Lecture 3: Genetic Linkage -Chromosome mapping -Tetrad analysis
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Perspective: Genetic elements "invisible"... only their expression can be monitored and their potential interactions inferred throug methematical analyses. Use of heterozygotes and "hemi" zygotes.

 

Sample Questions:

 

To Recap Aspects of Mendelian Genetics.

Ratios: 3:1 and 1:1 (test cross) for monihybrid crosses and 9:3:3:1 for dihybrid crosses.

Extension of Mendelian Genetics:-

Incomplete dominance: eg. Four o' clocks, carnations. BLENDING 1:2:1
Codominance: eg. M and N blood groups on chromosome 4 having specific antigens M and N

1:2:1 but where the heterozygotes (MN) gives rise to a distinct phenotype

Multiple alleles ABO blood types, A and B are dominant to O , but A and B codominant to each other
Lethal alleles eg. Yellow colouration in mice fur. 2 : 1 . Pleiotropy(?)
Several genes/same character Coat colour in mammals,

eg. mice

A (agouti), B (black/brown), C, (colour) D (intensity), S. (distribution) genes

Epistasis essentially "eliminates" or masks phenotypic expression of other genes,

eg. Labrador dogs fur colour, albinoism in mice

Complementary Genes eg. Pea plants, purple colouration. 9 : 7
Duplicate gene activity eg. Shepherd's purse, Round over narrow fruits, where

both A 1 - and A 2 - can cause heart shape 15 : 1

Recessive suppression eg. Case of purple eye colour in the fruit fly. 13 : 3

 

To Recap Linkage:

Discovery of Linkage: T. S. Morgan (1909) working on non-autosomal and sex chromosome -linked genes of Drosophila melanogaster -the commmon fruit fly.

   

  

 

But what about ALL the genes on the other autosomal chromosomes.

 b  b       
    

He thus defined the unit of genetic distance as being:

 

one crossover event/100 products of meiosis = one map unit or 1 centiMorgan (cM)

 

Morgan and his team then went on to demonstrate that the second kind of linkage could be used to define the order and relative location of these proximal genes.

 

One of Morgan's students, Sturtevant, quickly realized that such information could be used to map genes.

Three point test crosses. Mathematically, two independent events happening at the same time-

However, to observe these traits.....

1. Organism producing the crossover gametes must be heterozygous for all three traits.

2. The cross must be constructed so that each of the phenotypic traits must reflect a change in the genotype at each locus.

3. A sufficient number of offspring must be obtained to get statistically viable results.

 

The First question: would be... What is the gene order?  We find this out by looking at theNCO's and the DCO's.

X y z
x Y Z

 

The answer is (in this example, according to the numbers, remember)....   "y     x      z".

 

The Second question: Are the relative rates of SCO's and DCO's consistent? If not..... ?

Morgan's student, Sturtevant, observed this phenomenon in a number of his crosses, and rationally understood that frequency of recombinants increased with increasing distance, and that it was more accurate to add the intervening distances from multiple small intervals, rather than rely on a single cross between two fairly distant markers to determine the map distance.

 

  

Use of mathematical equation that was developed to predict the probability of random events... Poisson distribution.

RF (recombination frequency) = 0.5 x (1 - e -m ).

which relates mapping units "m" (or the mean number of exchanges) directly to RF.

b  

  

Balancer chromosomes, or rearranged chromosomes that can no longer line up appropriately in prophase 1 to allow for chiasmata to form.

  

 

Additional variables of gene "associations" that result from gene linkage and chromomal misbehaviour.

Meiotic Nondysjunction

     

Mitotic Nondysjunction.

    (a) Calvin Bridges noticed that a strain of flies M M+ (heterozygous for "bristle" configuration), sometimes expressed itself as a "local variant" of the dominant M allele (slender bristle) in which a region of the fly’s body exhibited wild-type (M+) bristle characteristics.

     

He attributed this phenomenon to a genetic event termed mitotic nondysjunction, where one half of the resulting progeny have only three copies of the chromosome (resulting from an aberrant mitotic division) and the other cell has only one.

Sometimes such a localized phenotypic variation can occur as a result of chromosome loss during mitosis of a somatic or non-germ-line cell.

Either way, the local variation was, and is, thought to result from a catastrophic loss or abnormal migration of whole chromosomes.

Segregation at the phenotypic level, i.e. heterozygotes exhibiting localized "patches" of variant phenotypes.

