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:
AB
Bivalent
ABAlleles in the CIS position (COUPLING phase)
4 chromatidsab
(tetrad)
ab
Ab
Bivalent
AbAlleles in the TRANS position (REPULSION phase)
4 chromatidsaB
(tetrad)
aB
Cross overs (Recombinants):further apart the genes – the greaterthe opportunity for crossing over to occur between gene loci (alleles)
Seen when the crossover forms between the gene loci under consideration
Example:
AB
AB
AB
AB
crossover
abab
ab
ab
no crossing over seen in the above example because the cross over did not occur between the two gene loci that were being observed.
Single Crossover:
AB
AB
Ab
AB
Double crossover
abaB
ab
ab
single crossover was detected in the above example as a change in the pair of alleles found on the same chromosome.
Multiple Crossovers:
AB
AB
AB
AB
Double crossover
abab
ab
ab
double crossover was not detected in the above example.
Example question:
A cross occurred between the following two organisms:
AB/abxab/ab( / - demonstrates linkage between alleles)
Of 100 progeny (offspring), the following were found:
34Ab/ab34aB/abtotal of 68%
16ab/ab16AB/abtotal of 32%
total = 100%
Answer:
The parental gametes always appear with greatest frequency
Therefore:parents:Ab/aBxadb/adb
Individual:Ab/aB (gametes produced)
Single crossover between A & b(20% - 20 offspring out of 100 show the new combination of traits)
Ab
Ab
AB
Ab
aBab
aB
aB
32% - 32 offspring out of 100 show the new combination of traits
Add a 3rd loci (trait – allele):
ADB
ADB
AdB
ADB
adbaDb
adb
adb
double crossovers between AB/ab can now be detected in the above example containing a 3rd loci.
Example question:
A cross occurred between the following two organisms:
ADB/adbxadb/adb( / - demonstrates linkage)
Of 100 progeny (offspring), the following were found:
34ADB/adb34adb/adbtotal of 68%
10Adb/adb10aDB/adbtotal of 20%
5Adb/adb5aDB/adbtotal of 10%
1AbB/adb1aDb./adbtotal of 2%
total = 100%
Answer:
The parental gametes always appear with greatest frequency
Therefore:parents:ADB/adbxadb/adb
Individual:ADB/adb (gametes produced)
Single crossover between A & D(20% - 20 offspring out of 100 show the new combination of traits)
ADB
ADB
Adb
ADB
adbaDB
adb
adb
Single crossover between D & B(10% - 10 offspring out of 100 show the new combination of traits)
ADB
ADB
ADb
ADB
adbadB
adb
adb
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)
ADB
ADB
AdB
ADB
adbaDb
adb
adb
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.
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.
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:Linear order of genes
2:relative distances between two loci on the chromosome
Map Distance: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
Example 1:
Ab/aB
8% AB8% ab42% Ab42% aB
Recombinant Parental
8% AB + 8% ab (recombinants) = 16% map distance (cM)A16 cMB
Example 2:
6% Bc + 6% cB (recombinants) = 12% map distance (cM)B12 cM C
BC/bc
6% Bc6% bC44% BC44% bc
Recombinant Parental
Mapping Trihybrid[2n gamete varieties (N = number of genes)
23 = 8 varieties]
Test cross of dataQqRrTtxqqrrtt
Written for linked genes or not linked genes – can differentiate by the number of offspring
Row / Genotype of Heterozygous Gametes / Number of Offspring Counted / % of Total Offspring1 / qrt / 12 / 0.1
2 / qrT / 4186 / 37.2
3 / qRt / 750 / 6.7
4 / Qrt / 575 / 5.1
5 / QRt / 4250 / 37.8
6 / QrT / 810 / 7.2
7 / qRT / 650 / 5.8
8 / QRT / 14 / 0.1
Totals / 11247 / 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)
Two Point Analysis:
Are q and r linked?% recombination?
Phenotypes / Data Rows / % Total Amongst Progenyqr / 1 & 2 / 0.1 + 37.2 = 37.3
qR / 3 & 7 / 6.7 + 5.8 = 12.5
Qr / 4 & 6 / 5.1 + 7.2 = 12.3
QR / 5 & 8 / 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 / Data Rows / % Total Amongst Progenyrt / 1 & 4 / 0.1 + 5.1 = 5.2
rT / 2 & 6 / 37.2 + 7.2 = 44.4
Rt / 3 & 5 / 6.7 + 37.8 = 44.5
RT / 7 & 8 / 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 / Data Rows / % Total Amongst Progenyqt / 1 & 3 / 0.1 + 6.7 = 6.8
qT / 2 & 7 / 37.2 + 5.8 = 43.0
Qt / 4 & 5 / 5.1 + 37.8 = 42.9
QT / 6 & 8 / 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)
Homologous pairing of chromosomes:
14.1 cM
qTr
qTr
24.8 cM
QtR
QtR
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.
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 vgfrom these experiments is 13.0 cM [(303/2335)*100].
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 thelinkage 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.
Genotype / Observed / Type of GameteABC / 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.
Genotype / Observed / Type of Gametev 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 ctcv.
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.
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 ctcv example described above, the recombination frequency was 0.132 between genes v and ct, and the recombination frequency between ct andcv 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.