Linkage and Chromosome Mapping Practice Test (page 4)
Review the following concepts if needed:
- Recombination Among Linked Genes for Genetics
- Genetic Mapping for Genetics
- Linkage Estimates from F2 Data for Genetics
- Use of Genetic Maps for Genetics
- Crossover Suppression for Genetics
- Tetrad Analysis in Fungi for Genetics
- Recombination Mapping with Tetrads for Genetics
Linkage and Chromosome Mapping Practice Test
For each of the following definitions, give the appropriate term and spell it correctly. Terms are single words unless indicated otherwise.
- The recovery and investigation of all the products of a single meiotic event (as occurs in ascomycete fungi).
- A cross-shaped cytological structure that is observed in paired homologous chromosomes during first meiotic prophase, and that is formed by the reciprocal breaking and rejoining of nonsister chromatids.
- The genetic event that recombines linked genes. (Two words.)
- The linkage arrangement in a dihybrid in which the two dominant genes are on one chromosome and their corresponding recessive alleles are on the homologous chromosome. (Two words.)
- The linkage arrangement in a dihybrid in which, on each chromosome of a homologous pair, there is one dominant and one recessive gene. (Two words.)
- A unit of map distance equivalent to 1% crossing over.
- The ratio of the double-crossover progeny observed in a three-point testcross relative to the double-crossover progeny expected on the basis of treating map units as if they were independent variables. (One to three words.)
- The complement of the coefficient of coincidence.
- A genetic system that can maintain dihybrid genotypes in repulsion-phase generation after generation. (Two words.)
- Investigation of the products of a given meiotic event in ascomycete fungi. (Two words.)
Choose the one best answer.
- If a chiasma forms between the loci of genes A and B in 20% of the tetrads of an individual of genotype AB/ab, the percentage of gametes expected to be Ab is (a) 40 (b) 20 (c) 10 (d) 5 (e) none of the above
- The average number of chiasmata that forms in one pair of homologous chromosomes is 1.2. The total length (in map units) for this linkage group is expected to be (a) 120 (b) 60 (c) 50 (d) 30 (e) none of the above
For questions 3–7, use the following information. Distances A–B=15 map units, B–C = 8 map units, and A–C = 23 map units.
- In an individual of genotype AbC/aBc, the percentage of gametes expected to be ABC is (a) 23 (b) 11.5 (c) 5.75 (d) 0.6 (e) none of the above
- In the cross Abc/abC × abc/abc, the percentage of Abc/abc progeny expected is (a) 39.7 (b) 23.4 (c) 77.1 (d) 48.6 (e) none of the above
- If from the cross ABC/abc × abc/abc we observe among 1000 progeny one ABC/abc and one aBc/abc, then the coefficient of coincidence is estimated to be approximately (within rounding error) (a) 0.054 (b) 0.167 (c) 0.296 (d) 0.333 (e) none of the above
- The percentage of progeny expected to be ABC/abc from the cross ABC/abc × abc/abc is (a) 15.0 (b) 7.5 (c) 6.9 (d) 23.0 (e) 18.7
- If the B–C distance of 8 map units is a weighted average of two experiments (experiment 1 = 7.3 map units based on a sample size of 100; experiment 2 = 8.4 map units), the sample size of experiment 2 was approximately (a) 116 (b) 144 (c) 167 (d) 175 (e) 190
- From the cross AB/ab × AB/ab we observe 9% of the progeny to be ab/ab. The estimate of the recombination percentage between these two loci is (a) 50 (b) 40 (c) 35 (d) 20 (e) none of the above
The following information applies to questions 9 and 10.Atrihybrid testcross (order of loci unknown) produced 35 AbC/abc, 37 aBc/abc, 8 ABc/abc, 10 abC/abc, 3 ABC/abc, 5 abc/abc, 1 Abc/abc, 1 aBC/abc
- The gene order (a) is CBA (b) is BAC (c) is BCA (d) cannot be determined from the information provided (e) is none of the above
- The distance (in map units) between the A and C loci is estimated to be (a) 30 (b) 25 (c) 20 (d) 15 (e) none of the above
Recombination Among Linked Genes Questions
- There is 21% crossing over between the locus of p and that of c in the rat. Suppose that 150 primary oocytes could be scored for chiasmata within this region of the chromosome. How many of these oocytes would be expected to have a chiasma between these two genes?