    (b) In 1936 Curt Sterns (another student of Morgan's working on D. melanogaster) undertook a cross of two other sex-linked traits y (yellow hair) and sg (singed hair).
y+ sn / y+ sn (singe-haired female) with y sn+ / Y (yellow singe-haired male)

Not too surprisingly the female progeny were mostly Wild-Type with grey bodies and normal bristles y+ sn / y sn+

Occasionally, however, Stern noticed that there were some localized twin spots of apparently "coupled" yellow and singed hair (y sn) phenotypes, which occurred as twin spots too often as to be mere coincidental juxtapositioning. How might these be explained?

   

He accounted for this by asserting the phenomenon of mitotic recombination, i.e. the chance occurrence of a crossing over event occurring between two homologous chromosomes during MITOSIS, whereby the event occurs between the two gene loci and the centromere.

N.B. Don't forget that diploid cells give rise to diploid progeny in mitotic divisions!!

Sterns' work was the first example of "mosaicism", which is the presene of two or more populations of cells with different genotypes. It was the birth of what is now a significnt branch of molecular genetics called "mosaic analysis", which is one of the major tools used today to analyze the effects of lethal or "specifying" genes in a number of organisms, by being able to compare the genetic heritage of two cell lines -with the only difference between the two being a somatic aberration, that can often be induced.

 

Additional concerns with genetic exchange through crossing over.

 

Consider a pair of homologous chromosomes prior to the MI division which contain a single crossover in the interval between markers B and D on chromatids 2 and 3, as shown in the Figure below. The outcome of the single crossover is two chromatids carrying a recombination of the phenotypic markers associated with the B and D genes. Now, consider that a second crossover occurs in this same interval. There are four ways this could occur, and each of these is equally probable.

     

• (Case 1):   As we have alrady dealt with.  The second crossover could occur again between chromatids 2 and 3, restoring the original configuration of markers B and D. If the investigator were only analyzing recombination of B and D, then the double crossover involving two strands would appear as though there had been no crossover at all: it would be indistinguishable from the 0 crossover class. Only if the investigator follows marker C in this example would the double crossover be apparent, providing a reciprocal recombination event of B-c-D and b-C-d.


• (Cases 2 and 3): The second crossover could involve only one of the original chromatids. The first crossover involved chromatids 2 and 3, so the second crossover could involve 1 and 3, or 2 and 4. (Recall that the meiotic crossover predominantly occurs between homologous chromosomes, and not between sister chromatids. Therefore, it is unlikely to occur between 1 and 2, or between 3 and 4.) This double crossover, involving three strands, gives the same recombination of the phenotypic markers as in the single crossover (only two chromatids show a recombination between B and D) and can only be distinguished from the single crossover class by analysis of the C marker.


• (Case 4): Finally, the second crossover could occur between chromatids 1 and 4, neither of which were involved in the first crossover. This case is unique, since it results in four recombinant chromosomes and no wild type associations. This feature is particularly powerful in identifying the frequency of double crossovers: this class can often be identified unambiguously, and represents 1/4 * the total number of double crossovers. Using tetrad analysis to determine gene-gene linkage for an example).

 

How do we analyze for this and asure ourselves that each non-sister chrmatid has an equal chance of crossing over with each of the other paired non-sister chromatids.

 

Yeast and fungi.

S. cerevisiae: and the potential POWER of yeast genetics -provides relatively immediate proof of all types of chiasmata and their consequences. Use of yeast and other similar fungi have many distinct advantages ...S. cerevisiae can grow mitotically in a stable fashion as either a haploid (with one copy of each chromosome) or a diploid (with two copies of each chromosome).  In essence, therefore, the consequences of meiosis can be "harvested" through analysis of both states, and the direct products of meoiotic events can be analyzed. 

Genetic Orderliness in Ordered and Unordered Tetrad analyses

Indeed, all fungi and algae form "haploid" spores, that contain 4 or 8 spores, resulting from a meiotic division.

These can be "ordered" or "unordered" within their respective asci (spore basample2 copysample1 copygs).

Linear tetrad analyses allows for "centromere mapping" or the distance from any given locus to the centromere (again using recombination as a tool to determine distance).

In choosing two linked genotypes that are easily discernible (phenotypes) this can give rise to a simple analysis of meiotic function.

