STUDY HINTS
The relative distances between pairs of loci can be estimated from a testcross. The map distance is defined as being equal to the percentage of recombination. The following is an example of the analysis of map distances in a three-point testcross.
In Drosophila melanogaster, three of the loci affecting the adult cuticle are
· h (hairy; extra hairs on the body)
· jv (javelin; bristies are cylindrical, rather than being tapered to a point)
· ve (veinlet; wing veins shortened)
The cross between the heterozygous (h+ jv+ ve+) female parent and the homozygous recessive (hh jvjv veve) male is a testcross. The phenotypes of the progeny represent the alleles inherited from the heterozygous parent, for any recessive allele from the heterozygous parent is automatically homozygous in the offspring. The phenotypes of the progeny are as follows:
|
+ |
+ |
+ |
362 |
|
jv |
+ |
+ |
4 |
|
+ |
+ |
ve |
90 |
|
jv |
+ |
ve |
39 |
|
jv |
h |
ve |
396 |
|
+ |
h |
ve |
2 |
|
+ |
h |
+ |
34 |
|
jv |
h |
+ |
100 |
|
1,027 |
Here are the steps you would follow to determine the map distances.
1. Determine whether the genes are linked or unlinked. They are linked in this example. If not, one would expect a 1:1:1:1:1:1:1:1 in such a testcross.
2. Determine the parental, or nonrecombinant, classes. Look for the most frequent classes: + + + and jv h ve. Therefore, the genotype of the heterozygous parent (without implying that this is the order of the alleles) is
Thus the alleles are in cis-linkages. If the parental classes had been jv h + and + + ve, then jv and h would be in cis-linkage, but jv and h would be in trans-linkage with ve.
3. Find the double recombinant class. Since having recombination occur in both of the regions flanked by heterozygous loci is less likely than having a single recombinant event, the double recombinant classes will be the smallest two classes. In this example, the double recombinant classes are jv + + (four individuals) and + h ve (two individuals).
4. Determine the gene order. Gene order can be found by comparing the parental classes and the double crossover classes. The linkages in the parent and the double crossover classes will be the same, except for one locus. The one that changes must be the middle gene in the sequence. For example, if the chromosomes are as shown in the figure,
the double crossover chromosomes will be a + c and + b +. The middle gene (b) has changed linkage relationships with the other loci. In the present example the classes are
The linkage of jv has changed with respect to the two other loci. The correct gene order is therefore ve jv h or h jv ve. These two alternatives are, in fact, equivalent, since the “left” and “right” ends of a chromosome are defined arbitrarily. The orientation is set when researchers make the first map of a species.
5. Determine the single crossover types:
6. Determine the frequencies of crossovers in each region.
The genetic map is therefore
where 19.1 and 7.7 are map units in centimorgans.
7. Determine the coefficient of coincidence. The coefficient of coincidence measures the difference between the observed and expected numbers of double crossovers. This reflects the degree to which one crossover event may enhance or interfere with another in an adjacent region.
The expected double crossovers will be the product of the likelihood of recombination in each of the independent regions, that is, the probability of two independent events occurring together is the product of their individual probabilities. Since the likelihood of recombination is the same as the map distance, for this problem:
The expected proportion of double crossovers is .191 × .077 = .014707, which we round to .015
The observed proportion of double crossovers is 6 out of 1,027 = .005842, which we round to .006.
The coefficient of coincidence (CC) is therefore the degree to which the observed data on double crossovers fit the theoretical expectation. The degree to which crossing over has been “interfered with” by a closely adjacent crossover event is
Interference = 1 − coefficient of coincidence.
If the CC is 1, then there is no interference. If the CC is less than 1, interference is positive and fewer double crossovers have occurred than expected. If the CC is greater than 1, interference is negative and more double crossovers have occurred than expected.
In this example, the interference is 1 − .4 = .6. Therefore, 60 percent of the expected double crossovers did not occur.
