Constructing a Genetic Map
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You should understand how recombination frequency is calculated from testcross data and recognise that recombination frequency reflects the physical distance between loci on a chromosome.
Why this matters
Genetic mapping allows us to order genes and estimate their distances without seeing the DNA sequence.
Before the advent of high-throughput sequencing, geneticists relied on recombination data to construct maps of chromosomes. By analyzing the frequency of recombinant offspring in crosses, we can determine the relative order of genes and the distance between them in map units. This approach, known as linkage mapping, remains fundamental in genetics, particularly for identifying disease genes and understanding chromosomal structure. In this lecture, we move beyond simple two-point crosses to three-point crosses, which allow us to determine gene order and detect double crossovers that would otherwise be missed.
The Logic of Gene Order
Imagine you are trying to determine the order of three towns on a straight road: Town A, Town B, and Town C. If you know the distance between A and B is 10 km, and the distance between B and C is 5 km, you still do not know if the order is A-B-C or C-B-A. However, if you also measure the distance between A and C, the logic becomes clear. If A to C is 15 km, then B must be in the middle, because 10 + 5 = 15. If A to C were only 5 km, then A and C would be close together, and B would be at one end. In genetics, we use recombination frequency as our "distance" measure. Recombination occurs when crossing over happens between two loci during meiosis. The closer two loci are, the less likely a crossover event will occur between them, and the lower the recombination frequency. The further apart they are, the higher the chance of a crossover, and thus the higher the recombination frequency. However, there is a limit: recombination frequency cannot exceed 50%, because even if loci are on opposite ends of a chromosome, multiple crossovers can restore the parental configuration. Therefore, recombination frequency is a proxy for distance, but it is not a direct linear measurement over long distances. For closely linked genes, where double crossovers are rare, the recombination frequency is approximately equal to the map distance in centimorgans (cM). This principle allows us to use three-point crosses to deduce the order of three genes by identifying which pair has the highest recombination frequency, which usually indicates the outermost genes, with the third gene lying in between.
Three-Point Crosses and Gene Order
To determine the order of three genes, we perform a trihybrid cross, typically using a heterozygous parent crossed with a homozygous recessive tester. The offspring phenotypes reveal the gamete types produced by the heterozygous parent. We categorize these gametes into parental types (most frequent), single crossovers (intermediate frequency), and double crossovers (least frequent). The key to determining gene order lies in the double crossover class. In a double crossover, the middle gene switches its allele relative to the two outer genes, while the outer genes retain their parental linkage. By comparing the double crossover phenotype to the parental phenotype, we can identify which gene is in the middle. For example, if the parental genotype is \( A B C / a b c \), and the double crossover offspring show the phenotype \( A b C \) (and \( a B c \)), then gene \( B \) is in the middle, because it has switched relative to \( A \) and \( C \). Once the order is established, we calculate the recombination frequencies for the two intervals. The distance between gene 1 and gene 2 is the sum of the single crossover class 1 and the double crossover class, divided by the total number of offspring, multiplied by 100. Similarly, the distance between gene 2 and gene 3 is the sum of the single crossover class 2 and the double crossover class, divided by the total, multiplied by 100. This gives us the map distances in centimorgans.
It is crucial to include the double crossover class in both interval calculations because these events represent crossovers in both intervals. Excluding them would underestimate the true map distance. This method relies on the assumption that crossovers in one interval do not influence crossovers in another, a concept known as independence, which is often modified by a phenomenon called interference.
Interference and Coefficient of Coincidence
In reality, crossovers are not entirely independent events. The occurrence of one crossover often reduces the probability of another crossover occurring nearby. This phenomenon is called interference. To quantify this, we calculate the expected number of double crossovers if the two intervals were independent. This is done by multiplying the recombination frequencies of the two intervals and the total number of offspring. We then compare this expected number to the observed number of double crossovers. The ratio of the observed double crossovers to the expected double crossovers is called the coefficient of coincidence (\( c \)). If \( c = 1 \), there is no interference; crossovers occur independently. If \( c < 1 \), there is positive interference, meaning fewer double crossovers were observed than expected. Interference is calculated as \( I = 1 - c \). A high value of interference indicates that a crossover in one region strongly suppresses crossovers in the adjacent region. This is biologically significant because it reflects the physical mechanics of the synaptonemal complex during meiosis, where the formation of one chiasma may mechanically hinder the formation of another nearby. Understanding interference is essential for accurate genetic mapping, especially when mapping genes that are moderately far apart, as it helps correct for the underestimation of map distances that occurs when double crossovers are missed or suppressed.