- The longest chromosome in the sweet pea has a minimum uncorrected map length (based on known genetic markers) of 118 units. Cytological observations of the longest chromosome in meiotic cells revealed an average chiasmata frequency of 2.96 per tetrad. Calculate the maximum number of crossover units remaining in this chromosome for the mapping of new genes outside the range already known.
Genetic Mapping Questions
- The distances between eight loci in the second chromosome of Drosophila are presented in the following table. Construct a genetic map to include these eight loci. The table is symmetrical above and below the diagonal.
- The recessive gene sh produces shrunken endosperm in corn kernels and its dominant allele sh+ produces full, plump kernels. The recessive gene c produces colorless endosperm and its dominant allele c+ produces colored endosperm. Two homozygous plants are crossed, producing an F1 all phenotypically plump and colored. The F1 plants are testcrossed and produce 149 shrunken, colored : 4035 shrunken, colorless : 152 plump, colorless : 4032 plump, colored. (a) What were the phenotypes and genotypes of the original parents? (b) How are the genes linked in the F1? (c) Estimate the map distance between sh and c.
- The presence of one of the Rh antigens on the surface of the red blood cells (Rh-positive) in humans is produced by a dominant gene R; Rh-negative cells are produced by the recessive genotype rr. Oval-shaped erythrocytes (elliptocytosis or ovalocytosis) are caused by a dominant gene E; its recessive allele e produces normal red blood cells. Both of these genes are linked approximately 20 map units apart on one of the autosomes. A man with elliptocytosis, whose mother had normally shaped erythrocytes and a homozygous Rh-positive genotype and whose father was Rh-negative and heterozygous for elliptocytosis, marries a normal Rh-negative woman. (a) What is the probability of their first child being Rh-negative and elliptocytotic? (b) If their first child is Rh-positive, what is the chance that it will also be elliptocytotic?
- The Rh genotypes, as discussed in Problem 6.15, are given for each individual in the pedigree shown below. Solid symbols represent elliptocytotic individuals. (a) List the E locus genotypes for each individual in the pedigree. (b) List the gametic contribution (for both loci) of the elliptocytotic individuals (of genotype Rr) beside each of their offspring in which it can be detected. (c) How often in part (b) did R segregate with E, and r with e? (d) On the basis of random assortment, in how many of the offspring in part (b) would we expect to find R segregating with e, or r with E? (e) If these genes assort independently, calculate the probability of R segregating with E, and r with e in all 10 cases. (f) Is the solution to part (c) suggestive of linkage between these two loci? (g) Calculate part (e) if the siblings II1 and II2 were identical twins (developed from a single egg). (h) How are these genes probably linked in I1?
- Two recessive genes in Drosophila (b and vg) produce black body and vestigial wings, respectively. When wild-type flies are testcrossed, the F1 are all dihybrid in coupling phase. Testcrossing the female F1 produced 1930 wild type : 1888 black and vestigial : 412 black : 370 vestigial. (a) Calculate the distance between b and vg. (b) Another recessive gene cn lies between the loci of b and vg, producing cinnabar eye color. When wild-type flies are testcrossed, the F1 are all trihybrid. Testcrossing the F1 females produced 664 wild type : 652 black, cinnabar, vestigial : 72 black, cinnabar : 68 vestigial : 70 black : 61 cinnabar, vestigial : 4 black, vestigial : 8 cinnabar. Calculate the map distances. (c) Do the b–vg distances calculated in parts (a) and (b) coincide? Explain. (d) What is the coefficient of coincidence?
- In corn, a dominant gene C produces colored aleurone; its recessive allele c produces colorless. Another dominant gene Sh produces full, plump kernels; its recessive allele sh produces shrunken kernels due to collapsing of the endosperm. A third dominant gene Wx produces normal starchy endosperm and its recessive allele produces waxy starch. A homozygous plant from a seed with colorless, plump, and waxy endosperm is crossed to a homozygous plant from a seed with colored, shrunken, and starchy endosperm. The F1 is testcrossed to a colorless, shrunken, waxy strain. The progeny seed exhibit the following phenotypes: 113 colorless, shrunken, starchy : 4 colored, plump, starchy : 2708 colorless, plump, waxy : 626 colorless, plump, starchy : 2 colorless, shrunken, waxy : 116 colored, plump, waxy : 2538 colored, shrunken, starchy : 601 colored, shrunken, waxy. (a) Construct a genetic map for this region of the chromosome. Round all calculations to the nearest tenth of a percent. (b) Calculate the interference in this region.