IMPORTANT TERMS
Centimorgan (cM)
Chiasma (Chiasmata)
Coefficient of coincidence
Gene conversion
Interference
Linkage
Locus (Loci)
Recombination
Tetrad analysis
Three-point testcross
Two-point testcross
PROBLEM SET 12
1. Crossover map distances determined by two-point crosses are P—C = 7, S—M = 10, C—M = 8, S—C = 2, and P—S = 5. The relative positions of these four linked loci are
(a) P S C M,
(b) S C P M,
(c) S C M P,
(d) P C S M,
(e) C S P M,
(f) none of the above.
link to answer
2. Assume that 6 percent of all meioses in the sugar beet have an exchange between the A and B loci. The crossover map distance separating these loci is
(a) 3 map units,
(b) 6 map units,
(c) 12 map units.
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3. An experimental setup that would permit you to correlate a genetic crossover with a physical exchange between the two chromosomes is indicated in which parts of the following figure?
(a) A, C, and E,
(b) B, C, and E,
(c) C, D, and E,
(d) C and E,
(e) D and E,
(f) E only.
link to answer
4. Diagram a three-strand double exchange in an ABC / abc trihybrid, with one exchange in the A—B region and the other in the B—C region. Give the genotypes of the four resulting chromosomes, and indicate whether the chromosome is a parental class (noncrossover), single-crossover, or double-crossover strand. Is it possible to diagram another three-strand double crossover that will give products that are genetically different from those that you have listed? If your answer is yes, give the genotypes and indicate whether they are noncrossover, single-crossover, or double-crossover strands.
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5. Diagram a four-strand double exchange in an ABC / abc trihybrid, with one exchange in the A—B region and the other in the B—C region. Give the genotypes of the four resulting chromosomes, and indicate whether the chromosome is a non-crossover, single-crossover, or double-crossover strand. Is it possible to diagram another four-strand double crossover that will give products that are genetically different from those you have listed? If your answer is yes, give the genotypes, and indicate whether they are noncrossover, single-crossover, or double-crossover strands.
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6. Trihybrid geranium plants with hairy leaves, blue flowers, and tall-growth habit are testcrossed with plants having smooth leaves, red flowers, and short-growth habit. The progeny are listed in the following table. What are the linkage relationships of these three loci?
|
Leaf |
Flower |
Height |
Number |
|
Hairy |
Blue |
Tall |
368 |
|
Hairy |
Blue |
Short |
20 |
|
Hairy |
Red |
Tall |
1,797 |
|
Hairy |
Red |
Short |
179 |
|
Smooth |
Blue |
Tall |
191 |
|
Smooth |
Blue |
Short |
1,752 |
|
Smooth |
Red |
Tall |
26 |
|
Smooth |
Red |
Short |
340 |
|
4,673 |
link to answer
7. Assume that in a self-fertilizing plant species an individual is a dihybrid, TY /ty. Plants that are T – are tall, whereas the tt plants are short. Plants that are Y – have yellow seeds, whereas yy plants have white seeds. We know, from testcrosses, that exchange occurs in the T—Y region with a frequency of 12 percent during egg formation and with a frequency of 8 percent during pollen formation. List the expected genotypes and phenotypes, together with their frequencies, for the progeny of the dihybrid.
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8. A trihybrid produces the following gametes:
(a) Which are the parental (noncrossover) types?
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(b) Which are the double-crossover types?
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(c) Which locus is in the middle?
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(d) Draw the two homologues as they would appear at synapsis.
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(e) Calculate the map distances.