                        e.g.                                     a       b          x          a+        b+

Three different assortment options no matter whether the two genes are linked or not

1 2 3
a b a b+ a b
a b a b+ a b+
a+ b+ a+ b a+ b
a+ b+ a+ b a+ b+

 

Parental Ditype (PD)... Non-Parental Ditype(NPD) ... Tetra-Type (TT)

 

 

The power of tetrad analysis comes from the fact that all four meiotic products are contained within a single ascus.

The most fundamental test by tetrad analysis is the one for single-gene segregation. This can determine whether a mutant phenotype is caused by a mutation in a single gene or not. If a single gene is the cause, the mutation should segregate within the tetrads in a predictable, Mendelian fashion for a single gene.

The ability to use tetrad analysis to map genes derives directly from the special information that can be obtained from having all four meiotic products in an ascus. As described below, this allows us to make and test specific predictions regarding the segregation patterns that will occur depending on whether two genes are linked or unlinked.

• A parental ditype (PD) is a tetrad in which two genetic markers are segregating, and two spores have one parental genotype and two spores have the other parental genotype.

• A nonparental ditype (NPD) is a tetrad in which two genetic markers are segregating, and two spores have one nonparental genotype and two spores have the other nonparental genotype.

 

The first class of tetrad is that in which the two parental sets of markers segregate away from each other at meiosis I.

  Meiosisblue ball

The resulting tetrads have two spores bearing the genotype of one parent (HIS4-trp1) and the other two spores bearing the genotype of the other parent .

This class of tetrad, having the two classes of parental spores, is called a parental ditype (PD),

and this arrangement of spores in the ascus can also be acheived if genes are on different chromosomes or on the same chromosome

  

The second class of tetrad is that in which segregation at meiosis I occurred in the opposite fashion. The result is a tetrad with two different classes of spores. This class of tetrad, having two classes of nonparental spores, is called a nonparental ditype (NPD)

Similarly, this arrangement of spores in the ascus can be acheived if genes are on different chromosomes or on the same chromosome.

  

Because segregation at meiosis I is "random" and occurs at equal frequency (in accordance with Mendel's First Law), for two unlinked genes, the frequency of PD tetrads should equal the frequency of NPD tetrads.

This prediction constitutes one of the fundamental tests of genetic linkage in yeast.

Therefore, when testing the possible linkage of two mutations with unknown map positions, the ffundamental question to answer is,

"Does the number of PD's equal the number of NPD's?"

If the answer is YES, the association of mutations (and hence the gene loci) are unlinked.

If the answer is NO, there is linkage.

The third type of tetrad arises as the result of a crossing over, or recombination, between one of the genes and its centromere. Crossing over during yeast meiosis occurs after replication, when there are four chromatids.

  

For two unlinked genes on different chromosomes the frequency of TT tetrads will depend on the linkage of each gene to its centromere, as it can result from a crossing over of either gene with respect to its centromere. This property can ultimately be used to determine whether a gene is linked to its centromere or not.

For two unlinked genes on the same chromosome (where PD = NPD), a TT tetrad can arise as the result of a crossover between one of the markers and its centromere. (We will see later that TT tetrads can also arise in another way).

---

Now let us consider a different type of linkage, in which two genes are now "LINKED" to each other.

By tetrad analysis we can determine the degree of genetic linkage between such markers.

In the extreme cases of two genes that are completely linked, with no detectable recombination between them, we expect that tetrad analysis will reveal that ALL the tetrads are PD tetrads.

What about cases where the two genes are linked but there is some recombination between them?

In those cases, the number of PD tetrads will be greater than the number of NPD tetrads (PD > NPD). This is because PD tetrads ONLY arise from no recombination.

In contrast, for two linked genes, NPD tetrads can also arise by a double crossover between the two linked markers in a three stranded exchange). As for centromere linkage, in the case of closely linked genes, we can calculate the degree of linkage based on the frequencies of the different types of tetrads produced.
..

A single crossover between linked markers produces a TT tetrad.

Recombination (SCO's and DCO's) between linked genes can also produce PD, NPD, and TT tetrads.

-Recall that for two unlinked genes, when PD = NPD, TT tetrads can also arise by a crossover between a gene and its centromere.

However, in the case of linked genes, when PD > NPD, TT tetrads can arise by a single crossover between the two genes and also by two of the four types of double crossovers, as shown in the multiple crossover figure.

    

It is, therefore, possible to relate T(etraTypes) and NPD' s (non-parental ditypes) to the potential single crossover events (SCO's) and double crossover events (DCO's).

      

With "linked" genes there is only ONE way to obtain an NPD -through a double cross over.

