To do this we need a table as given below number of recombinants between got-amy amy-sdh got-sdh --------------------------------------- 11 58 11 14 53 14 1 1 58 1 1 53 ---------------------------------------- totals 27 113 136To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100. 27distance (got-amy) = ------ x 100 = 2.7 m.u. 1000 113distance (amy-sdh) = ------ x 100 = 11.3 m.u. 1000distance (got-sdh) = 2.7 + 11.3 = 14.0To calculate the coefficient of coincidence we need to calculate the expected number of crossovers and then divide the number into the observed number of double crossovers. The expected frequency of double crossovers is given by multiplying the probability of a crossover between got-amy (0.027) by the probability of a crossover between amy-sdh (0.113). Thus the probability of a double crossover is 0.027 x 0.113 = 0.0031. Then multiplying this number by the total number of offspring (0.0031 x 1000) = 3.1, which is the expected number of offspring. The coefficient of coincidence is given by dividing the observed number of double crossovers by the expected number of double crossover (2/3.1) = 0.65.

P. 134, #5. The following three recessive markers are known in lab mice: h, hotfoot; o, obese; and wa, waved. A trihybrid of unknown origin is testcrossed, producing the following offspring:hotfoot, obese, waved 357hotfoot, obese 74waved 66obese 79wildtype 343hotfoot, waved 61obese, waved 11hotfoot 9a) If the genes are linked, determine the relative order and the map distance between them.b) What was the cis-trans allele arrangement in the trihybrid parent?c) Is there any crossover interference? If yes, how much?In this case only two classes are in highest frequency, which indicates that all three loci are linked. The next step is to determine the gene order. To do this we must compare the parental classes (highest frequency) to the double crossover classes (lowest frequency). As one parent is homozygous for the recessive alleles, we can easily determine the gamete donated to the offspring by the trihybrid parent.For the parental classes these are: h o wa and h+ o+ wa+For the double crossover classes these are: h+ o wa and h o+ wa+A comparison of the gametes from the parental classes with the double crossover classes shows that the locus hotfoot (h) is in the middle. In addition, an examination of the nonrecombinant gametes shows that all widltype alleles are on the same chromosome in the trihybrid and all the recessive alleles are on the other chromosome. This is a cis arrangement of the alleles in the trihybrid.The next step is to calculate the map distance among the loci. To do this we need a table as given below number of recombinants between obese-hotfoot hotfoot-waved obese-waved ----------------------------------------------- 79 74 79 61 66 61 11 11 74 9 9 66 ----------------------------------------------- totals 160 160 160To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100. 160distance (obese-hotfoot) = ------ x 100 = 16.0 m.u. 1000 160distance (hotfoot-waved) = ------ x 100 = 16.0 m.u. 1000distance (obese-waved) = 16.0 + 16.0 = 32.0To calculate the coefficient of coincidence we need to calculate the expected number of crossovers and then divide the number into the observed number of double crossovers. The expected frequency of double crossovers is given by multiplying the probability of a crossover between obese and hotfoot (0.160) by the probability of a crossover between hotfoot and waved (0.160). Thus the probability of a double crossover is 0.160 x 0.160 = 0.0256. Then multiplying this number by the total number of offspring (0.0256 x 1000) = 25.6, which is the expected number of offspring. The coefficient of coincidence is given by dividing the observed number of double crossovers by the expected number of double crossover (20/25.6) = 0.78. The percent of interference is given by (1 - cc) x 100. percent interference = (1 - 0.78) x 100 = 22%.