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9. Examine the pair of homologous chromosomes in the accompanying diagram. An exchange producing chromosomes, which can be shown to be crossovers, will occur
(a) in region 1 only,
(b) in region 2 only,
(c) in neither region 1 nor region 2,
(d) in either region 1 or region 2,
(e) in both regions.
link to answer
10. The accompanying figure shows the genotype of a trihybrid, together with the known map distances for the two regions. Assume no double crossovers, and determine the types and frequencies of gametes produced by this trihybrid. Then determine the expected genotypes in a sample of 1,000 progeny.
link to answer
11. The figure shows the genotype of a trihybrid, together with the known map distances for the two regions.
Assume that the coefficient of coincidence for the double crossover is 0.6. Determine the types and frequencies of the gametes produced by this trihybrid, and the genotypes expected in a sample of 1,000 progeny.
link to answer
12. A trihybrid produces the following gametes (total = 1,000).
Determine and thoroughly discuss the linkage relationships of these three loci.
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13. The following data come from a cross made to localize a recessive lethal mutation on the X chromosome of Drosophila. The lethal mutation is on a wild-type chromosome, and the cross (see the accompanying diagram) is designed to localize the lethal mutation relative to the genetic markers brought in by the y m car chromosome (which also carries the plus (+) allele for the lethal mutation). The Y chromosome is smaller and is cytologically distinguishable from the
X chromosome by its centromere’s position. The lethal mutation is somewhere on the + + + chromosome.
|
Class |
Females |
Males |
||
|
y |
m |
car |
427 |
459 |
|
+ |
+ |
+ |
486 |
0 |
|
y |
+ |
+ |
299 |
0 |
|
+ |
m |
car |
286 |
283 |
|
y |
m |
+ |
168 |
25 |
|
+ |
+ |
car |
181 |
158 |
|
y |
+ |
car |
69 |
74 |
|
+ |
m |
+ |
84 |
1 |
|
2,000 |
1,000 |
|||
(a) Determine the locations of m and car on the crossover map separately from the male and female data (assume that y is at 0).
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(b) Explain the source of the difference in map distances.
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(c) Localize the lethal locus.
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14. The two chromatids attached to a single centromere in early prophase of meiosis are derived by replication of a single original chromosome and are genetically identical. When, and under what circumstances, are the two chromatids that are attached to the same centromere not genetically identical?
link to answer
15. If the parental classes in a three-point testcross (RrYyBb × rryybb) are RYb and ryB, and the double crossover classes are rYb and RyB, which gene is in the middle?
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16. Please consider data from a two-point mapping testcross. This kind of cross involves mating an individual heterozygous at two loci (two genes) to an individual homozygous for the recessive alleles at both. In this case, the cross can be summarized as EeDd × eedd. Note, however, that this mating summary does not indicate the way the alleles are linked (if they are, indeed, linked) in the heterozygous parent.
|
E e D d |
22 |
|
E e d d |
116 |
|
e e D d |
109 |
|
e e d d |
17 |
(a) What frequencies of each type would you expect if the two genes assorted independently?
link to answer
(b) Are the alleles on the parental chromosomes linked in cis or in trans?
link to answer
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17. Consider the following data from a trihybrid testcross. Presence of a dominant allele from the trihybrid parent is denoted by +, and a recessive allele making the testcross progeny homozygous for the recessive trait is denoted by a lowercase letter.
|
+ |
+ |
m |
22 |
|
+ |
+ |
+ |
93 |
|
t |
b |
m |
98 |
|
t |
+ |
m |
383 |
|
+ |
b |
m |
5 |
|
t |
b |
+ |
29 |
|
+ |
b |
+ |
366 |
|
t |
+ |
+ |
4 |
|
1,000 |
(a) What is the correct gene order for these three genetic loci?
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(b) What is the map distance between the two most closely linked genes?
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(c) What is the value of the coefficient of coincidence?
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18. Assume you are studying the relationship between recombination of genes (alleles) and visible physical exchange in the chromosomes to repeat the classic experiment of Curt Stern. Chromosomes have allelic and physical differences as shown. Which of the following pairs of chromosomes would allow you to do the experiment properly?