Thus, with respect to the NPD’s .......... the number of DCO's obtained = 4NPD

In a similar way, SCO can be written as being = T - 2NPD, as there are three ways to get a Tetratype (2 of which are DCO's).

Now, from a previous understanding of crossover frequencies

NCO = 1(or unity) - (SCO + DCO)

(assume that triple, quadruple crosses etc can generally be ignored as having less than a significant affect upon the end results)

let m = the mean no. of crossovers, thus m = SCO + 2 x DCO (as the actual number of crossovers in a double cross is 2).

substitute appropriate values:

=> m = (T-2NPD) + 2(4NPD) = T + 6NPD

then (as always) one needs to convert to cM by multiplying by 50 =>

now RF (in cM) = 50 (T + 6NPD)

 

Chromatid interference.

Note: The theoretically maximal value for RF being 50 can only be true if and when there is NO chromatid interference.

Consequently, if you use the formula RF = 50 (T+6NPD) (from the above definition) then the occurrence of chromatid interference might conceivably overestimate the RF value, which could therefore (in practice) actually exceed 50%!!!!!

 

Centromere mapping. While unordered tetrads allow for a rapid and relatively simple evaluation of recombination frequencies (and therefore genetic linkage of proximal genes present on the same chromosome, analysis of ordered tetrads (or octads) allows for an additional evaluation of the proximity of genes to the centromere.

   

Neurospera crassa or Sordaria sp. are examples of fungi that gives rise to an ordered arrangement of its spores within their respective ascospore. Specific variations in the location of given trait within the ascospore leaves a tell-tale assortment of an allele that is due to a crossing over event between the centromere and a given gene locus. In a similar manner to regular crossing over phenomena the distance between the centromere and a gene can thus be estimated by analyzing the frequency of "second division" segregation within the ordered tetrad (octad) and dividing by 2 (as only half of the resulting allele locations are due to this crossing over event).

                  

 

e.g. in the case of a crossover between the A locus and the centromere.

      

 

A a A a A a
A a A a A a
A a a A a A
A a a A a A
a A A a a A
a A A a a A
a A a A A a
a A a A A a
_____ _____ _____ _____ _____ _____
126 132 9 11 10 12

 

Total = 300

The recombinant forms are: 9 + 11 + 10 + 12 = 42  =  (42 / 300) x 100 = 14% are recombinants.

Now divide by two to get cM

=> Locus A therefore lies 7cM from the centromere.

 

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So, we are now able to understand how genes on chromosomes are inherited and -using the recombination frequencies of "linked" genes- we are able to construct crude linkage maps of ther relative location, as well as being able to appreciate how these genes are arranged relative to each other on their respective chromosomes, and how meiosis -more so than mitosis- leads to genetic variation rather than genetic constancy..

We also. now, have some rudimentary understanding of the properties of chromosomes and how it can potentially provide the genetic blue print for gene expression, as well as how it can potentially be rearranged as it is replicated...

What are these genes, or genetic units...red ball. ..

In trying to unravel the makeup of genes and how they encode for functional gene products within their respective chromosomes, geneticists have investigated their role by isolating these units from one system/organism and have them expressed in "model systems" for well over 75 years now..

Perspectives!?

Newsweek Article.... "transgenic" monkees, humans?

 

Old news in prokaryotes, where DNA is actively taken up from various sources and used as cellular resources.

 

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While its great for geneticists, it does cause a problem for bacterial hosts..... and the acceptance or rejection of external sources of genetic elements boils down to a genetic definition of "self".

There are three basic fates of DNA as it enters the bacterial cell

1) It is degraded rapidly due to host defense mechanisms.
2) It is integrated into host chromosome by recombination (homologous or non-homologous).
3) The DNA is able to circularize and replicate independently (autonolmously) from the host chromosome (i.e. it contains an origin of replication that is recognized by host replicating enzymes).

R-factors provide just such an independent origin of replication, and also provide for a means of selection that maintains their presence in the transformed host. Therefore they provide a generic role as a vector for the "stable" cloning of genes into a bacterial host.

Uses

(a) for basic research, scientists?
(b) industrial purposes... BIG Pharma?
(c) for medicinal/medical purposes... Even BIGGER Pharma?

Are all plasmids or R-factors created equally? No!

Variables: Size, High copy and low copy plasmids, strict or relaxed control of plasmid replication, gives rise to an Incompatability of different plasmids.

Plasmids: Extrachromosomal, self replicating or autogenous replicating, covalently closed, circular pieces of dsDNA. They can, sometimes be integrated into the host chromosome, and if so they are often called and episome.

