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Help with gene linkage and mapping!

Help with gene linkage and mapping!


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Here's an interesting question I've been trying to figure out:

My annotations are in red. So far, I've figured out:

  • We know 630 colonies survived when plated with ACD antibiotics… meaning they picked up the a/c/d resistance alleles, but remain wild-type for the b allele. The genotype would be "a b+ c d". Because SO MANY (630!) colonies seemed to inherit a/c/d together, and b completely separate from them, I would say b is the gene quite distant from the 3 other genes.

However, I've unsure about how to proceed and determine the probably order of the 3 others genes: a, c, and d.

Here's my hunch: B--------------------C-D-A.

I'll explain:

  • Well, I looked and saw 942 (many!!!) picked up A/D together. So A/D are probably adjacent.

  • Then, 786 (many, but fewer) picked up C/D together. So C is further from D than A is far from D. The question becomes: is it "C---A-D" or is it "A-D---C"?

  • The insight comes in with 640 (still fewer) picked up A/C together. That means A is actually quite FAR from C, because more distance = more recombination events that could split them up. So between the "C---A-D" or "A-D---C" listed above, it's more likely "A-D---C".

  • Okay! We know B is far from the other ones, BUT is it like "B-------ADC", or is it like "ADC-------B"? Well, let's look: picked up B/C together is 51, but picked up A/B together is only 46. That means A is a little FURTHER from B than C is further from B, since 46 < 51.

    • This led me to my proposed answer: that the order is "ADC---------B", or if you just flip it around, its equivalent: "B---------CDA".

How's my reasoning? This stuff can be pretty confusing; am I approaching it from the right angles? Thanks so much; really appreciate it!


I think your reasoning is sound and your ordering very likely, but I wouldn't feel comfortable with deciding the side B is on. Actually, it's more than that: I think that by using basically the same number (~40) for everything that includes B, they want you to come to another conclusion.

Unlike eukaryotic cells undergoing meiosis, there is no crossing over in bacteria taking up DNA from the surroundings. Therefore, it shouldn't be possible for the cell to inherit A, B and C but not D, which is in the middle. Simply, from my understanding of the "experimental" setup, they added antibiotics A, B and C, so they have to assume the bacteria are wild type for D. However, it's basically for sure that those ABC-resistant colonies are also D-resistant.

Still, your reasoning is correct: comparing BC and AB should give you the side B is on, since the break could have been inside of the 3-gene cluster. However, the numbers are very close, and the "AB" and "BC" phenotypes are probably ABCD anyway, making the small difference even less meaningful (the true AB and BC are lost because they are mixed with the many ABCD).


You were never told that all the genes were on a single plasmid. I think you should treat all those counts < 50 as essentially zero; cells pretty much never get both the ACD and the B plasmid; they get one or the other.


Steps in Gene Tagging and Mapping: 8 Steps

In this article we will discuss about the eight main steps that are involved in gene tagging and mapping.

Step # 1. Selection of Target Traits:

This is the first step. Traits can be qualitatively or quantitatively inherited. Qualitatively inherited traits are controlled by one or few genes which have major effect on a particular trait and follow a typical Mendelian segregation. They are not influenced by environment and genetic background.

Examples include nematode resistance in tomato, TYLCV resistance in tomato, gall midge resistance and bacterial leaf blight resistance in rice. Quantitatively inherited traits are controlled by many genes/loci. Each locus has a small effect on the trait and cumulative effect of alleles at all loci controlling the trait determines the trait expression.

These traits show a continuous variation in segregating populations and are highly influenced by the environment and the genetic background. They are difficult to tag and map. Examples include several characters of economic importance like yield, drought tolerance, quality, etc.

Step # 2. Identification of Parents Differing in the Trait of Interest:

For success of gene tagging, one needs at-least two parents differing for the alternative forms of the trait of the interest. For example, for TYLCV gene tagging and mapping, there should be two parental lines one having gene for resistance to TYLCV and the other having gene for susceptibility for TYLCV.

For quantitatively inherited traits, a single cultivar can be selected as donor line with two or more recipient lines, which should essentially not possess the target trait for example drought tolerance.

Step # 3. Development of Appropriate Population Segregating for the Trait of Interest:

In case of quantitatively inherited traits, the early generation segregating populations like F2, F3, BC1F1 can be used. Geneticists also prefer to use advanced generation materials like F6, F7, recombinant inbred lines (RILs), near isogenic lines (NILs), or double haploid lines (DHLs) since they are homozygous for all the loci analyzed. RILs are usually, at F8/F9 generation onwards.

They are developed through single seed descent method or pedigree method. NILs are developed by repeated backcrossing of the F1S, BC1F1s, etc., with the recipient line also called as recurrent parent while simultaneously selecting for the trait of interest at each backcross generation. NILs will have only one or few genomic regions from the donor line and rest of the genome will be identical to the recipient/recurrent parent.

DHLs are a special class of population which are identical to RILs in their genetic constitution. DHLs are obtained by microspore culture of F1 anthers which give rise to haploid plants followed by induced doubling of chromosomes of haploid plants to yield double haploids. For simply inherited traits, F2 population size is 150-300 plants.