(a) chromosomes 1 and 2;
(b) chromosomes 1 and 3;
(c) chromosomes 1 and 4;
(d) chromosomes 2 and 5;
(e) chromosomes 3 and 4.
link to answer
19. The following sets of parental and double crossover progeny data come from three completely separate trihybrid testcrosses in which traits have been scored, but the order of the genes on the chromosome is not known. The symbol indicates the allele inherited from the trihybrid parent, as in the crosses summarized in earlier problems. Please determine the gene orders for each experiment.
link to answer
20. Consider the following data from an experiment like that in question 17.
|
+ |
+ |
e |
22 |
|
|
+ |
+ |
+ |
45 |
|
|
w |
d |
e |
53 |
|
|
+ |
d |
+ |
440 |
|
|
w |
+ |
e |
412 |
|
|
w |
+ |
+ |
1 |
|
|
w |
d |
+ |
25 |
|
|
+ |
d |
e |
2 |
Total progeny count = 1,000 |
(a) What was the linkage relationship of these genes in the trihybrid parent?
link to answer
(b) What is the gene order?
link to answer
(c) The distance between the two most closely linked genes is how many map units (centiMorgans)?
link to answer
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ANSWERS TO PROBLEM SET 12
1. Map orientation problems are simply puzzles that require that you put the distances together in a consistent way. To answer question 1, you could begin by drawing the S—M segment, which is 10 units long, as in the following figure. C could be either 8 units to the left of M or 8 units to the right, but of these two alternatives, only the order S C M predicts that S and C are 2 units apart. It remains only to place P in position.
For the first alternative (P to the left of S, a total of 5 units), the distance from P to C would be 7 units. This is the observed distance. With P to the right of S, as in the second alternative, the distance between P and C would be short (5−2 =3 units). The correct answer is, therefore (a), P S C M.
2. In each meiosis, 2 recombinant chromosomes are produced each time an exchange occurs. Two nonrecombinant chromosomes are also produced. Thus, since map distance is calculated as number of recombinants divided by total number of tested chromosomes, the map distance will be half as large as the percentage of meioses with an exchange (and vice versa). You can demonstrate this for yourself by drawing the meiotic products of 5 cells in which a single recombinant chromosome has occurred in only 1 of the cells. You will find that there will be 2 recombinant chromosomes and 18 nonrecombinant chromosomes. Although an exchange occurred in 20 percent of all meioses (1 in 5), the calculated map distance would be 10 map units (2 recombinant chromosomes divided by the total of 20 chromosomes, 4 chromosomes from each of the 5 cells). The answer to this question is therefore (a), 3 map units.
Relationships such as those described here between recombination rates and map distances may initially seem not only complex but trivial. They will seem less trivial if you realize that the relationship is important to an understanding of the process of recombination, and less complex if you learn to make a quick diagram or sketch of the products of such a problem so that you do not make a careless error by answering simply from memory.
3.
(d) C and E. What is needed here is the ability to detect both a genetic and a cytological (physical) exchange. The genetic exchange is easy. All it requires is heterozygosity at two loci, so that parental combinations can be distinguished from the recombinants. All setups except B satisfy this requirement. The cytological exchange is more difficult to detect, since homologous chromosomes, regardless of their allelic constitutions, are normally identical under the microscope. What is needed here is heterozygosity, to provide cytological markers at each end of the genetic region under test, so that parental chromosomes can be distinguished from the crossovers by the different morphological characteristics they exhibit when examined microscopically. If these conditions are met, crossovers detected genetically should have chromosome structures different from that of the parentals. It does not matter whether the parentals are ab and + +, or a + and + b, or whether one parent has both the knob and the additional length. Of the five setups, A, C, D, and E satisfy the genetic requirement. The requisite cytological heterozygosity is satisfied by setups B, C, and E. Both requirements are satisfied by setups C and E.