- A gene called "forked" (f) produces shortened, bent, or split bristles and hairs in Drosophila. Another gene called "outstretched" (od) results in wings being carried at right angles to the body. A third gene called "garnet" (g) produces pinkish eye color in young flies. Wild-type females heterozygous at all three loci were crossed to wild-type males. The F1 data appear below.
- Five sex-linked recessive genes of Drosophila (ec, sc, v, cv, and ct) produce traits called echinus, scute, vermilion, crossveinless, and cut, respectively. Echinus is a mutant producing rough eyes with large facets. Scute manifests itself by the absence or reduction in the number of bristles on certain parts of the body. Vermilion is a bright orange-red eye color. Crossveinless prevents the development of supporting structures in the wings. Cut produces scalloped and pointed wings with manifold (pleiotropic) effects in other parts of the body. At the beginning of our experiments we do not know the gene order. From the results of the following three experiments, construct a genetic map for this region of the X chromosome. Whenever possible use weighted averages.
(a) Which gene is in the middle? (b) What was the linkage relationship between alleles at the forked and outstretched loci in the maternal parent? (c) What was the linkage relationship between alleles at the forked and garnet loci in the maternal parent? (d) On what chromosome do these three genes reside? (e) Calculate the map distances. (f) How much interference is operative?
Experiment 1. Echinus females crossed to scute, crossveinless males produced all wild-type females and all echinus males in the F1. When the F1 females were testcrossed, the results (including both male and female progeny) were as follows:
Experiment 2. Crossveinless females crossed to echinus, cut males produced all wild-type females and all crossveinless males in the F1. When the F1 females were testcrossed, the results (including both male and female progeny) were as follows:
Experiment 3. Cut females crossed to vermilion, crossveinless males produced all wild-type females and cut males in the F1. When the F1 females were testcrossed, the results (including both male and female progeny) were as follows:
Linkage Estimates From F2 Data Questions
- Two recessive sex-linked genes are known in chickens (ZW method of sex determination; Chapter 5): rapid feathering (sl) and gold plumage (s). The dominant alleles produce slow feathering (Sl) and silver plumage (S), respectively. Females of the Silver Penciled Rock breed, with slow feathering and silver plumage, are crossed to males of the Brown Leghorn breed, with rapid feathering and gold plumage. The F2 progeny data appear below:
- Assume the genotype AB/AB is testcrossed and produces an F2 consisting of 37 A-B-, 11 A-bb, 12 aaB-, and 4 aabb. Estimate the percentage recombination between A and B by the square-root method and by the product ratio method.
- Two recessive genes in the third linkage group of corn produce crinkly leaves and dwarf plants, respectively. A pure crinkly plant is pollinated by a pure dwarf plant. The F2 progeny consist of 104 normal : 43 dwarf : 51 crinkly : 2 dwarf, crinkly. Using the square-root method, estimate the amount of recombination between these two loci.
- The duplicate recessive genes r1 and r2 produce a short, velvet like fur called "rex." Two rex rabbits of different homozygous genotypes were mated and produced an F1 that was then testcrossed to produce 64 rex and 6 normal testcross progeny. (a) Assuming independent assortment, how many normal and rex phenotypes would be expected among 70 progeny? (b)Do the data indicate linkage? (c) What is the genotype and the phenotype of the F1? (d) What is the genotype and phenotype of the testcross individuals? (e) Calculate the map distance.
(a) Determine the F1 genotypes and phenotypes. (b) In what linkage phase are the F1 males? (c) Calculate the amount of recombination expected to occur between these two loci in males.
Use of Genetic Maps Questions
- Two loci are known to be in linkage group IV of the rat. Kinky hairs in the coat and vibrissae (long nose "whiskers") are produced in response to the recessive genotype kk and a short, stubby tail is produced by the recessive genotype st/st. The dominant alleles at these loci produce normal hairs and tails, respectively. Given 30 map units between the loci of k and st, determine the expected F1 phenotypic proportions from heterozygous parents that are (a) both in coupling phase, (b) both in repulsion phase, (c) one in coupling and the other in repulsion phase.