The schematic illustration given below, depicts the possible Ways of developing different kinds of mapping populations. For mapping quantitatively inherited traits, advanced generation materials like NILs, RILs and DHLs are the most appropriate.

Breeders/geneticists also use a strategy called AB-QTL (advanced backcross-QTL) strategy wherein through backcrossing the population is developed and simultaneously phenotyped to identify co-segregating markers.

Step # 4. Screening the Population for the Target Trait (Phenotyping):

The method of phenotyping differs significantly between qualitatively and quantitatively inherited traits. For most of the qualitatively inherited traits like pest and disease resistance, phenotyping involves exposing the individuals of the population to a particular biotype/pathotype of the pest/disease and scoring the plants for resistance/susceptibility after a particular time interval.

Adequate care should be taken while phenotyping since the success of tagging and mapping efforts depends mainly on precise phenotyping. For quantitative traits, the process of phenotyping involves analysis of individual component characters that contribute towards the overall expression of the target trait.

For example when tagging and mapping QTLs for yield, it is necessary to phenotype individual components of yield. Performing the experiment in replicated multi-location trials helps to avoid the uncertainties induced by the environment.

Step # 5. Parental Polymorphism Survey with Markers and Identification of Markers that Co-segregate with Gene(s) of Interest in the Individuals Constituting the Population:

After developing the population and phenotyping, the next step is to identify markers that co-segregate with trait of interest. This requires analysing the polymorphism among the parental lines with molecular markers.

Usually, if mapped and co-dominant markers like SSRs are used, it is necessary to scan the parental lines with a set of uniformly spaces SSR markers (12-16 per chromosome) and identify at-least 6-8 polymorphic markers per chromosome.

Care should also be taken to ensure that the polymorphic markers on a chromosome are uniformly distributed. Just to give an example more than 20,000 SSR markers spread evenly across the rice genome are available and it is hoped that the process of tagging and mapping of agronomically important genes will become much easier due to the availability of a large number of markers and the time taken is also expected to reduce significantly. If dominant markers like RAPD and ISSRs are selected for the study, then the parental polymorphism survey should be done with as much markers as possible.

Once a set of markers polymorphic between the parental lines has been identified, the next step is to carry out co-segregation analysis for these markers. A simple strategy called ‘bulked- segregant analysis’ can be used to quickly identify markers, which co-segregate with trait of interest.

For example, A set of resistant and susceptible F2 lines (usually 10-15 lines in each case) are bulked separately and analysed with parental polymorphic markers. If a fragment (hereafter called as marker) is present in the resistant donor, absent in the susceptible recipient, present in the resistant bulk and absent in the susceptible bulk, then the marker is most probably associated with resistance.

The marker is then analysed individually in all the lines constituting the resistant and susceptible bulks. If the marker is present in a majority (>70%) of individuals constituting resistant bulk and absent in a majority of individuals constituting the susceptible bulk, then it can be assumed that the marker is linked to the resistance.

The next step is to perform co-segregation analysis with all the individuals constituting the population and then determine linkage distances based on the extent of resistant individuals showing amplification of the resistance linked marker. In a similar way, markers co-segregating with susceptibility can also be identified.

Once a fragment/band (marker) is observed to co-segregate with resistance or susceptibility in the individuals of the mapping population, then the next step is to undertake co-segregation analysis in all the individuals constituting the population, (i.e. to check for presence of the marker in the resistant or susceptible individuals of the population).

Once a marker is confirmed to clearly co-segregate with trait phenotype in a majority of individuals of the population, the next step is to identify its chromosomal location.

Thus, if an SSR marker is identified to tightly co- segregate with trait phenotype in a population, then the tentative chromosomal location of the gene controlling the trait can be easily identified and more SSR markers in the vicinity of the co-segregating marker can be used for locating the exact position of the gene.

Step # 6. Construction of Linkage Map:

Linkage maps are basically, a kind of “road map” of the chromosomes drawn based on segregation pattern of markers. They indicate the position and relative genetic distances between markers along chromosomes, which is quite analogous to signs or landmarks along a highway. A clear co-segregation data is required to draw linkage maps.

Based on the co-segregation patterns, percentage recombination is calculated for each pair of markers in terms of centimorgans (cM), which is the unit for linkage distance. Statistical software’s like ‘Mapmaker’, ‘Map manager’, ‘Join Map’, ‘Cartographer’ and ‘Linkage’ can be used for construction of linkage maps. The co-segregation data has to be fed to the computer to excel format and the software automatically, constructs linkage map after calculating ‘LOD’ score for each pair of markers.

For constructing linkage maps for quantitative traits, usually marker intervals showing association with trait phenotype are identified and based on the extent to which these intervals are showing association, linkage distances are calculated using the software ‘Mapmaker-QTL’ after taking into consideration the LOD scores.

However, reports of QTL mapping have tended to be based on individual small to moderately sized mapping populations screened with a relatively small number of markers providing relatively low resolution of marker-trait association. Very few of the QTLs reported have been used for MAS in plant breeding.

Step # 7. Test for Reliability of the Identified Markers in Predicting the Trait in Alternate Population (Marker Validation):

Once a marker or sets of markers are identified to be tightly linked to a particular gene, the next step is to validate the markers and their linkage distances in alternate populations.