4. There are two ways of drawing such a situation at synapsis of the first meiotic division.
5. Only one four-strand double exchange can be drawn, given the placement of the exchanges outlined in the problem. All are single crossovers.
6. This problem is essentially identical to that worked out in the Study Hints section of this chapter. Following that guide, determine whether the loci are linked or unlinked. There are clear pairs of phenotypic classes, which do not fit a 1:1:1:1:1:1:1:1 ratio. The parental (nonrecombinant) classes are the largest:
hairy, red, tall and smooth, blue, short
If the testcross was made to smooth, red, short plants, these three must be the recessive markers. Let us therefore denote them as
H, hairy h, smooth
R, blue r, red
S, tall s, short
Thus the genotype of the heterozygous parent (without implying that this is the order of the alleles) is
The double recombinant classes are the smallest: HRs and hrS. The middle gene is the one that differs from the parental linkage. Since R and s are linked in both the parental and double-crossover class (as is also true of r and S), the H locus must be the middle gene. The data can now be rewritten in the proper order, with the parental classes and single recombinants followed by the double recombinants.
|
Parental |
r |
H |
S |
1,797 |
|
R |
h |
s |
1,752 |
|
|
Crossover in region 1 |
r |
h |
s |
340 |
|
R |
H |
S |
368 |
|
|
Crossover in region 2 |
r |
H |
s |
179 |
|
R |
h |
S |
191 |
|
|
Double crossover |
r |
h |
S |
26 |
|
R |
H |
s |
20 |
|
|
4,673 |
7. The parental chromosomes are as follows:
The gametes produced by the males and females must be considered separately. If exchange occurs in the T—Y region at a frequency of 12 percent in eggs, then 6 percent of the eggs are Ty recombinants, and 6 percent are tYrecombinants. (Note the difference between the frequencies described here and those discussed in problem 2.) In pollen, 4 percent are Ty, and 4 percent are tY.
To calculate the frequencies in the dihybrid progeny, the most direct approach is to calculate the frequencies of each combination in a Punnett square.
|
Gametes of female |
Frequency |
Gametes of male |
Frequency |
|
TY |
.44 |
TY |
.46 |
|
ty |
.44 |
ty |
.46 |
|
Ty |
.06 |
Ty |
.04 |
|
tY |
.06 |
tY |
.04 |
|
1.00 |
1.00 |
The genotypes can be tallied from this Punnett square, and the phenotypes can be calculated by adding appropriate genotype frequencies, as follows:
|
Genotype |
Frequency |
Phenotype |
Frequency |
|
TTYY |
.202 |
Tall, yellow (T- Y-) |
.702 |
|
TTYy |
.046 |
Tall, white (T- yy) |
.048 |
|
TTyy |
.002 |
Short, yellow (tt -Y) |
.048 |
|
TtYY |
.046 |
Short, white (tt yy) |
.202 |
|
TtYy |
.408 |
1.000 |
|
|
Tt yy |
.046 |
||
|
tt YY |
.002 |
||
|
tt Yy |
.046 |
||
|
tt yy |
.202 |
||
|
1.000 |
8. This problem is set up and solved like problem 6. The answers are
(a) a T e and A t E are the parental types.
(b) A T e and a t E are the double-crossovers types.
(c) A is in the middle. The order in the parents is therefore T a e and t A E.
(d)
(e) Crossovers in region 1 = [(46 + 38 + 4 + 2/1000)] · 100 = 9 cM
Crossovers in region 2 = [(63 + 71 + 4 + 2)/1000] · 100 = 14 cM
9.
(d) Crossovers in region 1 and in region 2 cannot be distinguished since the b locus is homozygous and therefore does not change in a crossover. A crossover in region 1 produces AbD and abd strands, and a crossover in region 2 produces the same two strands. In no instance can a homozygous locus be of use in detecting linkage and mapping. Furthermore, an exchange in both regions (i.e., a double crossover) would produce recombinants that are indistinguishable from the parental linkages.
