Topic 10: Mapping Flashcards
Define the relationship between recombination frequency and the distance between genes on a homologous chromosome
You can use the RF to map the locations of different loci on chromosomes.
When two genes are located very close to each other on a chromosome, they are less likely to be separated by crossing over during meiosis. This results in a lower recombination frequency between them. On the other hand, if two genes are located far apart on a chromosome, they are more likely to be separated by crossing over during meiosis. This results in a higher recombination frequency between them.
Genetic Map
Generated using RF; more of a measure of relation relative between loci.
Physical Map
Normally measured in basepairs (bp); this is the actual physical distance measured in bp.
Map Units
Centimorgans (cM) = RF 1% = 1cM
How we do mapping using genetic maps
How do you map using genetic frequencies when recombination rates are additive?
NEED TO KNOW HOW TO GET THE RF, THEN DO THE MATH.
Here we have 3 loci, and we have the 3 RF.
Step 1. Draw a line representing the chromosome.
Step 2. Place the two loci with the smallest recombination frequency (i.e., the highest degree of linkage) at the ends of the chromosome. For example, if loci A and B have a recombination frequency of 5%, we can place them 5 cM apart on the chromosome.
Step 3. Place the third locus (C) relative to the first two loci based on their respective recombination frequencies. For example, if the recombination frequency between loci A and C is 15% and the recombination frequency between loci B and C is 10%, then we can place locus C closer to locus A, at a distance of 15 cM from A and 10 cM from B.
Step 4. The relative distances between the loci are generally additive, meaning that the distance between A and B plus the distance between B and C should equal the distance between A and C. If the distances do not add up, we may need to adjust the placement of the loci on the chromosome.
Step 5. Repeat this process for additional loci if necessary.
There are two ways to draw this, but the relative distances must be maintained.
Add another locus to this genetic map:
D is farthest from C, and closest to A.
We can not distinguish between unlinked genes on different chromosomes and unlinked genes that are just very far apart on the same chromosomes. Why?
When two genes are located on different chromosomes, they are considered unlinked, meaning they will assort independently during meiosis. As a result, the inheritance pattern of these genes will show a 50% parental (non-recombinant) and 50% recombinant ratio in their progeny.
- if they are very far apart, they will sort independently of each other
- when analyzing the inheritance pattern of two genes that are far apart on the same chromosome, they may appear to be unlinked, as if they were located on different chromosomes. This is because they will show a 50% parental and 50% recombinant ratio in their progeny, just like unlinked genes on different chromosomes. However, in reality, these genes are still physically located on the same chromosome, but are just very far apart from each other.
The mapping distance will be underestimated due to the double cross-overs between far-apart but linked genes. Why?
- we frequently have single crossing-over events
- but the farther apart, the more frequently we will have double crossing over
- in the image, we have recombinant chromosomes, but the loci did not get recombined
When two genes are far apart on a chromosome, it is more likely that multiple crossover events will occur between them during meiosis. A single crossover event between two genes results in a recombinant chromosome, where the alleles of the two genes are exchanged. However, if two crossover events occur between two genes, the alleles of the two genes will not be exchanged, resulting in a non-recombinant chromosome. This means that the two genes will appear to be linked and will not contribute to estimating their genetic distance.
The likelihood of double crossovers increases as the physical distance between two genes on the chromosome increases. This means that the frequency of recombination between the two genes will be underestimated, and the genetic distance between them will appear to be shorter than it actually is.
Example: make a genetic map with two-point test crosses
For gene loci in testcross, we are working with 4 loci and doing pairwise crosses with all the different loci. Two-point testcross is done with all the loci in a pairwise fashion.
Step 1: Determine the linkage groups by analyzing the recombination frequencies between the loci. Loci that have a recombination frequency of 50% or more are considered unlinked and are placed in different linkage groups. Loci that have a recombination frequency of less than 50% are considered linked and are placed in the same linkage group. In your example, loci A is unlinked with B, C, and D, while loci B, C, and D are linked.
Step 2: Draw a line representing the chromosome for each linkage group.
Step 3: Place one locus on each line based on the order of the linkage group. For example, in the first linkage group containing loci A, you would place loci A. In the second linkage group containing loci B, C, and D, you would place one of the linked loci, such as B.
Step 4: Place the other loci on the appropriate linkage group line, based on their distance from the first locus placed. For example, if loci B and D have a recombination frequency of 10%, you would place loci D closer to B on the second linkage group line.
Step 5: Determine the mapping distance between the loci by adding up the distances between them on the chromosome. However, keep in mind that the mapping distance may be underestimated if the loci are far apart, and it’s better to use the distances determined by test crosses between closer loci.
In this worked example, how would you determine how many progenies will have the Ac phenotype?
Step 1: In the parental cross, we have a homozygous parent for both dominant alleles and the other parent for both recessive alleles. This is going to give us an F1 generation that is dihybrid.
Step 2: In the test cross, we have a heterozygous parent for both alleles and the other for both recessive alleles. Normal for mapping and figuring out our recombination rates.
Step 3: We need to figure out the genotypes, do this by determining what the parental and recombinant progeny will be
- AC/ac (parental)
- ac/ac (parental)
- Ac/ac (recombinants)
- aC/ac (recombinants)
Step 4: Because we know that our recombination rate (RF) is 3%, we should have 3% of the recombinants in the progeny and 97% of the parental. Therefore, since 1000 testcross progeny are obtained, we have
- (0.97)*(1000) = 970 of the progeny will have the parental genotype
- (0.03)(1000) = 30 of the progeny will have the recombinant genotype
Step 5: The question is asking for the Ac phenotype. Therefore, looking at the recombinant progeny, we see that the Ac/ac will give us that phenotype. Therefore
- 30/2 = 15
- We expect 15 of the individuals from this cross to have that phenotype
Constructing a genetic map with the three-point testcross, we like this because:
- The more efficient mapping technique
- The order of the three genes can be established in a single set of progeny
- Some double crossovers can usually be detected
- Provides more accurate map distances
What is a three-point testcross, and what is the theory?