Alternate populations can be developed by selecting another donor line possessing the same resistance gene and crossing it with a susceptible parent. Before deploying markers in practical breeding, it is always ideal to validate the same in 1-2 alternate populations.

Step # 8. Utilisation of the Marker in Breeding Programmes:

Once closely linked markers (which are <2 cM from gene of interest) and/or flanking markers (which are <5 cM on either side of the gene) are identified and validated in alternate populations, they are ready for use in marker assisted breeding programmes. Flanking markers have distinct advantage compared to single marker since selection based on flanking markers will eliminate all false positives.

For e.g. if a single marker is

2 cM away from the target gene

2 out of 100 segregating plants will be false positives. But if flanking markers which are

5 cM apart from the target gene their combined recombinational values would be minimum (since their combined recombination will be 5/100 x 5/100 = 25/10000 = 0.0025 cM).

Finally, when a marker or a set of markers are to be deployed in practical breeding, it is necessary to verify whether they are polymorphic between the parental lines used in the particular breeding programme. If the markers are polymorphic, then they can be reliably deployed to track the introgression of the gene in breeding programmes.

It should be realised that the genetic basis of complex traits and the interaction between all related traits will become much better understood. This will allow accurate modeling of gent networks and development of robust simulation tools for designing target genomic ideotypes. With the availability of such knowledge and tools, early stages of plant breeding programmes will become much more efficient in a design-led way.

However, there will continue to be no substitute for multi-locational replicated evaluation trials for screening elite breeding lines for selection and validation of finished products before distribution to local breeding institutions, companies and farmer’s fields.


Study Questions

7.1 Compare recombination and crossover. How are these similar? How are they different?

7.2 Explain why it usually necessary to start with pure-breeding lines when measuring genetic linkage by the methods presented in this chapter.

7.3 If you knew that a locus that affected earlobe shape was tightly linked to a locus that affected susceptibility to cardiovascular disease human, under what circumstances would this information be clinically useful?

7.4 In a previous chapter, we said a 9:3:3:1 phenotypic ratio was expected among the progeny of a dihybrid cross, in absence of gene interaction.

a) What does this ratio assume about the linkage between the two loci in the dihybrid cross?

b) What ratio would be expected if the loci were completely linked? Be sure to consider every possible configuration of alleles in the dihybrids.

7.5 Given a dihybrid with the genotype CcEe:

a) If the alleles are in coupling (cis) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?

b) If the alleles are in repulsion (trans) configuration, what will be the genotypes of the parental and recombinant progeny from a test cross?

7.6 Imagine the white flowers are recessive to purple flowers, and yellow seeds are recessive to green seeds. If a green-seeded, purple-flowered dihybrid is testcrossed, and half of the progeny have yellow seeds, what can you conclude about linkage between these loci? What do you need to know about the parents of the dihybrid in this case?

7.7 In corn (i.e. maize, a diploid species), imagine that alleles for resistance to a particular pathogen are recessive and are linked to a locus that affects tassel length (short tassels are recessive to long tassels). Design a series of crosses to determine the map distance between these two loci. You can start with any genotypes you want, but be sure to specify the phenotypes of individuals at each stage of the process. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.

7.8 In a mutant screen in Drosophila, you identified a gene related to memory, as evidenced by the inability of recessive homozygotes to learn to associate a particular scent with the availability of food. Given another line of flies with an autosomal mutation that produces orange eyes, design a series of crosses to determine the map distance between these two loci. Outline the crosses similar to what is shown in Figure 7.8, and specify which progeny will be considered recombinant. You do not need to calculate recombination frequency.

7.9 Image that methionine heterotrophy, chlorosis (loss of chlorophyll), and absence of leaf hairs (trichomes) are each caused by recessive mutations at three different loci in Arabidopsis. Given a triple mutant, and assuming the loci are on the same chromosome, explain how you would determine the order of the loci relative to each other.

7.10 If the progeny of the cross aaBB x AAbb is testcrossed, and the following genotypes are observed among the progeny of the testcross, what is the frequency of recombination between these loci?

7.11 Three loci are linked in the order B-C-A. If the A-B map distance is 1cM, and the B-C map distance is 0.6cM, given the lines AaBbCc and aabbcc, what will be the frequency of Aabb genotypes among their progeny if one of the parents of the dihybrid had the genotypes AABBCC?

7.12 Genes for body color (B black dominant to b yellow) and wing shape (C straight dominant to c curved) are located on the same chromosome in flies. If single mutants for each of these traits are crossed (i.e. a yellow fly crossed to a curved-wing fly), and their progeny is testcrossed, the following phenotypic ratios are observed among their progeny.

b) Why are the frequencies of the two smallest classes not exactly the same?

7.13 Given the map distance you calculated between B-C in question 12, if you crossed a double mutant (i.e. yellow body and curved wing) with a wild-type fly, and testcrossed the progeny, what phenotypes in what proportions would you expect to observe among the F2 generation?

7.14 In a three-point cross, individuals AAbbcc and aaBBCC are crossed, and their F1 progeny is testcrossed. Answer the following questions based on these F2 frequency data.