10. In this problem you are asked to work backward from the map distances to the data that generated them. If there are no double crossovers to contend with, the four map units for the A—B region must come directly from 4 recombinant chromosomes (or progeny) in each 100 in the sample. The frequencies of the complementary chromosome types will be approximately equal (e.g., Ab and aB). A total of 3.5 in each 100 will be Bc, and 3.5 will be bC, totaling 7 percent for the B—C region. In the sample in this problem, there are 1,000 total progeny. Thus we multiply the expected numbers in each 100 progeny by 10. The parental genotypes will simply be whatever is left over after the single crossovers have been calculated.
|
Test cross |
|||
|
Gametes |
Frequencies |
Genotype |
Number |
|
ABC |
44.5 percent |
AaBbCc |
445 |
|
abc |
44.5 percent |
aabbcc |
445 |
|
Abc |
2.0 percent |
Aabbcc |
20 |
|
aBC |
2.0 percent |
aaBbCc |
20 |
|
ABc |
3.5 percent |
AaBbcc |
35 |
|
abC |
3.5 percent |
aabbCc |
35 |
|
100 percent |
1,000 |
||
11. The coefficient of coincidence (CC) compares the frequency of observed (O) and expected (E) double crossovers. In this problem, the expected frequency of double crossovers is .21 · .13 = .027. The CC is the frequency of observed crossovers divided by the frequency of expected crossovers: CC = O/E. The frequency of the observed double crossovers, therefore, is equal to the coefficient of coincidence multiplied by the frequency of the expected crossovers: CC · E = O. The frequency of the double crossovers is thus .6 · .027 − .016 (.008 of each type). This yields an expectation of 16 in the sample of 1,000 progeny. The frequency of single crossovers in region 1 = (21/100) − .016 = .194, or 194 per 1,000. The frequency of single crossovers in region 2 = (13/100) − .016 = .114, or 114 per 1,000.
Note: The double crossovers need to be subtracted from the calculation of singles, because this number will be added in when the map distances are tallied from the raw data. As before, the parental class simply includes the remaining individuals.
To simplify the following, the capital letter means they have the dominant allele from the testcross; the lowercase letter means they are homozygous recessive.
12. In these data there appears to be a total of four parental classes. The four recombinant classes are approximately equal. This indicates that one of the three loci is segregating independently of the other two (that is, two are linked, but the third is not linked). By considering the loci in pairs, it can be seen that the B locus is segregating at a 1:1 ratio with each of the others. Only I and D are linked.
Map distance = [(63 + 67)/1000] · 100 = 13 map units
13. This is a complex problem, which should definitely test how well you understand the concepts behind linkage analysis. The key is to recognize that a recessive lethal gene (l) will kill a hemizygous male. When such a lethal gene is located between two markers, only a proportion of the recombinants in that region will carry it, and the class will be reduced (but not eliminated) in males.
The data are in the proper order and are arranged for immediate analysis.
(a) Female data:
Crossover in region 1 = [(299 + 286 + 69 + 84)/2,000] · 100 = 36.9
Crossover in region 2 = [(168 + 181 + 69 + 84)/2,000] · 100 = 25.1
With y at 0.0, m is at 36.9, and car is at 62.0 (36.9 + 25.1) on the female crossover map.
Male data:
Crossover in region 1 = [(283 + 74 + 1)/1,000] · 100 = 35.8
Crossover in region 2 = [(25 + 158 + 74 + 1)/1,000] · 100 = 25.8
Thus m is at 35.8 and car is at 61.6 on the male crossover map.
(b) There is no significant difference between the groups.
(c) To localize the lethal mutation, one should look at the male data, since it is only there that the recessive lethal will not be masked.
In most of the classes, one or another of the complementary chromosomes is absent or reduced in frequency. They must have carried the lethal mutation. The most consistent are the parental and region 1 classes. When region 2 from the + + + chromosome is present, the fly dies. The lethal mutation must therefore be between m and car. For convenience, let us call the region to the left of the lethal 2a, and the region to the right 2b.