P. 136, #23. In people, the ABO system (A, B, Oalleles) is linked to the aldolase-B locus (al,al+alleles), a gene that functions in the liver. Deficiency, which is recessive, results in fructose intolerance. A man with blood type AB had a fructose intolerant, type B father and a normal type AB mother. He and a woman with blood type O and fructose intolerance had ten children, of which five were type A and normal, three were fructose intolerant and type B, and two were type A and intolerant to fructose. Draw a pedigree of this family and determine the map distance involved. (Calculate alodscore to determine the most likely recombination frequency between the loci.The first step is to create the pedigree for this cross and label the chromosomes with as much inofrmation as is possible. The man with type AB blood had a father who was homozygous for fructose intolerance, thus theBallele that the man has must have come from his father and is linked to a fructose intolerant allele. His mother had AB blood and was fructose tolerant and thus must have at least one fructose tolerant alleleal+. In this case theAallele from the mother is linked to anal+allele on the same chromosome. This makes the man with AB blood heterozygous for both loci. He marries a woman who had type O blood and is fructore intolerant, thus she is homozygous recessive at both loci. The cross (heterozygous dominant x homozygous recessive) is a testcross.

Of the ten children 5 inherit the chromsome with theAallele without any recombination and 3 inherit the chromosome with theBallele without any recombination. Two of the children inherited recombinant chromoses. To calculate the map distance between the two loci we use the formula below such that # recombinantsmap distance = ---------------- x 100 total # 2map distance = ---- x 100 = 20 cM 10To calculate thelodscore we must first calculate two probabilities. One is the probability of these offspring given a map distance of 20 cM and then the probability of these offspring given independant assortment. To determine the probability of these offspring given 20 cM distance between the loci we must first calculate the probability of any given child. Given a map distance of 20 cM, then 80% of the children will be nonrecombinants and 20% will be recombinats. Thus the probability of a type A and normal (nonrecombinant) is 0.40, and the probability of a type B and fructose intolerant (nonrecombinant) is 0.40, and the probability of a type A and fructose intolerant (recombinant) is 0.10. To get the probability of this family we then multiply the probabilities on each birth together.prob(given 20 cM distance) = 0.40^5 * 0.40^3 * 0.10^2 = 0.0000065536

To get the probability of this family given independant assortment, we assign the probabiliy of any given phenotype the value of 0.25. Given a dihybrid testcross, each phenotype is equally likely. To get the probability of this family we then multiply the probabilities on each birth together.prob(given independant assortment) = 0.25^10 = 0.0000009536743164062

To calculate the lod score we then take the logarithm of the ratio of these two probabilities.lod = log ((prob given 20 cM distance)/prob given independant assortment)) 0.00000655360lod = log ------------- = log 6.872 = 0.83 0.00000095368Anylodscore greater than zero indicates linkage, but it is considered strong evidence for linkage when thelodscore is greater than 3.

P. 136, #24. In the mouse, the autosomal alleles Trembling and Rex (short hair) are dominant to normal and long hair, respectively. Heterozygous Trembling, Rex females were crossed to normal, long-haired males and yielded the following offspring: Trembling, Rex 42 Trembling, long-haired 105 normal, Rex 109 normal, long-haired 44a) Are the two genes linked? How do you know?The cross is a dihybrid female testcrossed to a homozygous recessive male. If the two loci were not linked, then they would assort independently and you would see the four phenotypes in the offspring in a ratio of 1:1:1:1. In this case, two classes are in much higher frequency than the other two classes. This could be tested with a chi-square, which would reject the null hypothesis of independent assortment. Thus, we can see that the two loci do not assort independently and we must conclude that they are linked.b) In the heterozygous females, were Trembling and Rex in cis or trans position? Explain.The two classes in highest frequency represent the parental or nonrecombinant classes. This means that the parents of the dihybrid female were Trembling, long-haired and normal, Rex. The chromosomes of these parents would be homozygous for these traits. Trembling, long-haired normal, Rex Tr + + R ------------ x ---------- ------------ ---------- Tr + + RThe F1 dihybrid female (Trembling, Rex) would have received one of each chromosome from her parents, one chromosome from Trembling, long-haired and 1 from normal, Rex. Thus her chromosomes look like this Tr + --------------- --------------- + RIn this case, Trembling is on one chromosome and Rex is on the homologue. As they are on opposite chromosomes, they are in the trans position.c) Calculate the percent recombination between the two genes.We know that Trembling, long-haired and normal, Rex are the parentals thus, Trembling, Rex and normal, long-haired must be recombinants. To calculate the map distance, we use the formula number of recombinantsmap distance = --------------------------- x 100 total number individuals