- Deep-yellow hemolymph (blood) in silkworm larvae is the result of a dominant gene Y at locus 25.6 (i.e., 25.6 crossover units from the end of the chromosome). Another dominant mutation Rc, 6.2 map units from the Y locus, produces a yellowish-brown cocoon (rusty). Between these two loci is a recessive mutant oa governing mottled translucency in the larval skin, and mapping at locus 26.7. The loci Rc and oa are separated by 5.1 crossover units. An individual that is homozygous for yellow blood, mottled translucent larval skin, and wild-type cocoon color is crossed to an individual of genotype Y+ oa+ Rc/Y + oa+ RC that spins a rusty cocoon. The F1 males are then testcrossed to produce 3000 F2 progeny. Coincidence is assumed to be 10%. (a) Predict the numbers within each phenotypic class that will appear in the F2 (to the nearest hole numbers). (b)On the basis of probabilities, how many more F2 progeny would need to be produced in order to recover one each of the DCO phenotypes?
- The eyes of certain mutant Drosophila have a rough texture due to abnormal facet structure. Three of the mutants that produce approximately the samephenotype (mimics) are sex-linked recessives: roughest (rst), rugose (rg), and roughex (rux). The loci of these genes in terms of their distances from the end of the X chromosome are 2, 11, and 15 map units, respectively. (a) From testcrossing wild-type females of genotype rst + rux/ + rg + predict the number of wild-type and rough-eyed flies expected among 20,000 progeny. Assume no interference. (b) Approximately how many rougheyed progeny flies are expected for every wild-type individual? (c) If the females of part (a) were of genotype rst rq rux/ + + +, what would be the approximate ratio of wild-type: rough-eyed progeny?
- In Asiatic cotton, a pair of factors (R and r) controls the presence or absence, respectively, of anthocyanin pigmentation. Another gene, about 10 map units away from the R locus, controls chlorophyll production. The homozygous recessive genotype at this locus (yy) produces a yellow (chlorophyll-deficient) plant that dies early in the seedling stage. The heterozygote Yy is phenotypically green and indistinguishable from the dominant homozygote YY. Obviously, testcrosses are not possible for the Y locus. When dihybrids are crossed together, calculate the expected phenotypic proportions among the seedlings and among the mature F1 when parents are (a) both in coupling phase, (b) both in repulsion phase, (c) one in coupling and one in repulsion phase. (d) Which method [in parts (a), (b), or (c)] is expected to produce the greatest mortality?
Crossover Suppression Questions
- A black-bodied Drosophila is produced by a recessive gene b and vestigial wings by another recessive gene vg on the same chromosome. These two loci are approximately 20 map units apart. Predict the progeny phenotypic expectations from (a) the mating of repulsion phase females × coupling-phase males, (b) the reciprocal cross of part (a), (c) the mating where both parents are in repulsion phase.
- Poorly developed mucous glands in the female silkworm Bombyx mori cause eggs to be easily separated from the papers on which they are laid. This is a dominant genetic condition; its wild-type recessive allele Ng+ produces normally "glued" eggs. Another dominant gene C, 14 map units from Ng, produces a golden-yellow color on the outside of the cocoon and nearly white inside. Its recessive wild-type allele C+ produces normally pigmented or wild-type cocoon color. A pure "glueless" strain is crossed to a pure golden strain. The F1 females are then mated to their brothers to produce the F2. Predict the number of individuals of different phenotypes expected to be observed in a total of 500 F2 offspring. (Hint: Crossing over does not occur in female silkworms.)
- Two autosomal recessive genes, "dumpy" (dp, a reduction in wing size) and "net" (net, extra veins in the wing), are linked on chromosome 2 of Drosophila. Homozygous wild-type females are crossed to net, dumpy males. Among 800 F2 offspring were found: 574 wild type : 174 net, dumpy : 25 dumpy : 27 net. Estimate the map distance.
- Suppose that an abnormal genetic trait (mutation) appeared suddenly in a female of a pure culture of Drosophila melanogaster. We mate the mutant female to a male from a balanced lethal strain [Cy/Pm, D/Sb, where curly (Cy) and plum (Pm) are on chromosome 2 and dichaete (D) and stubble (Sb) are on chromosome 3]. About half of the F1 progeny (both males and females) exhibit the mutant phenotype. The F1 mutant males with curly wings and stubble bristles are then mated to unrelated virgin wild-type females. In the F2 the mutant trait never appears with stubble. Recall that this species of Drosophila has chromosomes X, 2, 3, and 4. Could the mutation be (a) an autosomal recessive, (b) a sex-linked recessive, (c) an autosomal dominant, (d) a sex-linked dominant? (e) In which chromosome does the mutant gene reside? (f) Suppose the mutant trait in the F2 appeared in equal association with curly and stubble. In which chromosome would the mutant gene reside? (g) Suppose the mutant trait in the F2 appeared only in females. In which chromosome would the mutant gene reside? (h) Suppose the mutant trait in the F2 never appeared with curly. In which chromosome would the mutant gene reside?
