Download Chromosome Linkage and Mapping

SBI3U – Assigned Homework
Monday, April 6, 2009
Linkage and Chromosome Mapping
Alleles may appear (in dihybrids) in either of two positions relative to one another on the chromosomes:
A
B
Bivalent
A
B
Alleles in the CIS position (COUPLING phase)
4 chromatids
a
b
(tetrad)
a
b
A
b
A
a
b
B
a
B
Bivalent
4 chromatids
(tetrad)
Alleles in the TRANS position (REPULSION phase)
Cross overs (Recombinants): further apart the genes – the greater the opportunity for crossing over to
occur between gene loci (alleles)
Seen when the crossover forms between the gene loci under consideration
Example:
A
A
crossover
a
a
A
B
A
B
a
b
a
b
B
B
b
b
no crossing over seen in the above example because the cross over did not occur between the two gene
loci that were being observed.
1
Single Crossover:
A
A
B
A
b
a
B
a
b
B
A
a
B
Double crossover
b
a
b
single crossover was detected in the above example as a change in the pair of alleles found on the same
chromosome.
Multiple Crossovers:
A
A
B
A
B
a
b
a
b
B
A
a
B
Double crossover
b
a
b
double crossover was not detected in the above example.
Example question:
A cross occurred between the following two organisms:
AB/ab
x
ab/ab
( / - demonstrates linkage between alleles)
Of 100 progeny (offspring), the following were found:
34
16
Ab/ab
ab/ab
34
16
aB/ab
AB/ab
total of 68%
total of 32%
total = 100%
2
Answer:
The parental gametes always appear with greatest frequency
Therefore:
parents:
Ab/aB
Individual:
Single crossover between A & b
A
b
A
b
a
B
a
B
x
adb/adb
Ab/aB (gametes produced)
(20% - 20 offspring out of 100 show the new combination of
traits)
A
b
A
B
a
b
a
B
32% - 32 offspring out of 100 show the new combination of traits
Add a 3rd loci (trait – allele):
A
D
B
A
D
B
a
d
b
a
d
b
A
D
B
A
d
B
a
D
b
a
d
b
double crossovers between AB/ab can now be detected in the above example containing a 3rd loci.
3
Example question:
A cross occurred between the following two organisms:
ADB/adb
x
adb/adb
( / - demonstrates linkage)
Of 100 progeny (offspring), the following were found:
34
10
5
1
ADB/adb
Adb/adb
Adb/adb
AbB/adb
34
10
5
1
adb/adb
aDB/adb
aDB/adb
aDb./adb
total of 68%
total of 20%
total of 10%
total of 2%
total = 100%
Answer:
The parental gametes always appear with greatest frequency
Therefore:
parents:
ADB/adb
Individual:
Single crossover between A & D
A
D
x
adb/adb
ADB/adb (gametes produced)
(20% - 20 offspring out of 100 show the new combination of
traits)
A
D
B
A
d
b
a
D
B
a
d
b
B
A
D
B
a
d
b
a
d
b
4
Single crossover between D & B
A
D
(10% - 10 offspring out of 100 show the new combination of
traits)
A
D
B
A
D
b
a
d
B
a
d
b
B
A
D
B
a
d
b
a
d
b
Double crossover (least occurring) (2% - 2 offspring out of 100 show the new combination of traits
– [multiply the two single rates together –
20% * 10% = 2% 2% * 100 individuals = 2 individuals)
A
D
A
D
B
A
d
B
a
D
b
a
d
b
B
A
D
B
a
d
b
a
d
b
Chromosome Linkage and Mapping
Sometimes, two traits are visible together. For instance, in corn, the traits for color and kernel fullness are usually
carried on one gene, and the recessive alleles for these traits can be carried on an another. When both the recessive or
the dominant alleles for two traits are on the same chromosome, it is called the cis phase. However, when a recessive
and dominant allele for the different traits are on the same chromosome, we call it the trans phase.
So, when two parents are crossed continually, very seldom do you see a mixing of this pattern. This is called linkage.
However, when colored, shrunken kernels apper less than the two homozygous types, we can assume that the two
homologous chromosomes crossed over and caused a variation. This is called crossing over.
5
Normally, when a person crosses a heterozygous colored, full kernel corn plant to a homozygous recessive, we would
expect to get:
The two types with the higher numbers represent the two parent-type gametes, since not all cells will crossover. The
other two represent the single crossovers (SXO).
Since we have figured out that crossing over has occurred, we can map the distance between these two traits on the
