11
map distance = ------ x 100 = 1.1 m.u. or 1.1 cM
1000
b) 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,ro and +,+) will occur with a combined frequency of 0.011, which when divided evenly between them means that the frequency of occurrence for gro,ro is 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 for gro,+ is 0.4945 and that the frequency of occurrence for +,ro is 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 0.5000
Based upon the expected frequency of occurrence of the flies and assuming 1000 offspring, we would expect to see in the offspring of a dihybrid cross
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 201 223
What is the linkage arrangement of these loci?
abnormal brown 416
abnormal 88
brown 72
wildtype 424
That 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
160
map distance = ------ x 100 = 16 cM
1000
P. 134, #3. In Drosophila, 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 total
inflated, warty 225 225 450
inflated 25 25 50
warty 25 25 50
wildtype 225 225 450
If 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?
For the parental classes these are:
got-s,amy-s,sdh-s and got-f,amy-f,sdh-f
For the double crossover classes these are:
got-f,amy-s,sdh-f and got-s,amy-f,sdh-s
A comparison of the gametes from the parental classes with the double crossover classes shows that the locus amy is in the middle.
number of recombinants between
got-amy amy-sdh got-sdh
---------------------------------------
11 58 11
14 53 14
1 1 58
1 1 53
----------------------------------------
totals 27 113 136
To 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.
27
distance (got-amy) = ------ x 100 = 2.7 m.u.
1000
113
distance (amy-sdh) = ------ x 100 = 11.3 m.u.
1000
distance (got-sdh) = 2.7 + 11.3 = 14.0
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 357 hotfoot, obese 74 waved 66 obese 79 wildtype 343 hotfoot, waved 61 obese, waved 11 hotfoot 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?
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. number of recombinants between
obese-hotfoot hotfoot-waved obese-waved
-----------------------------------------------
79 74 79
61 66 61
11 11 74
9 9 66
-----------------------------------------------
totals 160 160 160
To 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.
160
distance (obese-hotfoot) = ------ x 100 = 16.0 m.u.
1000
160
distance (hotfoot-waved) = ------ x 100 = 16.0 m.u.
1000
distance (obese-waved) = 16.0 + 16.0 = 32.0
P. 136, #23. In people, the ABO system (A, B, O alleles) 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 a lod score 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 the B allele 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 allele al+. In this case the A allele from the mother is linked to an al+ 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 the A allele without any recombination and 3 inherit the chromosome with the B allele 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
# recombinants
map distance = ---------------- x 100
total #
2
map distance = ---- x 100 = 20 cM
10
To calculate the lod score 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))
0.00000655360
lod = log ------------- = log 6.872 = 0.83
0.00000095368
Any lod score greater than zero indicates linkage, but it is considered strong evidence for linkage when the lod score 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 44
a) 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 + + R
The 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 +
---------------
---------------
+ R
In 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 recombinants
map distance = --------------------------- x 100
total number individuals
86
map distance = ----- x 100 = 28.7 m.u.
300
P. 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 59
a. Determine which genes are linked.
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 = 69
These 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. Tunicate, Glossy 58 + 55 = 113
Tunicate, nonglossy 15 + 13 = 28
nontunicate, Glossy 16 + 14 = 30
nontunicate, nonglossy 53 + 59 = 112
These 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.
T G L
---------------- ----------
---------------- ----------
t g l
The T and G alleles 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 = 112
From 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.
283P. 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 wildtype
a. Determine the chromosomal composition of the F1 females.
b. Derive a map of the three genes.
For the parental classes these are:
kidney, cardinal and ebony
For the double crossover classes these are:
kidney, ebony, cardinal and wildtype
A 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.
number of recombinants between
kidney-ebony ebony-cardinal kidney-cardinal
-----------------------------------------------------
64 49 49
67 46 46
3 3 64
4 4 67
------------------------------------------------------
totals 138 102 136
To 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.
138
distance (kidney-ebony) = ------ x 100 = 6.9 m.u.
2000
102
distance (ebony-cardinal) = ------ x 100 = 5.1 m.u.
2000
distance (kidney-cardinal) = 6.9 + 5.1 = 12.0
The 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:
Estimate 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 + - - + -
-------------------------------------------------
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 - - + - +
----------------------------------
Last updated on 22 August 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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