Help with gene linkage and mapping! - Biology

Genetic Mapping: Linkage and Recombination

Introduction
Genetic mapping is a key tool for classical genetics. When a phenotype or a disease is observed, it is always desirable to determine where the gene is located.

Linkage
Linkage means when gametes form, genes are not assorted independently, instead, they are linked together. This is because these genes are located close to each other on the same chromosome. Linkage can be tested by chi square.

Recombination
Genes located close enough can crossover and then recombination occurs. The recombination frequency can be used to measure the genetic distance between two markers: the longer distance, the more recombination events, the higher recombination frequency.

Genetic map
Genetic map is a chromosome map of a species or experimental population that shows the position of its known genes and/or markers relative to each other. It is based on the frequencies of recombination between markers. Genetic mapping can be carried out by two-point or three-point testcrosses.

A genetic map has all the known genes or genetic markers arranged on chromosome according to their genetic distance. The map is drawn according to the recombination frequencies of the linked genes. Genes on different chromosomes belong to different linkage group. Genetic mapping can also be achieved by advanced techniques such as tetrad analysis, somatic cell hybridization and genome project.

  • External concept map to describe the relationship of Mendelian genetics and other disciplines in biology
  • Internal concept map to reveal the relationships among the content within this tutorial
  • Step by step analysis of Mendel’s original experiments
  • Brief explanation of Mendelian genetics at chromosome level
  • Flow chart on calculating chi square
  • Diagrams and symbols for human pedigree reading

Introduction

Linkage theory

  • Discovery of linkage
  • Linkage theory development
  • Linkage and recombination
  • Genetic distance

Genetic mapping

  • Two-point testcross
  • Genetic map
  • Three-point testcross
  • Interference
  • Coefficient of coincidence

Advanced mapping technique

See all 24 lessons in Genetics, including concept tutorials, problem drills and cheat sheets:
Teach Yourself Genetics Visually in 24 Hours


Gene Mapping by Three-Point Test Cross | Cell Biology

Recombination frequencies are directly pro­portional to distances between genes in question and these values can be used in preparation of linkage maps. A three-point test cross (involving three genes) gives us information regarding rela­tive distances between the genes and tells us the linear order in which these genes are present on the chromosome.

An important feature of all linkage maps is their linearity i.e., all genes in a given linkage group can be shown to map in a linear array. Let us presume that there are three genes A, B and C present on the same chromosome (i.e., they are linked).

There could be three possible linear orders in which these genes may be present on a chromosome. These are A-B-C, A-C-B or B-A-C. In one case, B is in the middle and in the other two, C and A respectively are in the middle.

Therefore, in finding out the linear order, the gene present in the centre should be found out.

For this purpose, a three point test- cross is made, which involves crossing of a tri-hybrid ABC/abc (obtained from a cross ABC/ABC X abc/abc) with triple homozygous recessive abc/abc. The progeny obtained will represent the gametes formed by the hybrid. Presuming A-B-C as the order of genes, the results expected can be diagrammatically represented as in the given Fig. 8.16.

Hypothetical frequencies of eight types of progenies are listed in the Fig. 8.17 and can be used for preparation of linkage map.

Linkage Gene Mapping Construction:

Linkage maps are prepared with the help of recombination frequencies.

Let us consider an example from maize involving three endosperm characters. These three characters are coloured aleurone (C) versus colourless aleurone (c), full endosperm (Sh) ver­sus shrunken endosperm (sh) and non-waxy endosperm (Wx) versus waxy endosperm (wx). The data presented by C.B. Hutchinson in 1922 are given in Fig. 8.18.

The three recombination values, i.e., C-Sh, Sh-Wx, C-Wx should be worked out in order to find out the linear order of the three genes, C, Sh and Wx. In the data presented, the progeny of parental types are present in higher frequencies.

C and sh are present together in P1, therefore, the progeny showing their separation would be recorded as recombination between C and Sh. Similarly recombination between sh and Wx as well as between C and Wx could be recorded.

The mathematical relationship among the recombination values of three genes may be utilized for determining the gene order. From the values of X, Y and Z of the example (Fig. 8.17), the order of genes can be worked out:

if Z = X -I- Y, the order of genes is A-B-C

if Z = X – Y, the order of genes is A-C-B

if Z = Y – X, the order of genes is B-A-C.

In the example (Fig. 8-18), the recombina­tion value C-Wx (21.7%) is nearly equal to recombination value of (C-sh) + (sh – Wx) = 3.5 + 18.4 = 21.9%. Therefore, sh should be located between C and Wx. Another way of determining gene order is comparison of the allelic combination of paren­tal and double crossover recombinant classes of progeny.

Out of the eight (4 pairs) phenotypic classes of progeny, one pair has the highest fre­quency representing parental (non-recombinant) class and one pair has the lowest frequency representing double crossover recombinant class.

In the example (Fig. 8.18), the highest frequency progeny class develops from non- recombinant gametes, C sh Wx and c Sh wx and the lowest frequency progeny class develops from double crossover recombinant gametes, C Sh Wx and c sh wx.