86map distance = ----- x 100 = 28.7 m.u. 300P. 136, #25. In corn, a trihybrid Tunicate (T), Glossy (G), Liguled (L) plant was crossed with a nontunicate, nonglossy, liguleless one and produced the following offspring. Tunicate, liguleless, Glossy 58 Tunicate, liguleless, nonglossy 15 Tunicate, Liguled, Glossy 55 Tunicate, Liguled, nonglossy 13 nontunicate, Liguled, Glossy 16 nontunicate, Liguled, nonglossy 53 nontunicate, liguleless, Glossy 14 nontunicate, liguleless, nonglossy 59a. Determine which genes are linked.In this case, an examination of the numbers of each phenotype shows that four classes are in essentially equal frequency. This indicates that not all of the loci are linked. The problem now becomes one of identifying which loci are linked. The next step is to look at the phenotypic ratios taking the data two loci at a time. The first pair will be the tunicate and ligule loci.Ignore the phenotype at the glossy locus. Tunicate, liguless 58 + 15 = 73 Tunicate, Liguled 55 + 13 = 68 nontunicate, liguless 14 + 59 = 73 nontunicate, Liguled 16 + 53 = 69These four phenotypes are in about equal frequency, which indicates independent assortment between the tunicate and liguled loci, and thus these two loci are not linked.The second pair will be the tunicate and glossy loci. Ignore the phenotype at the liguled locus. Tunicate, Glossy 58 + 55 = 113 Tunicate, nonglossy 15 + 13 = 28 nontunicate, Glossy 16 + 14 = 30 nontunicate, nonglossy 53 + 59 = 112These four phenotypes are not in equal frequency, in fact two classes are at much higher frequency (these are nonrecombinants) and two classes are at much lower frequency (these are recombinants). Thus, these data indicate that these two loci are linked. As we know that the liguled locus is not linked to the tunicate locus and that the tunicate locus is linked to the glossy locus, then we can assume that the liguled locus is not linked to the glossy locus. We could show this by the above method, if necessary.b. Determine the genotypes of the heterozygote; be sure to indicate which alleles are on which chromosome.At the tunicate, glossy loci, the nonrecombinants were tunicate, glossy and nontunicate, nonglossy (see table above). This indicates that the chromosomes in the heterozygote had the following arrangement. T G L ---------------- ---------- ---------------- ---------- t g lTheTandGalleles are cis on the same chromosome, however, the liguled locus is on a separate chromosome.c. Calculate the map distance between the linked genes. Tunicate, Glossy 58 + 55 = 113 Tunicate, nonglossy 15 + 13 = 28 nontunicate, Glossy 16 + 14 = 30 nontunicate, nonglossy 53 + 59 = 112From this table, the number of recombinants is 58 and the total numbers of individuals is 283. 58 map distance = ----- x 100 = 20.5 m.u. 283

