Tetrad Analysis

I am a meiosis - analyze me: All the four haploid cells of an individual meiosis are referred to as a tetrad. In most species we can not collect or analyse individual tetrads because the haploid gametes of many different meioses are issued from numerous meioses. However, in some species such as the ascomycete fungi, tetrads are enclosed in a sac called an ascus. The ascospores, haploid meiotic spores of individual asci, can be isolated and analyzed for the segregation of genetic markers.

 

Unordered Tetrads in Yeast

When the four products of a meiosis are found in an ascus, ascospores of sister nuclei (containing sister chromatids) usually can not be identified. Rather all four haploid gametes are randomly distributed within the ascus as an unordered tetrad.

Consider the life cycle and genetic map of the common bakers' yeast Saccharomyces cerevisiae - an ascomycete with an unordered tetrad.

 

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The number of chromosomes in haploid cells is ? , diploid cells ? ; the two mating types of haploid cells are alpha and ? ; the spindle orientations at anaphase I and II ? , result in any specific placement of the four ascospores in the ascus, meiosis produces ? haploid gametes (spores) - two of genotype alpha, and two of genotype ? ; cells in the haploid cycle are either alpha or ? genotype whereas cells in the diploid cycle are of alpha / a genotype.

Tetrad analysis of unlinked genes: The summation of data from numerous tetrads provided the basis for Mendelian genetics.Thus the segregation of alleles in individual tetrads is governed by Mendel's principles of segregation and independent assortment.

Consider a cross in coupling between two unlinked genes. Haploid yeast strains, AB and ab, of opposite mating type are crossed to form the diploid AB / ab. Following meiosis, there are three possible tetrads. They are parental ditype (PD), non parental ditype (NPD) and tetratype (TT). As seen below PD and NPD result from ? and thus occur in equal frequency. That is, the frequency of PD = frequency of NPD. The TT tetrad results from ? . The frequency of TT will increase as the distance between genes A or B and their centromeres. If there are no crossovers between alleles and their centomeres the frequency of TT will be ? . As genes A and B are further removed from their centomeres, the upper limit of TT is 2 /3. Thus for two genes, unlinked to each other, or to their centromeres, the expected distribution is PD: NPD: TT in a 1: 1: 4 ratio.----> Explanation 1 (Student Answer)

Explanation 2 (Taken from " An Introduction to Genetic Analysis " by David T. Suzuki / Anthony J.F. Griffiths. CHAPTER 5, pages 114 - 115.)

 

Sample problems:


The following genomic map is taken from the Saccharomyces Genome Database(SGD). Mutant pho11 is deficient in acid phosphatase and rad16 is a radiation sensitive mutant. Their respective wild-type alleles are PHO11 and RAD16. pho11 is 48 cM from CENI and rad16 is 58 cM from CENII. In the cross pho11 by rad16, which of the following tetrad ratios would be expected?

10 [2 pho11 RAD16, 2 PHO11 rad16]; _10 [2 PHO11 RAD16, 2 pho11 rad16]; _40 [pho11 RAD16, PHO11 rad16; PHO11 RAD16, pho11 rad16]
10 [2 pho11 RAD16, 2 PHO11 rad16]; _40 [2 PHO11 RAD16, 2 pho11 rad16]; _10 [pho11 RAD16, PHO11 rad16; PHO11 RAD16, pho11 rad16]
40 [2 pho11 RAD16, 2 PHO11 rad16]; _10 [2 PHO11 RAD16, 2 pho11 rad16]; _10 [pho11 RAD16, PHO11 rad16; PHO11 RAD16, pho11 rad16]
30 [2 pho11 RAD16, 2 PHO11 rad16]; _30 [2 PHO11 RAD16, 2 pho11 rad16]; _0 [pho11 RAD16, PHO11 rad16; PHO11 RAD16, pho11 rad16]

cdc10 results in abnormal bud growth. In the cross of pho11 by cdc10, which of the following tetrad ratios would be expected? [Recall that pho11 is ca at map positon 50 on the right arm of chromosome I.]