In the parental cross, we have a homozygous dominant ABC/ABC individual crossed with a homozygous recessive abc/abc individual. This will give us F1 progeny that are all heterozygous AbC/aBc.
We have three different options for how crossing can happen between these loci, therefore three different groups of recombinants in the progeny.
- Option 1: crossing over between A and B; this shuffles the A allele
- Option 2: crossing over between C and B; this shuffles the C allele
- Option 3: the only way to get a shuffling between the B loci against the A and C loci (i..e, just the middle allele) is a double cross over
Single crossover events are more common and will occur more frequently in the same meiosis than double crossing-over events. When we look at the progeny in F2, and we look at all the recombinant progeny, the class of recombinant progeny that is the rarest, are the ones that are going to be representing the double crossover events that are happening by analyzing those ones, we can figure out which allele is in the middle.
For example, with genetic mapping using a three-point testcross:
- st = eye colour
- e = body colour
- ss = body spines
Stepping up the cross:
- P1 = homozygous for dominant alleles
- P2 - homozygous for all recessive alleles
- When we cross P1 and P2, we know in our F1, we will have one chromosome that has all the wildtype alleles (denoted by +) linked together. And on the other chromosome, we will have all the
recessive alleles linked together.
- We do not know the actual order of the loci, so they are written in random order.
- In our F1, we know that the alleles at the three different loci are linked with each other. Therefore we have easily identified our recombinants by looking at their phenotypes.
When looking ar our F1 progeny, we see that there is a group presenting a phenotype that is happening far less than the others, therefore it must be the one representing the double cross. THE RAREST COMBINATION.
Here are the steps to determine genetic mapping using a three-point testcross / Steps in determining gene order in a three-point cross:
The steps look mostly correct, but here are some suggestions to make them more clear and accurate:
Step 1: Cross a homozygous dominant parent with a homozygous recessive parent and observe the phenotype of the F1 progeny. The F1 progeny will be heterozygous for all three loci.
Step 2: Cross the F1 progeny with a homozygous recessive parent to perform a three-point testcross.
Step 3: Observe the phenotype of the testcross progeny and record the number of each phenotype.
- Identify the nonrecombinant progeny (the two most numerous phenotypes) and the double-crossover progeny (the two least numerous phenotypes).
Step 4: Compare the phenotypes of the double-crossover progeny with the phenotypes of the nonrecombinant progeny. The characteristic that differs between the double-crossover and nonrecombinant progeny is the characteristic encoded by the middle gene.
Step 5: Determine the recombination frequency (RF) between each pair of loci by calculating the percentage of recombinant progeny for each pair. Use the formula RF = (number of recombinant progeny/total number of progeny) x 100.
Step 6: Use the double-crossover progeny to establish the order of the loci. The locus that differs between the double-crossover and nonrecombinant progeny is in the middle.
Step 7: Calculate the genetic distances (in cM) between the loci using the RF values. Use the formula genetic distance = (RF between two adjacent loci/2) x 100.
Note: Using the correct terminology when describing the results is important. The “rarest phenotype” is not always the double-crossover progeny; it’s the phenotype that represents the middle locus that is different between the double-crossover and nonrecombinant progeny.
Here is an example using the A, B, and C loci - with steps to determine genetic mapping using a three-point testcross:
Step 1: The F1 progeny from the cross between the homozygous dominant and homozygous recessive parents all have the wildtype phenotype (ABC/abc).
Step 2: Cross the F1 progeny with the homozygous recessive parent (abc/abc).
Step 3: Observe the phenotype of the testcross progeny:
ABC/abc: 80
AbC/abc: 9
aBC/abc: 14
Abc/abc: 174
ABc/abc: 15
Abc/aBc: 173
aBC/aBc: 11
abc/aBc: 514
Step 4: Determine the RF between each pair of loci:
A and B: (9+15)/1000 = 0.024 or 2.4%
B and C: (9+11)/1000 = 0.02 or 2%
A and C: (15+11)/1000 = 0.026 or 2.6%
Step 5: Use the double-crossover progeny (Abc/aBc and aBC/aBc) to establish the order of the loci. In this case, Abc/aBc represents a double crossover between A and B, and aBC/aBc represents a double crossover between B and C. Therefore, the order of the loci is A-B-C.
Step 6: Calculate the genetic distances between the loci using the RF values.
The distance between A and B is 2.4 cM.
The distance between B and C is 2.0 cM.
The distance between A and C is the sum of the distances between A and B and between B and C, which is 4.4 cM.
Therefore, the genetic map for these three loci is:
A —- 2.4 cM —- B —- 2.0 cM —- C
For example, with genetic mapping using a three-point testcross:
- st = eye colour
- e = body colour
- ss = body spines
What does this picture represent?
It represents all the potential double cross events, the one that matches the more rarest RF represented the actual double cross over event.
The rarest phenotype represents the double crossover event because it is the least likely to occur due to the two recombination events required for it to happen. In a three-point testcross, we are trying to determine the order of three loci on a chromosome, and we know that there will be a higher frequency of the parental (non-recombinant) progeny compared to the recombinant progeny. The double crossover events require two recombination events to happen, and thus are less frequent than single crossover events, which only require one recombination event. As a result, the double crossover progeny will be the least frequent among all the recombinant progeny, and will therefore represent the middle locus in the genetic map.