P. 136, #26. In Drosophila, kidney bean eye (k), cardinal eye (cd), and ebony body (e) are three recessive alleles. If homozygous kidney, cardinal females are crossed to homozygous ebony males, the F1 offspring are all wildtype. If heterozygous F1 females are mated with kidney, cardinal, ebony males, the following two thousand progeny appear: 880 kidney, cardinal 887 ebony 64 kidney, ebony 67 cardinal 49 kidney 46 ebony, cardinal 3 kidney, ebony, cardinal 4 wildtypea. Determine the chromosomal composition of the F1 females.From the flies in the P generation ("kidney, cardinal" and "ebony"), we know that in the F1 one chromosome will carry the allelesk,cd,+while the other chromosome will carry the alleles+,+,e.b. Derive a map of the three genes.In this case only two classes are in highest frequency, which indicates that all three loci are linked. That next step is to determine the gene order. To do this we must compare the parental classes (highest frequency) to the double crossover classes (lowest frequency).For the parental classes these are: kidney, cardinal and ebonyFor the double crossover classes these are: kidney, ebony, cardinal and wildtypeA comparison of the phenotypes from the parental classes(for example, "kidney, cardinal") with the double crossover classes (for example "kidney, ebony, cardinal") shows that the locus ebony is in the middle, because it is the one locus that does not match.The next step is to calculate the map distance among the loci. To do this we need a table as given below number of recombinants between kidney-ebony ebony-cardinal kidney-cardinal ----------------------------------------------------- 64 49 49 67 46 46 3 3 64 4 4 67 ------------------------------------------------------ totals 138 102 136To calculate the distance between each locus, we use the same formula as before: the number of recombinants between the loci divided by the total number of flies and then times 100. 138distance (kidney-ebony) = ------ x 100 = 6.9 m.u. 2000 102distance (ebony-cardinal) = ------ x 100 = 5.1 m.u. 2000distance (kidney-cardinal) = 6.9 + 5.1 = 12.0The map is this. k e cd ---+------------+------------+------- 6.9 m.u. 5.1 m.u.

P. 137, #30. Hemophilia and color blindness are X-linked recessive traits. A normal woman whose mother was color-blind and whose father was a hemophiliac mates with a normal man whose father was color-blind. They have the following children:4 normal duaghters1 normal son2 color-blind sons2 hemophiliac sons1 color-blind, hemophiliac sonEstimate the distance between the two genes.

P. 137, #31. The results of an anlysis of five human-mouse hybrids for five enzymes are given in the following table along with the human chromosomal content of each clone (+ = enzyme of chromosome present; - = absence). Dedue which chromosome carries which gene. see your book for the large table (Table 1)

Table 2. Clone --------------------- human enzyme A B C D E ------------------------------------------------- gluthathione reductase + + - - - malate dehydrogenase - + - - - adenosine deaminase - + - + + galactokinase - + + - - hexosaminidase + - - + - -------------------------------------------------The problem is to determine which clones have which enzymes. For example, gluthathione reductase is present in clones A and B, but absent from clones C, D, and E. The next step is to determine from Table 1 which chromosome or chromosomes are in clones A and B and also absent from clones C, D, and E. An examination of Table 1 shows that only chromosome 8 is present in clones A and B while being absent from clones C, D, and E. Thus gluthatione reductase is on chromosome 8.For malate dehydrogenase we need to determine which chromosome is present in clone B and absent from clones A, C, D and E. The only chromosome that fits this criterion (present in B and absent from all others) is chromosome 2. Thus malate dehydrogenase is on chromosome 2.By the same reasoning as outlined above; adenosine deaminase is on chromosome 20, galactokinase is on chromosome 17, and hexosaminidase is on chromosome 5.

P. 138, #32. You have selected three mouse-human hybrid clones and analyzed them for the presence of human chromosomes. You then analyze each clone for the presence or absence of particular human enzymes (+ = presence of human chromosome or enzyme activity). Based on the following results, indicate the probable chromosomal location for each enzyme.Table 1. Human Chromosomes -------------------------------Clone 3 7 9 11 15 18 20--------------------------------------------- X - + - + + - + Y + + - + - + - Z - + + - - + +---------------------------------------------

Table 2. Enzyme --------------------Clone A B C D E---------------------------------- X + + - - + Y + - + + + Z - - + - +----------------------------------Enzyme A is on chromosome 11. We know this because enzyme A is found in clones X and Y but not in clone Z (Table 2). An examination of Table 1 reveals that chromosome 11 is the only chromosome in clones X and Y but not in clone Z. As the presence and absence of enzyme A matches that of chromosome 11, we conclude that enzyme A in on chromsomes 11.Enzyme B is on chromosome 15Enzyme C is on chromosome 18Enzyme D is on chromosome 3Enzyme E is on chromosome 7

Last updated on 22 August 1996.Provide comments toDwight Mooreatmooredwi@esumail.emporia.edu.Return to theGeneral Genetics Home PageatEmporia State University.