10 [2 pho11 CDC10, 2 PHO11 cdc10]; _10 [2 PHO11 CDC10, 2 pho11 cdc10]; _40 [pho11 CDC10, PHO11 cdc10; PHO11 CDC10, pho11 cdc10]
10 [2 pho11 CDC10, 2 PHO11 cdc10]; _40 [2 PHO11 CDC10, 2 pho11 cdc10]; _10 [pho11 CDC10, PHO11 cdc10; PHO11 CDC10, pho11 cdc10]
60 [2 pho11 CDC10, 2 PHO11 cdc10]; _0 [2 PHO11 CDC10, 2 pho11 cdc10]; _0 [pho11 CDC10, PHO11 cdc10; PHO11 CDC10, pho11 cdc10]
30 [2 pho11 CDC10, 2 PHO11 cdc10]; _30 [2 PHO11 CDC10, 2 pho11 cdc10]; _0 [pho11 CDC10, PHO11 cdc10; PHO11 CDC10, pho11 cdc10

Tetrad analysis of linked genes: If genes A and B are linked, the map distance can be calculated in the usual manner as the % of recombinant gametes. Alternatively, the relative frequencies of PD, NPD and TT can be used to advantage in determining gene distance.

Tetrad types. Consider a cross in coupling - AB X ab. PD results from the absence of a cross over between genes A and B. A single cross over between the two loci produces a TT tetrad. Two-strand double crossovers produce ? , three-strand double crossovers produce ? and four-strand double crossovers produce ? . Note that only two of the four spores in TT are recombinant for genetic markers - regardless of the number of cross-overs between the markers.

% recombinant gametes assuming: (i) TT results from single cross-overs, and (ii) only NPD results from double cross-overs. Under these assumptions calculation of map distance as % recombinant gametes by the formula
cM = {1/2 [%TT]} + [%NPD]
gives a low estimate of distance. Thus if
PD = 15, TT = 4, NPD = 1;
% recombinant gametes = 1/2 [20] + [5] = 15 or cM = 15.

NOTE: PD contains only parental gametes and NDP conains four recombinant gametes. Thus a key feature of genetic linkage is that the frequency of PD > NPD and TT < 2/3.

% crossover gametes. A better estimate of distance would acknowledge that TT and PD can result from double crossovers. That is, the distance between two markers is defined as the % crossover gametes between the two markers.

If two genes are linked to each other, the map distance between them can be calculated according to the following formula.
cM = 50 {[ (TT + 6 NPD) ] / [ ( PD + NPD + TT ] }

Thus  if  PD = 15, TT = 4, NPD = 1;       50{[ 4 + 6 ] / 20} = 25 cM


This formula assumes only zero, one or two crossovers in a map interval. The distances calculated by this formula vary slightly from SGD values due to SGD's use of a mathematical model. Consider a cross between two haploid strains carrying two different right-arm chromosome 1 linked genes: strain1:  flo1;  strain2: cdc15.

 

SGD Home Maps Physical and Genetic Maps I     Evaluation from the map gives a flo1 - cdc15 distance of approximately ? cM.

The diploid genotype would be flo1 CDC15 / FLO1 cdc15. For linked genes,  PD tetrads ? are in excess of NPD tetrads ? ,  and the frequency of TT tetrads ? varies with the extent of marker linkage.

SGD data for the cross cdc15 x flo1 is 61 PD, 1 NPD and 78 TT. Map distance assuming TT and PD can result from double crossovers is ? cM; whereas cM = ? when assuming that TT results from single cross-overs, and only NPD results from double cross-overs. The latter value is lower because we ignore that ? can result from double cross-overs.

Ordered Tetrads

First and Second Division Segregation

Review meiosis paying particular attention to;

  1. separation of two homologous centromeres at anaphase I ( stage of first division segregation of CEN-linked genes) and
  2. splitting, and subsequent separation, of each of the homologous centromeres at anaphase II ( stage of second division segregation of CEN-linked genes).

In Saccharomyces cerevisiae the four products of meiosis, ascospores, are enclosed within a sac-like structure called an ascus. [SGD-life cycle] When these four ascospores are isolated, and the genotypes of each determined, the resultant tetrad analysis can be used to map genes. In the case of a single heterozygote, such as ADE1 / ade1, the expectation is 2 ADE1 : 2 ? spores. This 2 : 2 segregation is evidence that a single genetic locus is responsible for the phenotype being studied.

Normally the four ascospores of the first and second meiotic divisions are not constrained (placed) in the ascus in any recognizable way. Thus it is not possible to determine if alleles separated (segregated) from each other at the first or second meiotic division. When alleles separate from each other at the first meiotic division there has not been a cross-over of sister chromatids between the allele and the centromere (CEN) - called first division segregation.

Thus first division tetrads contain four non cross-over gametes. If there has been a cross-over between the allelic pair and CEN, the two alleles will segregate from each other at the second meiotic division - called second division segregation. Second division tetrads contain two non cross-over gametes and two cross-over gametes.