Comparison of allelic arrangements of non-recombinant gametes with double crossover recombinant gametes [(C sh Wx and C Sh Wx) or (c Sh wx and c sh wx)] shows that Sh or sh stands out as the changed locus indicating its position in the middle. Therefore, the gene order will be C-sh-Wx.

Gene Mapping Distance and Unit:

The map distance is given by recombination frequency, represented in map unit (m.u.), also called centi-Morgan (cM).

1% recombination = 1 m.u. = 1 cM.

In case of the above example (Fig. 8.18), the linkage map will thus look like the following:

Linkage map may be prepared by another calculation when the gene order is detected directly from the data by comparing the allelic combinations of parental and the double cross­over recombinant classes of progeny. In the example of Fig. 8.18, the gene order determined is C-sh-Wx.

In three-point test cross, some parental com­binations are resulted from double crossovers (products of some multiple crossovers are not recombinant). These crossovers could not be included to determine recombinant frequency between terminal genes. Thus all the map distances based on recombinant frequency might be underestimated of physical map distances.

When the two loci are farther apart on a chromosome, double crossovers occur between them will tend to mask the recombinants. So the distantly linked loci usually appear closer than they really are. Thus more accurate map dis­tances are those established on very closely linked loci.

Therefore, summed short distances are more accurate than directly measured long distances. In the example (Fig. 18.8), the difference between the summed short distance values (3.5 + 18.4 = 21.9) and longer distance value (21.7) is 21.9 – 21.7 = 0.2, which is due to the fact that in the longer distance value double crossovers are not included.

Thus over long distances on a map, the mea­sured map distance (recombination frequency) and actual map distance do not correspond and the linear relationship between the two does not hold good. The recombination frequency between any two genes will never exceed 50, but the map distance may exceed 50.

At lower values there is linear relationship, but as recom­bination value approaches towards 50%, the linear relationship is gradually lost and the recombination frequency is always less than the map distance. This is due to presence of greater number of double and even numbered multiple crossovers which tend to mask recombinants.

Gene Mapping Function:

Actual map distances should not be con­fused with measured map distance based on recombination frequencies. Recombination frequencies need a mathematical treatment (correc­tion) to get relatively precise estimates of map distance between loci. True map distance will actually be obtained by making use of a mapping function as worked out by Haldane (1919).

It reflects the relationship between the real map distance and recombination frequency (RF) which assumes the form of a curve (Fig. 8.19A), showing anticipated increased map distances for longer recombination frequencies. An accurate measure of physical distance is the mean number (m) of crossovers that occur in that segment per meiosis.

Thus the mapping function approach is to find a function that relates RF to ‘m’. In any chromosomal region the various crossover possi­bilities are 0, 1, 2, 3, 4 or more. Any number of crossover produces a frequency of 50% recom­binants (Fig. 8.19B). Thus the only class which is crucial is zero class.

Hence the true determinant of RF is the relative sizes of the classes with no crossovers, versus the classes with any non-zero number of crossovers.

The occurrence of crossovers along a chro­mosome which is responsible for recombination can be described by a statistical distribution called the Poisson distribution, used for events having low mean value. Over a small region of a chromosome, crossing over will take place in a small number of ceils out of the total number of cells undergoing meiosis. If we know the mean value (m) of crossovers in this small region of

M =mean number of crossovers,

I = actual number of crossovers.

For example, if there is an average of one physical crossover event for a particular interval, some cells may have no crossover for that inter­val, whereas others, have one, two, three, etc.

The frequency of cells to have no crossover for that interval may be calculated using Poisson distribution:

Therefore, frequency of cells with at least one crossover in that region in the total popu­lation (i.e., 1.0) will be:

1 – e -m = 1 – 0.37 = 0.63. These cells will have 50% products recom­binants, so the recombination frequency, RF = ½(1 – e -m ) = ½ X 0.63 = 0.315. Thus a linkage interval of one physical crossover event will show only 31.5% recombinant products whereas 50% recombination would have been expected.

If RF is known, m can be calculated:

RF = ½(1 – e -m ), or, 2RF = 1 – e -m ,

or, e -m = 1 – 2RF, or, -m = logn(1 – 2RF),

where logn = natural logarithm,

or, m = -logn(1 – 2RF) = Mapping function.

For example, if in a test cross RF is 27.5%, then e -m = 1 – 2 X 0.275 = 0.45 using calculator we can deduce m = 0.8, i.e. there are 0.8 crossovers per meiosis in that chromosomal region.

The final step is to convert this measure of physical map distance to corrected map unit. In very small genetic regions, RF is expected to be an accurate measure of physical distance because there are not any multiple crossovers. In fact, meiosis will show either no crossovers or one crossover.

The frequency of crossovers (m) will then be translatable into a correct recombinant fraction of m /2 because the recombinants will be 1 /2 of the chromatids arising from the single crossover class. This defines a general relation­ship between ‘m’ and a corrected recombinant fraction, can be thought of as m /2.

Hence, in the example above, ‘m’ value of 0.8 can be conver­ted into a corrected recombinant fraction of 0.8 /2=0.4(40%) or 40 map units, which is sub­stantially larger than 27.5 m.u. deduced from observed RF.