For example, if the recombination frequency between genes A and B is 0.10 and between B and C is 0.20, the expected frequency of double crossovers is \( 0.10 \times 0.20 = 0.02 \). If the observed frequency is 0.01, then \( c = 0.01 / 0.02 = 0.5 \), and interference \( I = 0.5 \). This means 50% of the expected double crossovers were suppressed.
Limitations of Recombination Mapping
While recombination mapping is powerful, it has inherent limitations. The most significant is the 50% recombination frequency ceiling. Loci that are far apart on the same chromosome, or on different chromosomes, will both show a recombination frequency of approximately 50%. This makes it impossible to distinguish between unlinked genes on different chromosomes and unlinked genes on the same chromosome using recombination data alone. To resolve this, geneticists use additional markers or physical mapping techniques. Another limitation is the non-linearity of recombination frequency with physical distance. Over short distances, recombination frequency is roughly proportional to physical distance. However, over longer distances, the relationship saturates due to multiple crossovers. This means that a recombination frequency of 50% does not imply an infinite distance, but rather a distance where crossovers are so frequent that the parental and recombinant classes are equally likely. Furthermore, recombination rates vary across the genome. Some regions are "hotspots" with high recombination rates, while others are "coldspots" with low rates. This variation means that map distance in centimorgans does not always correlate directly with physical distance in base pairs. Despite these limitations, genetic maps remain a crucial tool for identifying the chromosomal location of genes associated with specific traits or diseases, providing a framework for subsequent molecular analysis.
The concept
Genetic linkage and recombination
Nearby genes tend to be inherited together.
A common misconception is that a recombination frequency of 50% always means the genes are on different chromosomes. In fact, genes on the same chromosome can also show 50% recombination if they are far enough apart that multiple crossovers occur frequently. The key distinction is that unlinked genes on different chromosomes always show 50% recombination, while linked genes on the same chromosome show 50% recombination only when the distance is large enough to allow for independent assortment due to multiple crossovers. C
Worked example
Problem. In a three-point cross, the following offspring phenotypes and counts were observed: Parental types: \( A B C \) (400), \( a b c \) (400). Single crossovers between 1 and 2: \( A b c \) (50), \( a B C \) (50). Single crossovers between 2 and 3: \( A B c \) (30), \( a b C \) (30). Double crossovers: \( A b C \) (5), \( a B c \) (5). Total offspring: 970. Determine the gene order and map distances.
- Identify the parental and double crossover classes. The parental classes are \( A B C \) and \( a b c \). The double crossover classes are \( A b C \) and \( a B c \).
- Determine the gene order by comparing the double crossover phenotype to the parental phenotype. In the double crossover class \( A b C \), the middle allele \( b \) is different from the parental \( B \), while the outer alleles \( A \) and \( C \) are the same. This indicates that gene \( B \) is in the middle. The order is \( A - B - C \).
- Calculate the recombination frequency for the interval between \( A \) and \( B \). This includes single crossovers between 1 and 2 and double crossovers. \( RF_{A-B} = \frac{50 + 50 + 5 + 5}{970} = \frac{110}{970} \approx 0.1134 \). The map distance is 11.34 cM.
- Calculate the recombination frequency for the interval between \( B \) and \( C \). This includes single crossovers between 2 and 3 and double crossovers. \( RF_{B-C} = \frac{30 + 30 + 5 + 5}{970} = \frac{70}{970} \approx 0.0722 \). The map distance is 7.22 cM.
- Calculate the coefficient of coincidence. Expected DCO frequency = \( 0.1134 \times 0.0722 \approx 0.00819 \). Expected DCO count = \( 0.00819 \times 970 \approx 7.94 \). Observed DCO count = 10. \( c = \frac{10}{7.94} \approx 1.26 \). Since \( c > 1 \), there is negative interference, which is unusual but possible in some organisms or regions. Interference \( I = 1 - 1.26 = -0.26 \).
Units check. The map distances are expressed in centimorgans (cM), which is the standard unit for genetic map distance, representing 1% recombination frequency.
Three quick tests
- What is the maximum recombination frequency between two genes on the same chromosome?
Answer
50%, because multiple crossovers can restore the parental configuration, making the recombination frequency appear as if the genes are unlinked. - If the coefficient of coincidence is 0, what does this imply about interference?
Answer
It implies complete positive interference, meaning that a crossover in one interval completely prevents a crossover in the adjacent interval, so no double crossovers are observed. - Why is it necessary to include double crossover classes in the calculation of map distances?
Answer
Because double crossovers represent recombination events in both intervals. Excluding them would underestimate the true map distance between the outer genes and the middle gene.
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Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Concepts cited: T-015 · Threads: information, evolution