chromosome. The closer the traits are on the chromosome, the less likely crossing over will occur.
To figure map distance, we can look at the numbers to determine the parent crosses, and whether or not the genes are
in trans or cis formation. Then we can use the equation:
This percent can be used as the number of map units apart the two genes are. If the two genes are further than 50 units,
crossing over will not be a factor.
To determine if crossing over exists, it is best to cross a heterozygous plant to a homozygous recessive plant. Then
once can calculate the possible gametes from the heterozygous parent. Remember, the larger numbers of the results
will show the parental genotypes.
For example:
If we know the map units between two traits, we can figure out the outcome of a cross of a heterozygous individual to
a homozygous recessive. If two traits are 10 map units apart, we know that this also can be placed as a percent and
represent the amount of crossovers occurring. Since there are two possible crossover gametes that occur 10% of the
time, we can divide .10 by 2 to get .05 of a chance per crossover gamete. To figure the parental occurrence, we know
that 100 - 10 = 90% occurrence, and that there are two possible gametes, so we get a .45 chance for each.
6
Genetic Mapping
Loci:
places where genes reside on chromosomes (considering genetic mapping, these genes reside on
the same chromosome)
Two aspects to genetic mapping: 1:
2:
Map Distance:
Linear order of genes
relative distances between two loci on the chromosome
is the expression of the probability that crossing over will occur between two genes (loci)
Measured as a Centimorgan (cM)
1 Centimorgan is equivalent to 1% crossing over
7
Example 1:
Ab/aB
8% AB
8% ab
42% Ab
Recombinant
42% aB
Parental
8% AB + 8% ab (recombinants) = 16% map distance (cM)
A
16 cM
B
12 cM
B
Example 2:
6% Bc + 6% cB (recombinants) = 12% map distance (cM)
C
BC/bc
6% Bc
6% bC
Recombinant
Mapping Trihybrid
44% BC
44% bc
Parental
[2n gamete varieties (N = number of genes)
23 = 8 varieties]
8
Test cross of data
QqRrTt
x
qqrrtt
Written for linked genes or not linked genes – can differentiate by the number of
offspring
Row
1
2
3
4
5
6
7
8
Totals
Genotype of
Heterozygous
Gametes
qrt
qrT
qRt
Qrt
QRt
QrT
qRT
QRT
Number of Offspring
Counted
% of Total Offspring
12
4186
750
575
4250
810
650
14
11247
0.1
37.2
6.7
5.1
37.8
7.2
5.8
0.1
100
Problem:
1: construct a gene map
2: show distance between loci
3: genotype of trihybrid to show allelic combinations of the homologous chromosomes (CIS or TRANS)
9
Two Point Analysis:
Are q and r linked?
% recombination?
Phenotypes
qr
qR
Qr
QR
Data Rows
1 & 2
3 & 7
4 & 6
5 & 8
% Total Amongst Progeny
0.1 + 37.2 = 37.3
6.7 + 5.8 = 12.5
5.1 + 7.2 = 12.3
37.8 + 0.1 = 37.9
Not independent assortment (would be 1:1:1:1 ratio) – therefore genes are linked in the CIS position
Recombinants are: qR + Qr (lowest numbers) = 12.5 + 12.3 = 24.8% (24.8 cM distance)
Are r and t linked?
% recombination?
Phenotypes
rt
rT
Rt
RT
Data Rows
1 & 4
2 & 6
3 & 5
7 & 8
% Total Amongst Progeny
0.1 + 5.1 = 5.2
37.2 + 7.2 = 44.4
6.7 + 37.8 = 44.5
5.8 + 0.1 = 5.9
Not independent assortment (would be 1:1:1:1 ratio) – therefore genes are linked in the TRANS position
Recombinants are: rt + RT (lowest numbers) = 5.2 + 5.9 = 11.1% (11.1 cM distance)
Are q and t linked?
% recombination?
Phenotypes
qt
qT
Qt
QT
Data Rows
1 & 3
2 & 7
4 & 5
6 & 8
% Total Amongst Progeny
0.1 + 6.7 = 6.8
37.2 + 5.8 = 43.0
5.1 + 37.8 = 42.9
7.2 + 0.1 = 7.3
Not independent assortment (would be 1:1:1:1 ratio) – therefore genes are linked in the CIS position
Recombinants are: qt + QT (lowest numbers) = 6.8 + 7.3 = 14.1% (14.1 cM distance)
10
Homologous pairing of chromosomes:
14.1 cM
q
T
r
q
T
r
Q
t
R
Q
t
R
24.8 cM
11.1 cM
Recombination and Estimating the Distance Between Genes
Physical crossing over during meiosis I is a normal event. The effect of this event is to rearrange
heterozygous homologous chromsomes into new combinations. The term used for crossing over is