The ‘m’ value or actual average number of crossovers will be an index of true map distance in a region (m = 1 =50 cM, m = 2 = 100 cM, m = 3 = 150 cM, m = 4 = 200 cM). ‘m’ value derived through the use of mapping function when plotted against RF values, RF values up to 10% will be linear with map units, but for higher values, the relationship does not hold good, e.g., for RF = 27.5%, m = 0.8 = 40 m.u. = 40 cM.

So if RF values are used for mapping directly without employing mapping function, the distance will be underestimated.


Three factor crosses

Distances between multiple loci can be determined using three factor testcrosses. Again, we will cross a heterozygous parent to a parent homozygous recessive for all three genes. When solving three factor test crosses, remember that in the heterozygote the dominant and recessive alleles can be on the same or different chromosomes. You will know which chromosomes are parental because they will be the most abundant offspring from the testcross. In contract, the double crossover offspring will be the least abundant, because the double crossover events between the genes of interest are more rare than single crossovers.

Note: Steps for solving a three factor cross

  • Identify the parental offspring (the most abundant).
  • Identify the double crossover offspring (the least abundant).
  • Determine which locus is in the middle (the one that "swaps places" in the double crossover).
  • Determine the recombination frequency between one locus and the middle locus.
  • Determine the recombination frequency between the other locus and the middle locus.
  • Draw the genetic map.

Note that other factors that might influence recombination and double crossovers, such as the position along the chromosome or whether recombination at one site suppresses recombination nearby. These might make the number of observed recombinants different from expected, but we will not consider these factors at this time.


7: Linkage and Mapping

  • Contributed by Todd Nickle and Isabelle Barrette-Ng
  • Professors (Biology) at Mount Royal University & University of Calgary

As we learned in Chapter 6, Mendel reported that the pairs of loci he observed behaved independently of each other for example, the segregation of seed color alleles was independent from the segregation of alleles for seed shape. This observation was the basis for his Second Law (Independent Assortment), and contributed greatly to our understanding of heredity. However, further research showed that Mendel&rsquos Second Law did not apply to every pair of genes that could be studied. In fact, we now know that alleles of loci that are located close together on the same chromosome tend to be inherited together. This phenomenon is called linkage, and is a major exception to Mendel&rsquos Second Law of Independent Assortment. Researchers use linkage to determine the location of genes along chromosomes in a process called genetic mapping. The concept of gene linkage is important to the natural processes of heredity and evolution.

Figure (PageIndex<1>): Linkage affects the frequency at which some combinations of traits are observed. (Wikipedia-Abiyoyo-CC:AN)

  • 7.1: Linkage Mendel&rsquos Second Law does not apply to every pair of genes that could be studied. In fact, we now know that alleles of loci that are located close together on the same chromosome tend to be inherited together. This phenomenon is called linkage, and is a major exception to Mendel&rsquos Second Law of Independent Assortment. Researchers use linkage to determine the location of genes along chromosomes in a process called genetic mapping and is important to natural processes of heredity and evolution.
  • 7.2: Recombination In heredity, recombination is any process that results in gametes with combinations of alleles that were not present in the gametes of a previous generation. Interchromosomal recombination occurs either through independent assortment of alleles whose loci are on different chromosomes. Intrachromosomal recombination occurs through crossovers between loci on the same chromosomes . In both cases, recombination is a process that occurs during meiosis (mitotic recombination is relatively rare).
  • 7.3: Linkage Reduces Recombination Frequency
  • 7.4: Crossovers Allow Recombination of Linked Loci
  • 7.5: Inferring Recombination From Genetic Data
  • 7.6: Genetic Mapping Because the frequency of recombination between two loci (up to 50%) is roughly proportional to the chromosomal distance between them, we can use recombination frequencies to produce genetic maps of all the loci along a chromosome and ultimately in the whole genome.
  • 7.7: Mapping With Three-Point Crosses A particularly efficient method of mapping three genes at once is the three-point cross, which allows the order and distance between three potentially linked genes to be determined in a single cross experiment. This is particularly useful when mapping a new mutation with an unknown location to two previously mapped loci.
  • 7.E: Linkage and Mapping (Exercises)
  • 7.S: Linkage and Mapping (Summary)

7.5: Genetic Mapping

  • Contributed by Todd Nickle and Isabelle Barrette-Ng
  • Professors (Biology) at Mount Royal University & University of Calgary

Because the frequency of recombination between two loci (up to 50%) is roughly proportional to the chromosomal distance between them, we can use recombination frequencies to produce genetic maps of all the loci along a chromosome and ultimately in the whole genome. The units of genetic distance are called map units (mu) or centiMorgans (cM), in honor of Thomas Hunt Morgan by his student, Alfred Sturtevant, who developed the concept. Geneticists routinely convert recombination frequencies into cM: the recombination frequency in percent is approximately the same as the map distance in cM. For example, if two loci have a recombination frequency of 25% they are said to be

25cM apart on a chromosome (Figure (PageIndex<9>)). Note: this approximation works well for small distances (RF<30%) but progressively fails at longer distances because the RF reaches a maximum at 50%. Some chromosomes are >100 cM long but loci at the tips only have an RF of 50%. The method for mapping of these long chromosomes is shown below.

