1

Lecture 7 Three point cross

2

A geneticist isolates two mutations:

ｷ A = tall ｷ a = short

ｷ H = hairy ｷ h = no hair

and constructs the following pure-breeding stocks:

AAhh and aaHH

Tall short

No hair hairy

Linkage

3

These individuals are mated and the F1 progeny are mated to the double recessive. The following results are obtained in the

F2: indep assortment linked loci

Tall, no hair

Short, hairy

Tall, hairy

Short, no hair

total

Do these genes reside on the same or different chromosomes?

Answer-

If on the same chromosome, what is the distance between them?

We simply identify the parental and recombinant classes and determine the recombinant frequency

#Recombinants/Total progeny

Which class is the recombinant?

997

1007

410

415

2827

700

710

690

700

2800

4

P AAhh x aaHHA h x a HA h a H

F1 AaHh x aahhA h x a ha H a h

ah

A h

a H

A H

a h

Aahh

aaHh

AaHh

aahh

Parental

Recomb

410

997

1007

415

Which of these are parental and which are recombinants?

Tall, No hair short, hairy

5

Which are the parental and which are the recombinant classes?

What is the recombination frequency?

So the map distance between the A and H genes is

410+415 825 =29%

410+1007+997+415 2829

6

Another mutation C (crinkled) is isolated and recombination frequencies between this gene and the A and H genes are determined

% recombinantsA to H 29A to C 15H to C 16

A HC

29

15 16

15+16=31

31 is close to 29!

7

What is going on? The map is not very accurate

There is a small error in all our results

8

What is going on? The map is not internally consistent?

---A-------h---------------------A-------h------------------

---a-------H---------------------a-------H------------------

A-h parentalA-H Recomb

a-h Recomba-H parental

---A--------------------h--------A--------------------h-----

---a--------------------H--------a--------------------H-----

A-h parentalA-h parental

a-H parentala-H parental

Single cross-over

What if we get two crossovers between A and H A double cross-over

Now the parental class is over counted for progeny with 2 crossoversThe recombinant class is under counted for the progeny with two crossovers

Over large distance there will be a significant number of double crossovers that go undetected - the genetic distances are underestimated

9

The double crossovers go undetected and therefore over large distances the genetic distances are underestimated

The solution is to include additional markers between A and H to greatly reduce the probability of undetected doubles:

For instance with the intervening C marker the double crossovers can be separated:

---A--------C------------H--------A--------C------------H-----

---a--------c------------h--------a--------c------------h-----

A-C-H parentalA-c-H db Recomb

a-C-h db Recomba-c-h parental

A 15 C 16 H

29+ undetected double (2)

10

Three point cross

Because of the problem of undetected double crossovers, geneticists try to map genes that are closely linked (LESS than 10 m.u.) when constructing a map. This is one of the reasons behind a mapping technique known as

The Three-Point Testcross

To map three genes with respect to one another, we have used a series of pair-wise matings between double heterozygotes

A more efficient method is to perform a single cross using individuals triply heterozygous for the three genes

11

First example

P sc ec vg x Sc+ Ec+ Vg+sc ec vg Sc+ Ec+ Vg+

F1 Sc+ Ec+ Vg+ x sc ec vg

sc ec vg sc ec vg

F2

If these genes were on separate chromosomes, they should be assorting independently and all the classes should be equally frequent.

ARE ALL THREE GENES ON THE SAME CHROMOSOME?

sc ec vg

sc ec vg

Sc+ Ec+ Vg+

sc ec Vg+

Sc+ Ec+ vg

sc Ec+ vg

Sc+ ec Vg+

sc Ec+ Vg+

Sc+ ec vg

Sc= scutellar bristleEc= echinus rough eyeVg= vestigial wing

235

241

243

233

12

14

14

16

12

Are sc and vg linked/not linked???

To map them, we simply examine the pair-wise combinations and identify the parental and recombinant classes:

To determine the distance between sc vg we remove ec

sc vgsc ec vg 235Sc+ Ec+ Vg+ 241sc ec Vg+ 243Sc+ Ec+ vg 233sc Ec+ vg 12Sc+ ec Vg+ 14sc Ec+ Vg+ 14Sc+ ec vg 16

247

255

257

249

# recombinant/total progeny = 506/1008 = 50%

Therefore sc and vg are NOT LINKED!

sc

vgNext: What about sc and ec?

sc vg Sc+ Vg+ sc Vg+ Sc+ vg sc vgSc+ Vg+sc Vg+Sc+ vg

13

What about sc and ec? Are they linked not linked?

sc ecSc+ Ec+sc ecSc+ Ec+sc Ec+Sc+ ecsc Ec+Sc+ ec

478

474

26

30

# recombinant/total progeny = 56/1008 =5.5%

Sc and Ec are Linked

scec

vg

sc vgsc ec vg 235Sc+ Ec+ Vg+ 241sc ec Vg+ 243Sc+ Ec+ vg 233sc Ec+ vg 12Sc+ ec Vg+ 14sc Ec+ Vg+ 14Sc+ ec vg 16

14

Are ec and vg linked? In theory they should not be.

ec vgEc+ Vg+ec Vg+Ec+ vgEc+ vgec Vg+Ec+ Vg+ec vg

251

255

257

245

# recombinant/total progeny = 502/1008 = 50%

Therefore ec and vg are NOT LINKED!