recombination. Recombination can occur between any two genes on a chromosome, the amount of
crossing over is a function of how close the genes are to each other on the chromosome. If two genes are
far apart, for example at opposite ends of the chromosome, crossover and non-crossover events will occur
in equal frequency. Genes that are closer together undergo fewer crossing over events and non-crossover
gametes will exceed than the number of crossover gametes. The figure below shows this concept.
Finally, for two genes are right next to each other on the chromosome crossing over will be a very rare event.
Two types of gametes are possible when following genes on the same chromosomes. If crossing over does not occur,
the products are parental gametes. If crossing over occurs, the products are recombinant gametes. The allelic
composition of parental and recombinant gametes depends upon whether the original cross involved genes in
coupling or repulsion phase. The figure below depicts the gamete composition for linked genes from coupling and
repulsion crosses.
11
It is usually a simple matter to determine which of the gametes are recombinant. These are the gametes that are found
in the lowest frequency. This is the direct result of the reduced recombination that occurs between two genes that are
located close to each other on the same chromosome. Also by looking at the gametes that are most abundant you will
be able to determine if the original cross was a coupling or repulsion phase cross. For a coupling phase cross, the
most prevalent gametes will be those with two dominant alleles or those with two recessive alleles. For repulsion
phase crosses, gametes containing one dominant and one recessive allele will be most abundant. Understanding this
fact will be important when you actually calculate a linkage distance estimate from your data.
The important question is how many recombinant chromosomes will be produced. If the genes are far apart on the
chromosome a cross over will occur every time that pairing occurs and an equal number of parental and recombinant
chromosomes will be produced. Test cross data will then generate a 1:1:1:1 ratio. But as two genes are closer and
closer on the chromosome, fewer cross over events will occur between them and thus fewer recombinant
chromosomes will be derived. We then see a deviation from the expected 1:1:1:1 ratio.
How can we decide how close two genes are on a chromosome? Because fewer crossover events are seen between
two genes physically close together on a chromosome, the lower the percentage of recombinant phenotypes will be
seen in the testcross data. By definition, one map unit (m.u.) is equal to one percent recombinant phenotypes.
In honour of the work performed by Morgan, one m.u. is also called one centimorgan (cM).
Now let's determine the linkage distance between the genes pr and vg. We can actually make two estimates because
we have the results from coupling and repulsion phases crosses. The coupling phase analyzed a total of 2839
gametes, and of these gametes 305 (151 pr+ vg+ 154 pr vg+) gametes were recombinant. To determine the linkage
distance simply divide the number of recombinant gametes into the total gametes analyzed. So the linkage distance is
equal to 10.7 cM [(305/2839)*100)].
We can also perform the same calculations with the results from the repulsion phase cross. For this experiment, a
total of 2335 gametes were analyzed, and 303 (151 pr+ vg++ 154 pr vg) of these were the result of recombination.
The estimate of the linkage distance between pr and vg from these experiments is 13.0 cM [(303/2335)*100].
12
Obviously, we can conclude that the two genes are linked on the same chromosome. But what is the true linkage
distance, the 10.7 cM value from the coupling experiment or the 13.0 value from the repulsion experiment? Actually
neither is correct or wrong. These again are two estimates. Only by repeating this experiments many times using a
number of different independent crosses can we settle on a value.