Note that the map distance of two loci alone does not tell us anything about the orientation of these loci relative to other features, such as centromeres or telomeres, on the chromosome.

Figure (PageIndex<9>): Two genetic maps consistent with a recombination frequency of 25% between A and B. Note the location of the centromere. (Original-Deyholos-CC:AN)

Map distances are always calculated for one pair of loci at a time. However, by combining the results of multiple pairwise calculations, a genetic map of many loci on a chromosome can be produced (Figure (PageIndex<10>)). A genetic map shows the map distance, in cM, that separates any two loci, and the position of these loci relative to all other mapped loci. The genetic map distance is roughly proportional to the physical distance, i.e. the amount of DNA between two loci. For example, in Arabidopsis, 1.0 cM corresponds to approximately 150,000bp and contains approximately 50 genes. The exact number of DNA bases in a cM depends on the organism, and on the particular position in the chromosome some parts of chromosomes (&ldquocrossover hot spots&rdquo) have higher rates of recombination than others, while other regions have reduced crossing over and often correspond to large regions of heterochromatin.

Figure (PageIndex<10>): Genetic maps for regions of two chromosomes from two species of the moth, Bombyx. The scale at left shows distance in cM, and the position of various loci is indicated on each chromosome. Diagonal lines connecting loci on different chromosomes show the position of corresponding loci in different species. This is referred to as regions of conserved synteny. (NCBI-NIH-PD)

When a novel gene or locus is identified by mutation or polymorphism, its approximate position on a chromosome can be determined by crossing it with previously mapped genes, and then calculating the recombination frequency. If the novel gene and the previously mapped genes show complete or partial linkage, the recombination frequency will indicate the approximate position of the novel gene within the genetic map. This information is useful in isolating (i.e. cloning) the specific fragment of DNA that encodes the novel gene, through a process called map-based cloning.

Genetic maps are also useful to track genes/alleles in breeding crops and animals, in studying evolutionary relationships between species, and in determining the causes and individual susceptibility of some human diseases.

Figure (PageIndex<11>): A double crossover between two loci will produce gametes with parental genotypes. (Original-Deyholos-CC:AN)

Genetic maps are useful for showing the order of loci along a chromosome, but the distances are only an approximation. The correlation between recombination frequency and actual chromosomal distance is more accurate for short distances (low RF values) than long distances. Observed recombination frequencies between two relatively distant markers tend to underestimate the actual number of crossovers that occurred. This is because as the distance between loci increases, so does the possibility of having a second (or more) crossovers occur between the loci. This is a problem for geneticists, because with respect to the loci being studied, these double-crossovers produce gametes with the same genotypes as if no recombination events had occurred (Figure (PageIndex<11>)) &ndash they have parental genotypes. Thus a double crossover will appear to be a parental type and not be counted as a recombinant, despite having two (or more) crossovers. Geneticists will sometimes use specific mathematical formulae to adjust large recombination frequencies to account for the possibility of multiple crossovers and thus get a better estimate of the actual distance between two loci.


Neurospora: Recombination, Gene Mapping and Genotypic Control | Genetics

In this article we will discuss about:- 1. Recombination in Neurospora 2. Gene Mapping in Neurospora 3. Genotypic Control.

Recombination in Neurospora:

The fungus Neurospora crassa has some advantages for study of crossing over. In the vegetative phase of its life cycle, there is growth of filamentous, coenocytic hyphae to form the haploid mycelium. Sexual reproduction is achieved by coming together of two hyphae of proper mating type and their nuclei fuse to form a diploid zygote (Fig. 8.11).

The zygote starts enlarging into an elongated structure called ascus and the zygote nucleus starts dividing meiotically. The four products of meiosis are arranged in a single linear row inside the ascus. Each one of these undergoes a mitotic division so that 8 nuclei are formed which develop a wall and become ascospores.

What makes Neurospora interesting for a geneticist is the fact that the ascospores are linearly ordered in the same sequence as the chromatids were on the meiotic metaphase plate. It is therefore possible to recover all the four products of a single meiosis and analyse them, as also each chromatid from a tetrad. This is called tetrad analysis.

Neurospora thus presents a direct way of demonstrating recombination which in higher organisms can be inferred on a statistical basis from progeny counts. A still further advantage is the use of centromere as a marker for determining map distances.

When the segregation of a single pair of alleles is analysed, the ascus of Neurospora shows two arrangements of the eight ascospores: 4: 4 ratio resulting from first division segregation and 2:2:2:2 ratio resulting from second division segregation (Fig. 8.12).

First Division Segregation:

The results of a cross between a normal (al + ) and an albino (al) strain of Neurospora are shown in Fig. 8.12. If the 8 ascospores are removed from the ascus and each is grown separately, it is found that 4 ascospores produce mycelium of normal type and 4 of albino type.

This is due to first division segregation explained as follows: during anaphase I of meiosis in the zygote, one homologue carrying al + al + segregates from the second homologue which has the other allele alal and both move to opposite poles.

Their products are recovered in a 4: 4 ratio after meiosis II (and after mitosis) because no crossing over has occurred between the gene and the centromere. This situation results when the gene is located close to the centromere. Therefore, if a large number of asci are found to exhibit 4: 4 ratio, it indicates that the gene locus in question is close to the centromere.