Gene - Centromere Distance:

The cross-over (recombinant) gametes result from a single cross-over between the gene and its CEN. Note that the first division segregation of the two homologous CEN 1s vs the second division segregation of ‘ADE1 / ade1' is dependent on a cross-over between ‘ADE1 / ade1' and the centromere. The one cross-over of a second division tetrad results in ? % cross-over (recombinant) and ? % non cross-over (parental) gametes.

One tetrad of an ADE1 / ade1 heterozygote showed first division segregation and one tetrad showed second division segregation. The number of non cross-over gametes is ? and the number of cross-over gametes is ? . The % cross-over gametes = ? .

The fundamental rule of mapping states that 1 % cross-over gametes = 1 map unit (centiMorgan - cM). Gene-CEN distance can be determined as 1/2 the % of second division tetrads.

cM = 1/2 ( % of second division segregation.)

One tetrad from ADE1 / ade1 showed first division segregation and one tetrad showed second division segregation. Using cM = 1/2 ( % of second division segregation), the ADE1-CEN 1 distance = 1/2 ( %) ? or cM ?

CEN Mapping via Linear Asci in Yeast:

In a founding contribution to yeast genetics, Donald Hawthorne selected special yeast strains with linear asci.

Hawthorne, D.C. 1955. The use of linear asci for chromosome mapping in Saccharomyces. Genetics 40: 511-518.

"On the basis of genetic evidence, it was concluded that the nuclear distribution in the linear four-spored asci of a diploid Saccharomyces hybrid follows a definite pattern with the alteration of spores with non-sister nuclei in the ascus. A cytological investigation of the distribution during meiosis also indicated that there was an alternation of non-sister nuclei in the elongated ascogenous cell . Thus, a deviation from the alternation of the dominant and recessive characters in the scoring of these asci also indicates the occurrence of a crossover and gives a simple and direct means to determine the linkage of a gene to the centromere of its chromosome.

From the analysis of 74 linear asci of the diploid hybrid, the gene for tryptophane independence was mapped at about 2.5 units from the centromere, a value which was confirmed independently in tetraploid material. Evidence is also given that the mating type locus is approximately 27 units from the centromere. The genes for melibiose and galactose ( G-2) fermentation, histidine, uracil, and methionine independence did not show linkage."

He found that the spores in the linear ascus were placed in the ascus in a particular order related to the first and second meiotic divisions. The two homologous centromeres of the first division, CEN i and CEN i i, come to rest in the centre of the ascus.

At this stage the two homologous CENs have been partitioned into two different nuclei. In the absence of a cross-over between CEN and an heterozygous site (e.g., TRP1 / trp1), the two alleles would also have undergone first division segregation.

 

 

 

 

 

 

 

 

 

 

 

 

The spindles of the second meiotic division overlap so that the final distribution of daughter CENs is CEN i , CEN i i, CEN i, CEN i i.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

First division segregation of CEN - linked alleles would be:  trp1 TRP1 trp1 TRP1 or ? , whereas patterns of:  trp1 tpr1 TRP1 TRP1, ? or ? or ? or are patterns of second division segregation.

The latter three patterns result from a cross-over between TRP1/ trp1 and CEN4. The map distance can be determined by cM = ? ( % of 2nd div.seg).

Hawthorne observed one TRP1, TRP1, trp1, trp1; one TRP1, trp1, trp1, TRP1; two trp1, TRP1, TRP1, trp1 and seventy TRP1, trp1, TRP1, trp1 asci. The distance between TPR1and CEN4 is ?   cM.

 

Gene Mapping using CEN-linked Markers

Observe the centromere linked markers of yeast by clicking on the 16 centromeres of the SGD map. TRP1 resides on chromosome IV and is immediately flanked by markers SOK1 and ? .

Subsequent detailed trisomic analyses showed a trp1 second division segregation frequency of 0.0050 or a CEN4-trp1 distance of 0.0025 cM. Open SGD's TRP1/YDR007W and link to Locus History at the bottom of the page.

For all practical purposes the close linkage of trp1 to CEN4 precludes a cross-over in the trp1 - CEN4 interval. Other closely linked CEN markers can be used in the same way.

When one marker, such as trp1, is essentially completely linked to its CEN, a second CEN-linked marker can be mapped without the necessity of linear tetrads. Several such analyses resulted in an ade1- CEN distance calculation of 4 cM. In these analyses the second division segregation of ade1 is identified by the observation of an ascus containing 4 spore types ( tetratype ascus). In a TRP1 ade1 by trp1 ADE1 cross there are three possible types of asci:
parental ditype PD TRP1 ade1, trp1ADE1
non parental genotype NPD [ ] ? ;
tetratype TT [ ] ? .
PD and NPD result from first division segregation whereas TT results from second division segregation. See problem 4 below.