Once we have settled on a value, these genes can then be graphically displayed. Let's say that the true distance
between the pr and vg genes is 11.8 cM, that is the average of our two estimates. We can next display them along a
chromosome in the manner shown below. (Note that it is customary to use the allelic designations of the mutant
phenotype when drawing these maps.)
The final point that we need to make regards the maximum distance that we can measure. Because of the way in
which the calculations are performed, we can never have more that 50% recombinant gametes. Therefore the
maximum distance that two genes can be apart and still measure that distance is just less that 50 cM. If two genes are
greater than 50 cM apart, then we can not determine if they reside on the same chromosome or are on different
chromosomes. In practice though, when experimental error is considered, as distances approach 50 cM it is difficult
to determine if two genes are linked on the same chromosome. Therefore, other mapping techniques must be used to
determine the linkage relationship among distantly associated genes. One method that allows us to deal with distantly
related genes and to order genes is the three-point cross.
Deriving Linkage Distance and Gene Order From Three-Point Crosses
By adding a third gene, we now have several different types of crossing over products that can be obtained. The
following figure shows the different recombinant products that are possible.
Now if we were to perform a testcross with F1, we would expect a 1:1:1:1:1:1:1:1 ratio. As with the two-point
analyzes described above, deviation from this expected ratio indicates that linkage is occurring. The best way to
become familiar with the analysis of three-point test cross data is to go through an example. We will use the arbitrary
example of genes A, B, and C. We first make a cross between individuals that are AABBCC and aabbcc. Next the F1
is test-crossed to an individual that is aabbcc. We will use the following data to determine the gene order and linkage
distances. As with the two-point data, we will consider the F1 gamete composition.
13
Genotype Observed
Type of Gamete
ABC
390
Parental
abc
374
Parental
AbC
27
Single-crossover between genes C and B
aBc
30
Single-crossover between genes C and B
ABc
5
Double-crossover
abC
8
Double-crossover
Abc
81
Single-crossover between genes A and C
aBC
85
Single-crossover between genes A and C
Total
1000
The best way to solve these problems is to develop a systematic approach. First, determine which of the the
genotypes are the parental genotypes. The genotypes found most frequently are the parental genotypes. From
the table it is clear that the ABC and abc genotypes were the parental genotypes.
Next we need to determine the order of the genes. Once we have determined the parental genotypes, we use that
information along with the information obtained from the double-crossover. The double-crossover gametes are
always in the lowest frequency. From the table the ABc and abC genotypes are in the lowest frequency. The next
important point is that a double-crossover event moves the middle allele from one sister chromatid to the
other. This effectively places the non-parental allele of the middle gene onto a chromosome with the parental alleles
of the two flanking genes. We can see from the table that the C gene must be in the middle because the recessive c
allele is now on the same chromosome as the A and B alleles, and the dominant C allele is on the same chromosome
as the recessive a and b alleles.
Now that we know the gene order is ACB, we can go about determining the linkage distances between A and C, and
C and B. The linkage distance is calculated by dividing the total number of recombinant gametes into the total
number of gametes. This is the same approach we used with the two-point analyses that we performed earlier. What
is different is that we must now also consider the double-crossover events. For these calculations we include those
double-crossovers in the calculations of both interval distances.