Recombination Mapping With Tetrads Questions
- Given the adjoining meiotic metaphase orientation in Neurospora, determine the simplest explanation to account for the following spore patterns.
- A certain strain of Neurospora cannot grow unless adenine is in the culture medium. Adenineless is a recessive mutation (ad). Another strain produces yellow conidia (ylo). Below are shown the results from crossing these two strains. Calculate the map distance between these two genes.
- A riboflavineless strain (r) of Neurospora is crossed with a tryptophaneless strain (t) to give
- Two of the genes s, t, and u are linked; the third assorts independently and is very tightly linked to its centromere. Analyze the unordered tetrads produced by the cross (stu) × ( + + + ). (Hint: See Problem 6.10.)
Construct a map that includes these two genes.
- tetrad analysis
- chiasma, or X (chi)
- crossing over
- coupling phase
- repulsion phase
- coincidence, coincidence coefficient, or coefficient of coincidence
- balanced lethals
- tetrad analysis
Recombination Among Linked Genes
- 30 units
- net (4) ho (7) ed(1) ft(1) dp (3.5) cl (14.5) d (10) J
- (a) sh c/sh c(shrunken; colorless) × sh+ c+/sh+ c+ (plump; colored) (b) sh+ c+/sh c(coupling phase) (c) 3.6 map units
- (a) 2/5 (b) 1/5
- (a) All open symbols = ee, all solid symbols = Ee (b) re (II1, II2, III6), RE (II5, II9, III2-5, (7) (c) All 10 cases (d) 5 each (e) 1/1024 (f) Yes (g) 1/512 (h) RE/re (coupling phase)
- (a) 17 map units (b) b-cn = 8.9 map units; cn-vg = 9.5 map units (c) No. In part (a), two-point testcross cannot detect double crossovers that then appear as parental types, thus underestimating the true map distance. (d) 0.89
- (a) c (3.5) sh (18.4) wx (b) 86.1%
- (a) f (b)Repulsion phase (c)Coupling phase (d)X chromosome (sex-linked) (e) g–f = 12.0, f–od = 2.5 (f) None
Linkage Estimates From F2 Data
- (a) Sl S/sl s (slow, silver males) × sl s/W (rapid, gold females) (b) Coupling phase (c) 14.2%
- (a) 52.5 rex : 17.5 normal (b) Yes (c) R1r2/r1R2, normal (d) r1r2/r1r2, rex (e) 17.14 map units
Use of Genetic Maps
- (a) 1408 yellow, mottled : 1408 rusty : 16 yellow, rusty : 16 mottled : 76 yellow, mottled, rusty : 76 wild type. Note: Rounding errors may allow one whole individual difference in each of these phenotypic classes. (b) 32,651
- (a) 19,964 rough-eyed : 36 wild type (b) Approximately 1 : 555 (c) Approximately 1 : 1.289
- (a) (b) (c)
- (d) All three crosses = 25% mortality
- (a) 55% wild type : 20% black : 20% vestigial : 5% black, vestigial (b), (c) 50% wild type : 25% black : 25% vestigial
- 250 glueless, golden : 125 glueless : 125 golden
- 13 map units
Recombination Mapping With Tetrads
- A dominant trait may appear suddenly in a population by mutation of a recessive wild-type gene to a dominant allelic form. Such a mutant individual would be heterozygous. (a) No (b) No (c) Yes (d) No (e) Chromosome 3 (f) Chromosome 4 (g) X chromosome (sex-linked) (h) Chromosome 2
- (a) (1, 3) (b) (2, 4) (c) (1, 3)(1, 3) or (1, 3)(2, 4) (d) (1, 3)(2, 3) or (1, 3)(1, 4) (e) (2, 3)(1, 3) or (1, 4)(1, 3) (f) (2, 3)(2, 3) or (1, 4)(2, 3) (g) (2, 4)(1, 3) or (2, 4)(2, 4) (h) (2, 4)(2, 3) or (2, 4)(1, 4) (i) (1, 3) (j) (2, 3) (k) (2, 4) (l) (1, 4)
- 5.83 map units
- (1.8) r (16.4) t
- Genes u and s are linked on the same side of the centromere with u closest to the centromere. Centromere to u = 16 units, u to s - 14 units. Gene t is on another chromosome and very closely linked to its centromere.