Sample problem:

Use SGD to solve the following: LEU2 , a leucine requiring mutant, is known to be linked to CEN3 at approximately 5 cM. In a cross of cdc10 by leu2, which tetrad ratios would be expected? Assume that CDC10 is completely linked to CEN3. Then use the map distance to calculate the expected frequency of PD, NPD and TT]

10 [2 cdc10 LEU2, 2 CDC10 leu2]; __10 [2 CDC10 LEU2, 2 cdc10 leu2; __ 40 [cdc10 LEU2, CDC10 leu2,CDC10 LEU2, cdc10 leu2]
45 [2 cdc10 LEU2, 2 CDC10 leu2]; __0 [2 CDC10 LEU2, 2 cdc10 leu2 ;__ 5 [cdc10 LEU2, CDC10 leu2,CDC10 LEU2, cdc10 leu2]
50 [2 cdc10 LEU2, 2 CDC10 leu2]; __0 [2 CDC10 LEU2, 2 cdc10 leu2; __ 10 [cdc10 LEU2, CDC10 leu2,CDC10 LEU2, cdc10 leu2]
30 [2 cdc10 LEU2, 2 CDC10 leu2]; __30 [2 CDC10 LEU2, 2 cdc10 leu2; __ 0 [cdc10 LEU2, CDC10 leu2,CDC10 LEU2, cdc10 leu2]

 

Ordered Tetrads with Eight Spores:

Several species of ascomycete fungi have linear tetrads with eight ascospores. Unlike yeast, linear tetrads of Neurosporsa crassa and Sordaria fimicola occur naturally, so it was not necessary to select genetic variants with linear tetrads. As seen below, the two meiotic divisions are constrained by the ascus so that  sister nuclei   ( containing sister chromatids) are adjacent to each other in the ascus. Thus nuclei ? contain sister chromatids. The four haploid products of meiosis then divide mitotically to produce eight ascospores. Since the mitotic divisions are also constrained by the ascus, the genotype of couplets of spores is identical.

Full Screen Image

In 1932, Lindegren analyzed N. crassa tetrads for the segregation of mating type A and a. A summary of his data on 273 tetrads indicated first and second division segregation for this trait. The minor component of ascus types I, II, IV and VI were interpreted to result from a mitotic division spindle overlap, or a spore displacement during dissection of the ascus. Tetrads ? below show first division segregation and tetrads ? show second division segregation.

Ascus Type
I
II
III
IV
V
VI
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 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
102 3 123 6 14 8 1 5 10 1

 

Events that lead to Lindegren's asci types I and II (left )and III - VI (right) are depicted below. The map distance between the mating type locus and its centromere is ? cM

 

Gene Conversion    

All the tetrads in Lindegren's mating type data had four A alleles and four a alleles. This is the expectation of an Aa heterozygote segregating for two alleles with equal frequency. Mendel's Law of Segregation and the Chromosomal theory of Heredity predict that both alleles pass into gametes with equal frequency.

However, in 1930 the German scientist Winkler observed tetrads in yeast that deviated from the expected 2:2 meiotic segregation of alleles. Winkler called this phenomenon gene conversion since it seemed that one of the four alleles was converted into the complementary allele. He proposed that one A allele could be converted into an a allele to give a 3 a : 1 A tetrad, or one a allele could be converted into an A allele to give a ? tetrad. Although several scientists observed conversion ratios the unexpected nature of gene conversion prevented its general acceptance.

 

Mitchell (1955) studied tetrads from a pyroxidine marker in N. crassa that was flanked by two other genetic markers. S pyr T / s PYR t triple heterozygotes produced 6 pyr : 2 PYR and 6 PYR : 2 pyr conversion tetrads that segregated 4 : 4 for the flanking markers "S" and "T". Mendelian segregation for flanking markers ruled out the possibility of abnormal chromosomal or nuclear segregation and gave credence to the conversion hypothesis. Rather conversion resulted from abnormal meiotic segregation of a specific site in the chromosome.