The recombinants between m and car will be of two types: those with the lethal mutation (which die) and those without the lethal mutation:
Region 2a: y m l + which die, and + + + car which survive
Region 2b: y m+ + which survive, and + + l car which die
The relative numbers of these can be determined by identifying those that have a lethal mutation (and are thus reduced) and those that do not have the lethal mutation (and are thus about the same magnitude as in the female). The recombinants that survive (and the region they represent) are
|
ym + |
(region 2b) |
25 |
|
+ +car |
(region 2a) |
158 |
|
y + car |
(region 2a) |
74 |
|
+ m + |
(region 2b) |
1 |
Crossovers in region 2a = [(158 + 74)/1,000] · 100 = 23.2 map units from m
Crossovers in region 2b = [(25 + 1)/1,000] · 100 = 2.6 map units from car
The map is
14. One explanation is that a mutation has occurred during or shortly after the replication producing the two chromatids. This would be expected, but very infrequently. A much more likely mechanism is a nonsister strand exchange at the four-strand stage in a heterozygote (Aa). The exchange would have to occur between the heterozygous locus and its centromere:
This results in two centromeres, each with its attached chromatids being partly “sister” (and homozygous for genes between the exchange and the centromere) and partly “nonsister” (and heterozygous for genes distal to the exchange) in makeup.
15. The gene order is Y R B (do not worry about dominant versus recessive alleles when discussing order; a gene is a gene, and its alleles are just different forms at the same location). To answer this question, you look for the allele that has changed linkage relations with the others. To check your conclusion, work backward from your answer. The proposed gene order in the parental chromosomes is YRb and yrB. The double crossover classes should, therefore, be yRB and Yrb. Compare these to the linkages in the original problem (rearranging the presented order) to check that the predicted order is consistent with the original data. It is.
16.
(a) If the two genes were unlinked and assorted independently, you would expect a 1:1:1:1 ratio in a testcross. Since there are 264 progeny, you would expect 66 of each type. Clearly that is not what was observed. One could test the deviation using a chi-square test, but observation is enough to show these data fail to meet the predications of independent assortment.
(b) In a testcross, the homozygous recessive parent contributes a recessive allele for both traits. Mark through these, and the remaining alleles must have come from the dihybrid parent. The parental classes are the largest. Thus, one parental chromosome carried alleles E and d, and the other carried alleles e and D. They were linked in trans.
17.
(a) The order is t m b.
(b) 6 cM. Remember that there are four categories of offspring: the pair of parental linkages, two pairs of single-crossover genotypes, and the double-crossover class. You can quickly see which are the most closely linked genes, since they have the smallest number of individuals in the pair of single-crossover genotypes. The other pair of genes is 20 cM apart.
(c) CC = 0.75. There are nine double-crossover individuals (9/1,000 = 0.009). The expected number is 0.06 × 0.20 = 0.012. Thus, 0.009/0.012 = 0.75.
18. To correlate a change in allele linkage with a physical event at the chromosome level, one requires two heterozygous genes and heterozygous markers at both ends of the chromosome. The only suggested pair that meets both requirements is choice (e), chromosomes 3 and 4. Although they were not offered as choices for this question, one could also have done the experiment with other combinations, such as chromosomes 1 and 6.
19. By comparing the parental and DCO linkages, find which gene has changed its linkage with the others. That must be the gene in the middle.
(a) s b a
(b) h a f
(c) c k t
20.
(a) The parental classes show the original linkage is + d + and w + e.
(b) Gene order is w e d.
(c) The two map distances are w to e = (22 + 25 + 1 + 2)/1,000 = 0.050 = 5% = 5 cM; e to d = (45 + 53 + 1 + 2)/1000 = 0.101 = 10.1% =10.1 cM.