Plasmids of 3,000 - 5,000 bp, often have a high copy number (15 - 100 copies per cell).
Plasmids of 4,000 - 300,000 bp (300 kbp), are as common in nature, but less highly copied per cell(one or two per cell) and (due to these factors) are less easily manipulatable.

Conjugative plasmids invariably contain tra and mob genes, which are necessary to promote cell to cell interaction and and also to promote movement of the DNA through the "conjugative bridge".

eg. F-factor in E. coli, which is ~94.5 kbp in size and present in any given cell at approximately one copy per cell.

                  

As a rule of thumb, therefore, these conjugable plasmids are "large" as they have to encode a number of genes that undertake the tra and mob functions.(even though, some of these functions can be chromosomally encoded... merodiploid)


Plasmids are not named and grouped by size, however, or even by DNA homology, but by.......their "incompatibility".

eg. IncP plasmids have a broad-host range and include the IncQ or IncP4 group of plasmids.

            

Incompatabilities occur in any number of ways, but normally affect either the initiation of replication or the control of the attachment of plasmids to the bacterial membrane (which, for some, is required for efficient segregation of low copy number plasmids into the two daughter cells).

So, what is necessary for genetic manipulation of DNA in bacteria

(a) Suitable vector
(b) Restriction enzymes.
(c) DNA Ligases.

 

Restriction Endonucleases: Restriction endonuclease provide an additional tool to facilitate the creation of physical maps of DNA

 

Type I restriction modification enzymes (first identified by Werner Arber and Dussoix in the1960's using lamda phage infection of E. coli) initially defined two different strains of E. coli -E. coliB and E. coli K12 (two E. coli strains that encode for slight, but specific variants of their HSD system (Host Specificity Determinant) -encoded by the hsdR, hsdMand hsdSgenes). These enymes are expressed togther and generally require interactions with cofactors, such as S-Adenosyl methionine (AdoMet), hydrolyzed adenosine triphosphate (ATP), and magnesium (Mg2+) ions.

           

The enzymes are bi-functional, multimeric (multi-subunited) enzymes and ara capable of two incompatiable functions -either methylating or restricting DNA (i.e. EcoB strain will restrict and EcoK DNA and vice versa). Essentially, the enzymes are very similar, but distinct in their ability to recognize different DNA sequences.

The enzyme system is comprised of three distinct subunits R (135kDa) M (62kDa)and S (55kDa) subunits. Restriction and methylation are mutually exclusive.


         
        

 

eg. EcoB recognizes  TGA (N8) TGCT, EcoK12 recognizes  AAC(N6)GTGC

5' TGANNNNNNNNTGCT3'
3' ACTNNNNNNNNACGA
5'

 

Type II restriction enzymes (most commonly used in Biotechnology) are only able to restrict DNA any methylase activity (if any) is present on a separate protein. Type II enzymes are usually dimeric proteins, and have a variety of digest patterns.

Restriction characteristics. Blunt- , 5' and 3' "sticky- ends". DpnI (meth) or DpnII

 

  

         

 

Usually the site of restriction is found WITHIN the palindromic region, but not always, Type IIS

 

eg. FokI

5' GGATGNNNNNNNNN|N3'
3' CCTACNNNNNNNNNNNNNN|N
5'


In using these enzymes to clone fragments of DNA into cloning vectors there are number of variables that need to be considered.

 

  

Type III restriction enzymes are similar to Type I enzymes, they also have an ATPase requirement and differ mainly in that their M and S subunits are combined into one ~75kDa subunit, with the additional R subunit being ~108kDa. Again these enzymes are BI-functional enzymes, normally as heterodimers, which can methylate and/or restrict simultaneously, although the methylase subunits can often work on its own. Methylation only occurs on one strand.

Usually the site of restriction is removed from the recognition site. with the enzyme cutting often cuttingh some 24-28 bases down from recognition site, eg. EcoP1 and EcoP15, and Hinf in Haemophilus influenzae.

 

5' AGACC - 23-NNN-|- N 3'
3' TGTGG - 2
3-|-NNNNN 5' 

 

In using these enzymes to clon fragments of DNA into cloning vectors there are number of variables that need to be considered.


Size of restriction recognition site, will affect frequency of site within any given DNA sequence.

G/C content of restriction site vs. G/C content of DNA to be restricted.
Time

Compatability of ends

 

 

 

 

 


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