Kitani (1962) examined more than 200,000 asci for the segregation of two alleles determining spore color in Sordaria fimicola. The numbers expected per 100,000 tetrads showed five types of conversion tetrad. Type a, cartooned as a second division segregant, is a non conversion tetrad. There are ? additional non conversion ascospore patterns - see table above.  The cross used was RDS by rds where D resulted in a dark spore and the allele d resulted in a light spore. R / r and S / s refer to markers that flank D / d - similar to the S and T markers used by Mitchell in 1955 ( see above). Type b was also shown to be a non conversion tetrad by following the segregation of flanking markers, R ( dark spore D allele) S / r (light spore d allele) s, which indicated spindle overlap - not gene conversion. Thus spores ? result from spindle overlap or mistakes made during spore dissection. In contrast spores ? of type c segregated for non parental  flanking markers ( e.g. R d S and r D s) and type c was defined as an aberrant 4 : 4 conversion tetrad. The five types of conversion tetrad are: c [ aberrant 4 : 4 ]; d [ 5 : 3 ]; e [ ? ]; f [ ? ]; g [ ? ].

Gene conversion is clearly an observable phenomenon that occurs with a low but appreciable frequency. In Saccharomyces cerevisiae the average conversion frequency per locus is approximately 1 %.

What is the biological significance of gene conversion? A hint comes from the observation that chromosomal regions flanking the converted allele are often (30-50%) involved in reciprocal recombination with non sister chromatids. Thus gene conversion appears to be intimately involved in the mechanism of general chromosomal recombination. The next question is how? - what is the mechanism of gene conversion? how does this facilitate general reciprocal recombination? We take this topic up under "Mechanism of Genetic Recombination".

Review: for an overview / review of tetrad analysis work through the following Visual Genetics animation.

 

PROBLEMS

1. Consider the e-text data above PD = 15, TT = 4, NPD = 1. Recall the the fundamental rule of mapping is that
1cM = 1% recombinant gametes.Calculate linkage assuming the ascospores came from a random spore analysis. The total number of ascospore (gametes) is 20 X 4 = 80. The number of recombinant gametes is ? cM = ? . Note that cM = {1/2 [%TT]} + [%NPD] also gives a map distance of 15 cM which is an overestimate ? underestimate ? of map distance compared to cM = 50 {[ (TT + 6 NPD) ] / [ ( PD + NPD + TT ] }.

2. Use SGD's genomic map view to identify genes in the CEN area of chromosomes 1 and 16. In a genetic analysis of the cross, nup60  X   hat1, what type of tetrads would you expect to see ? .

The ascospore genotypes of the observed tetrads are ?

This result is a demonstration of ? .

 

 

 

3. Review CEN Mapping via Linear Asci above. Use SGD's summary of CEN1-ade1 data to get the average map distance from nine different analyses ? Recall from above that in yeast linear tetrads there was an alternation of non-sister nuclei in the elongated ascogenous cell. Assume you only had an ade1 marked strain and that no other markers were available. How would you determine the CEN1-ade1 distance? Give the tetrad type and the expected frequency of tetrads ?

4.  Review Gene Mapping using CEN-linked Markers above. Open SGD's TRP1/YDR007W and link to Locus History at the bottom of the page. Note the trp1 second division segregation frequency of 0.0050 or a CEN4-trp1 distance of 0.0025 cM. For all practical mapping purposes trp1 is completely linked to CEN4. Use the trp1 marker, in a cross in coupling, to map the ade1 CEN1 interval. In your answer provide the expected tetrads and their frequencies. ?

ASCI TYPES
1 2 3 4 5 6
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
37 35 7 5 6 10

5. The segregation of the mating type locus in ordered tetrads of Neurospora crassa is shown in the table. First division tetrads are ? and second division tetrads are ? Why do asci types 1 and 2 differ from each other ? What is the relationship of tetrads 3 to 6 ? Cartoon meiotic events that lead to tetrads 3 , 4, 5, and 6.

See CEN Mapping via Linear Asci in Yeast: above. If this table described yeast linear tetrads (where couplets such as A  A were only one spore A) which asci would result from first division segregation ?

What mapping information can you derive from this table ?

 

 

 

Ascus Type
1 2 3 4 5 6 7
c+ c+ cd +d c+ cd c+
c+ cd cd c+ ++ ++ +d
+d ++ ++ c+ cd cd cd
+d +d ++ +d +d ++ ++
1 17 41 1 5 3 1

6.Sixty-nine ordered tetrads from a Neurospora cross cd X ++ produced seven different types of asci.

PD asci are ? , NPD asci are ? , and TT asci are ? .

What are the gene-centromere distances for the two markers ? Are c and d linked ?

 

7. Why is tetrad c a conversion tetrad while tetrad b is said to be a non conversion tetrad ?

STUDY GUIDE FOR FINAL EXAM - 2005

1. Compare and contrast first and second division in yeast and show how second division segregation can be used to map yeast genes.

2. Study the problem set and be prepared to answer similar questions.