So the distance between genes A and C is 17.9 cM [100*((81+85+5+8)/1000)], and the distance between C and B is
7.0 cM [100*((27+30+5+8)/1000)].
Now let's try a problem from Drosophila, by applying the principles we used in the above example. The following
table gives the results we will analyze.
14
Genotype Observed
Type of Gamete
v cv+ ct+
580
Parental
v+ cv ct
592
Parental
v cv ct+
45
Single-crossover between genes ct and cv
v+ cv+ ct
40
Single-crossover between genes ct and cv
v cv ct
89
Single-crossover between genes v and ct
v+ cv+ ct+
94
Single-crossover between genes v and ct
v cv+ ct
3
Double-crossover
v+ cv ct+
5
Double-crossover
Total
1448
Step 1: Determine the parental genotypes.
The most abundant genotypes are the parental types. These genotypes are v cv+ ct+ and v+ cv ct. What is different from our first
three-point cross is that one parent did not contain all of the dominant alleles and the other all of the recessive alleles.
Step 2: Determine the gene order
To determine the gene order, we need the parental genotypes as well as the double crossover genotypes As we mentioned
above, the least frequent genotypes are the double-crossover genotypes. These genotypes are v cv+ ct and v+ cv ct+. From this
information we can determine the order by asking the question: In the double-crossover genotypes, which parental allele is not
associated with the two parental alleles it was associated with in the original parental cross. From the first double crossover, v
cv+ ct, the ct allele is associated with the v and cv+ alleles, two alleles it was not associated with in the original cross. Therefore,
ct is in the middle, and the gene order is v ct cv.
Step 3: Determing the linkage distances.
v - ct distance caluculation. This distance is derived as follows: 100*((89+94+3+5)/1448) = 13.2 cM
ct - cv distance calculation. This distance is derived as follows: 100*((45+40+3+5)/1448) = 6.4 cM
Step 4. Draw the map.
15
Three-point crosses also allows one to measure interference (I) among crossover events within a given region of a
chromosome. Specifically, the amount of double crossover gives an indication if interference occurs. The concept is
that given specific recombination rates in two adjacent chromosomal intervals, the rate of double-crossovers in this
region should be equal to the product of the single crossovers. In the v ct cv example described above, the
recombination frequency was 0.132 between genes v and ct, and the recombination frequency between ct and cv was
0.064. Therefore, we would expect 0.84% [100*(0.132 x 0.64)] double recombinants. With a sample size of 1448,
this would amount to 12 double recombinants. We actually only detected 8.
To measure interference, we first calculate the coefficient of coincidence (c.o.c.) which is the ratio of observed to
expected double recombinants. Interference is then calculated as 1 - c.o.c. The formula is as follows:
For the v ct cvdata, the interference value is 33% [100*(8/12)].
Most often, interference values fall between 0 and 1. Values less than one indicate that interference is occurring in
this region of the chromosome.
16
In Class - Homework Assignment
Answer the following questions:
1. For each of the following crosses, explain with the aid of diagrams the genotype of the parents,
gametes, and offspring. Use Punnett squares where appropriate.
i.
A homozygous, gray bodied (G), long winged (vg+) fruit fly (Gvg+/Gvg+) was crossed with a black
(g), vestigial winged (vg) fly (gvg/gvg).
ii.
If a fruit fly from the F1 generation above was backcrossed to a black, vestigial winged fly, what
phenotypic ratio would you expect, assuming no crossing over.
iii.
If the offspring of the backcross in part ii above consisted of 400 flies, and if crossing over
occurred in 10% of the cases, what ratio of phenotypes would you expect in the offspring?
2. Three recessive genes in linkage group V of the tomato are 'a' producing absences of anthocyanin
pigment, 'hl' producing hairless plants, and 'j' producing jointless fruit stems. Among 3000 progeny
from a trihybrid testcross, the following phenotypes were observed:

259 hairless

268 anthocyaninless, jointless, hairless

40

941 anthocyaninless, hairless

931 jointless

32

260 normal

269 anthocyaninless, jointless
jointless, hairless
anthocyaninless
i.
How were the genes originally linked in the trihybrid parent?
ii.
Estimate the distance between the genes.
17
3. The genes r, s and z are known to be linked in some organism. A testcross was performed between
+++/rsz (writing the genes in arbitrary alphabetical order). The following progeny were obtained.
(since one allele at each locus is always recessive it has been omitted).

rsz - 396

+++ - 406

r++ - 73

+sz - 73

+s+ - 25

r+z - 25

++z - 1
Mapping a Trihybrid
Test cross data Are the genes linked or not linked?
ctct+/gg+/vv+
x
ctct/gg/vv
Gametes:
where ct = wing shape
g = eye colour
v = eye colour
F1:
ctgv
ctgv+
ctg+v
ctg+v+
ct+gv
ct+gv+
ct+g+v
ct+g+v+
ctgv
ctgv
ctgv
ctgv
ctgv
ctgv
ctgv
ctgv
Genotypes
Number
ctgv
ctgv+
ctg+v
ctg+v+
ct+gv
ct+gv+
ct+g+v
ct+g+v+
Total
196
202
51
49
52
49
197
204
1000
ctc/tgg/vv
ctct/gg/vv+
ctct/gg+/vv
ctct/gg+/vv+
ctct+/gg/vv
ctct+/gg/vv+
ctct+/gg+/vv
ctct/gg+/vv+
Percentage of Total
Offspring
19.6
20.2
5.1
4.9
5.2
4.9
19.7
20.4
100
18
Problem:
1:
2:
Construct a gene map;
Give the genotype of the heterozygous trihybrid to show allelic combinations of the two
homologous chromosomes (CIS or TRANS).
Two Point Analysis
Are genes ct & g linked?
Look at % recombinations.
Genotypes/phenotypes
ct/g
ct+/g
ct/g+
ct+/g+
% among total progeny
19.6
5.1
5.2
19.7
+
+
+
+
20.2 = 39.8
4.9 = 10.0
4.9 = 10.1
20.4 = 40.1
Genes do not assort independently (would see a 1:1:1:1 ratio) – therefore, the genes are linked.
Parent types (greatest numbers):
ct/g & ct+/g+
in the CIS configuration
Recombinants (lesser numbers):
ct/g+ & ct+/g
10.1 + 10.0 = 20.1%
Therefore:
distance between ct & g is 20.1 cM.
Are genes ct & v linked?
Look at % recombinations.
Genotypes/phenotypes
ct/v
ct+/v
ct/v+
ct+/v+
% among total progeny
19.6
19.7
20.2
20.4
+
+
+
+
5.1
5.2
4.9
4.9
=
=
=
=
24.7
24.9
25.1
25.3
Genes do assort independently (see a 1:1:1:1 ratio) – therefore, the genes are not linked.
No Recombinants found.
Are genes g & v linked?
Look at % recombinations.
Genotypes/phenotypes
g/v
g/v+
g+/v
g+/v+
% among total progeny
19.6
20.2
19.7
20.4
+
+
+
+
5.2 =
4.9. =
5.1 =
4.9 =
24.8
25.1
24.8
25.3
Genes do assort independently (see a 1:1:1:1 ratio) – therefore, the genes are not linked.
No Recombinants found.
19
Chromosome Mapping
Add a 3rd loci (trait – allele):
ct
g
v
ct
g
v
ct+
g+
v+
ct
g+
v+
20.1 cM
Two different chromosomes
In Class - Homework Assignment
Answer the following questions:
1. A kidney-bean shaped eye is produced by a recessive gene k on the third chromosome of Drosophila.
Orange eye color, called "cardinal", is produced by the recessive gene cd on the same chromosome. Between
these two loci is a third locus with a recessive allele e producing ebony body color. Homozygous kidney,
cardinal females are mated to homozygous ebony males. The trihybrid F1 females are then testcrossed to
produce the F2. Among 4000 F2 progeny are the following:

1761 kidney, cardinal

97 kidney

128 kidney, ebony

8 wild type

1773 ebony

89 ebony, cardinal

138 cardinal

6 kidney, ebony, cardinal
a.
Determine the linkage relationships in the parents and the F1 trihybrids
b.
Estimate the map distances.
20
2. The map distance for six genes in the second linkage group of the silkworm Bombyx mori are shown in the
following table. Construct a genetic map which includes all of these genes.
Gr
Rc
S
Y
P
oa
Gr
-
25
1
19
7
20
Rc
25
-
26
6
32
5
S
1
26
-
20
6
21
Y
19
6
20
-
26
1
P
7
32
6
26
-
27
oa
20
5
21
1
27
-
3. The recessive mutation called "lemon" (le) produces a pale yellow body color in the parasitic wasp Bracon
hebetor. This locus exhibits 12% recombination with a recessive eye mutation called "canteloupe" (c).
Canteloupe shows 14% recombination with a recessive mutation called "long" (l), causing antennal and leg
segments to elongate. Canteloupe is the locus in the middle. A homozygous lemon female is crossed with a
hemizygous long male (males are haploid). The F1 females are then testcrossed to produce the F2. (a)
Diagram the crosses and the expected F1 and F2 female genotypes and phenotypes. (b) Calculate the amount
of wild types expected among the F2 females.
4. The distances between 8 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.
5. 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 heterozygous in cis 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 heterozygous. 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.
6. 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 heterozygous in cis 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 heterozygous. 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.
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7. 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 wx 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 exhibits 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.
8. 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:
Females:
Males:
all wild type
57 garnet, outstretched
60 forked
2 garnet
13 wild type
419 garnet, forked
1 outstretched, forked
439 outstretched
9 outstretched, garnet, forked
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.
9. Maize plants homozygous for the recessive gene "variable sterile" (va) exhibit irregular distribution of
chromosomes during meiosis. Yellowish-green seedlings are the result of another recessive gene called
"virescent" (v). A third recessive called "glossy" (gl) produces shiny leaves. All three of these genes are
linked. Two homozygous plants were crossed and produced an all normal F1. When the F1 was testcrossed,
progeny phenotypes appeared as follows:

60 virescent
4 variable sterile, virescent

48 virescent, glossy
40 variable sterile

7 glossy
62 variable sterile, glossy

270 variable sterile, virescent, glossy
235 wild type
a. What were the genotypes and phenotypes of the original parents?
b. Diagram the linkage relationships in the F1.
c. Determine the gene order.
d. Calculate the amount of recombination observed.
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10. 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.
11. 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:

810 echinus
828 scute, crossveinless

88 crossveinless
89 scute

62 echinus, scute
103 echinus, crossveinless
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:

2207 crossveinless
2125 echinus, cut

273 echinus, crossveinless
265 cut

223 crossveinless, cut
217 echinus

5 wild type
3 echinus, crossveinless, cut
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:

766 vermilion, crossveinless
759 cut

140 vermilion, cut
158 crossveinless

73 vermilion
85, crossveinless, cut

2 wild type
2 vermilion, crossveinless, cut
12. Two loci are known to be in linkage group IV of the rat. Kinky hairs in the coat and vibrissae (long nose
"whickers") 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 the
heterozygous parents which are (a) both in cis phase, (b) both in trans phase, (c) one in cis phase and the
other in trans phase.
13. In mice, the genes frizzy (fr) and albino (c) are linked on chromosome 1 at a distance of 20 map units.
Heterozygous wild type females in trans phase are mated to heterozygous wild type males in cis phase.
Predict the offspring phenotypic expectations.
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