Second Division Segregation:

As shown in Fig. 8.12 if crossing over occurs between the said gene and the centromere, then one chromatid of each homologue will carry al + and the other al. Therefore al + and al will not be able to segregate from each other at anaphase of first meiosis when the two homologues separate to the two poles.

It is only at anaphase of second meiosis when centromeres divide and chromatids separate from each other to the two poles that al + and al will segregate from each other. This is called second division segregation and produces a ratio of 2al + : 2al: 2al + : 2al, which also indicates that crossing over has occurred between the gene locus and centromere.

Gene Mapping in Neurospora:

In Neurospora the centromere is a marker for determining map distances. For detecting linkage and map distances, the frequency of crossing over is determined from the number of asci showing second division segregation. If there is one crossover, the resulting ascus shows 50% of ascospores with parental combinations and 50% with re-combinations.

Suppose in a cross involving a pair of alleles 30% of asci show second division segregation. This shows that 30% of zygotes had crossing over during meiosis and 70% did not. Since there are four chromatids in each tetrad, the 50% asci have resulted from 30 x 4 = 120 original chromatids in meiosis. When there is crossing over only two of the four chromatids are involved in an exchange.

Therefore only half of the 120 chromatids i.e., 60 are crossover chromatids, the remaining 60 chromatids are non-crossover chromatids. It was also stated above that 70% of zygotes did not have crossing over, which means that 70 x 4 = 280 are non-crossover chromatids. The actual number of non-crossover chromatids is larger because the 30% asci showing second division segregation also have 60 non-crossover chromatids.

The exact number is therefore 280 + 60 = 340. Therefore, of the original 100 tetrads or asci 340 are non-crossover chromatids and 60 are crossover chromatids. Since 100 tetrads or asci also mean 400 chromatids, the percentage of crossover chromatids is (60/400) x 100 = 15%.

From this we can conclude that there was 15% crossing over between the gene and the centromere. We can also say that the gene in question is 15 map units apart from the centromere. Because the centromere itself serves as a marker, in Neurospora it is possible to map a single gene pair. It is also called a two-point cross.

It follows that the method of detecting gene linkage in fungi is basically similar to that for diploids. The main feature in all cases consists in comparing the frequency of parental types to recombinant types. If there is a significant reduction in the frequency of recombinant types from the frequencies expected on the basis of independent assortment, we can consider linkage.

Whereas in Neurospora the meiotic products occur in an ascus in a linear order (ordered tetrads) this is not true for other fungi in Ascomycetes which have unordered tetrads. Let us examine a cross involving two linked genes + +/ab in a fungus having unordered tetrads.

The zygote (++/ab) undergoes meiosis, and depending upon the occurrence of crossover, the tetrads are of following three types:

(a) Parental Ditype (PD) (Fig. 8.13a):

If a crossover does not occur between these two loci, or if a two strand double crossover occurs between them, the resulting meiotic products will be of two kinds, both resembling parental combinations, and appear in equal frequency (1 + + 1 ab). Such a tetrad is called parental ditype (PD).

(b) Non-parental Ditype (NPD) (Fig. 8.13 b):

If a four strand double crossover occurs between the two genes, two kinds of products are formed, both being re-combinations. Such a tetrad is called a non-parental ditype (NPD).

(c) Tetra-type (TT) (Fig. 8.13 c):

This is produced either by a single crossover or a three- strand double crossover (of two types) between the two genes.

Whenever the number of parental di-types and non-parental di-types are significantly unequal, linkage between the two genes must be considered.

For determining the amount of recombination between the two genes, the following formula is used:

Recombination Frequency = NPD + 1/2 TT/Total number of tetrads

Genotypic Control of Neurospora:

In some cases at least recombination itself is under the control of genes. In Neurospora crassa several recombination genes control frequency of recombination either between genes or within genes. Thus a recessive gene rec-1 controls frequency of recombination at the his-1 locus. Both rec-1 and his-1 are on the same chromosome.

In maize chiasma formation can be entirely suppressed by a recessive gene as present on chromosome 1. There are quite a few examples known of single genes as well as polygenes which can cause variation in frequency and distribution of chiasma in one or more chromosomes.


Genetic versus Physical Maps

Chromosome mapping by counting recombinant phenotypes produces a genetic map of the chromosome. But all the genes on the chromosome are incorporated in a single molecule of DNA. Genes are simply portions of the molecule (open reading frames or ORFs) encoding products that create the observed trait (phenotype). The rapid progress in DNA sequencing has produced complete genomes for hundreds of microbes and several eukaryotes.

Having the complete sequence makes it possible to determine directly the order and spacing of the genes. Maps drawn in this way are called physical maps.

What is the relationship between the genetic map and the physical map of a chromosome? As a very rough rule of thumb, 1 cM on a chromosome encompasses 1 megabase (1 Mb = 10 6 bp) of DNA. But for the reasons mentioned above, this relationship is only approximate. Although the genetic maps of human females average 90% longer than the same maps in males, their chromosomes contain the same number of base pairs. So their physical maps are identical.


Watch the video: Gene Linkage and Genetic Maps (July 2022).


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