From these observations what is the map distance between ec and vg?

scec

vg

sc vgsc ec vg 235Sc+ Ec+ Vg+ 241sc ec Vg+ 243Sc+ Ec+ vg 233sc Ec+ vg 12Sc+ ec Vg+ 14sc Ec+ Vg+ 14Sc+ ec vg 16

15

More three point crosses

Here is another example involving three linked genes:

v - vermilion eyes

cv - crossveinless

ct - cut wings

To determine linkage, gene order and distance, we examine the data in pair-wise combinations

When doing this, you must first identify the Parental and recombinant classes!

P

F1

F2

16

More three point crosses

P V+ cv ct x v Cv+ Ct+V+ cv ct v Cv+ Ct+

F1 v Cv+ Ct+ x v cv ct

V+ cv ct v cv ct

F2

v cv ct

v Cv+ Ct+ 580

V+ cv ct 592

v cv Ct+ 45

V+ Cv+ ct 40

v cv ct 89

V+ Cv+ Ct+ 94

v Cv+ ct 3

V+ cv Ct+ 5

P

P

R

R

R

R

R

R

v - vermilion eyes

cv - crossveinless

ct - cut wings

Vermilion, norm vein, norm wing

Norm eye, crossvein, cutwing

Vermilion, crossvein, norm wing

Norm eye, norm vein, cutwing

Vermilion, crossvein, cutwing

Norm eye, norm vein, norm wing

Vermilion, norm vein, cutwing

Norm eye, crossvein, norm wing

17

Distance between v and cv

v to cv

v cv ctv Cv+ Ct+ v Cv+ 580V+ cv ct V+ cv 592v cv Ct+ v cv 45V+ Cv+ ct V+ Cv+ 40v cv ct v cv 89V+ Cv+ Ct+ V+ Cv+ 94v Cv+ ct v Cv+ 3V+ cv Ct+ V+ cv 5

Parental

V Cv+ 583

V+ cv 597

Recombinant

V+ Cv+ 134

v cv 134268/1448 = 18.5%

The genes are linked!

18

Distance between ct and cv

ct to cv

v cv ctv Cv+ Ct+ Cv+ Ct+ 580V+ cv ct cv ct 592v cv Ct+ cv Ct+ 45V+ Cv+ ct Cv+ ct 40v cv ct cv ct 89V+ Cv+ Ct+ Cv+ Ct+ 94v Cv+ ct Cv+ ct 3V+ cv Ct+ cv Ct+ 5

Parental

Cv+ Ct+ 674

cv ct 681

Recombinant

Cv+ ct 43

cv Ct+ 50

93/1448 = 6.4%

The genes are linked!

19

Distance between v and ct

v to ct

v cv ctv Cv+ Ct+ v Ct+ 580V+ cv ct V+ ct 592v cv Ct+ v Ct+ 45V+ Cv+ ct V+ ct 40v cv ct v ct 89V+ Cv+ Ct+ V+ Ct+ 94v Cv+ ct v ct 3V+ cv ct+ V+ Ct+ 5

Parental

v Ct+ 625

V+ ct 632

Recombinant

V+ Ct+ 99

v ct 92

191/1448 = 13.2%

The genes are linked

v cv

18.5

v ct

13.2

ct cv6.4

v cv

18.5

ct13.2 6.4

The map is not very accurate

It is internally inconsistent -Undetected DCO

A more accurate map is:

v cvct13.2 6.4

21

21

San Francisco is 100 miles from Santa CruzSan Francisco is 40 miles from Half moon bay

Is Santa Cruz 60 miles from Half moon bay or140 miles from Half moon bay

San Francisco Santa Cruz

Half Moon San Francisco

100

40

Half Moon 60Santa Cruz

22

DCO

Notice if one focuses on the v and cv markers, they will be scored as non-recombinant (parental).

However if one also scores v-ct and ct-cv the double recombination event from which they arose can be detected.

In fact, when scored, a number of recombinations occur between v and cv. These classes should be counted. By including these double recombinants the map is internally consistent.

Parental chromosomes

v----Ct+-----cv+ & V+----ct----cvv----Ct+-----Cv+ V+----ct----cv

v Ct+ Cv+

v Ct+ Cv+

V+ ct cv

V+ ct cv

The parental homologs will pair in meiosisI. Crossing over will occur and a Double crossover produces:

v Ct+ Cv+

v ct Cv+

V+ Ct+ cv

V+ ct cv

23

Another method to solve a three point cross

Solving three-point crosses

1. Identify the two parental combinations of alleles

2. The two most rare classes represent the product of double crossover.

v cv ctv Cv+ Ct+ 580V+ cv ct 592v cv Ct+ 45V+ Cv+ ct 40v cv ct 89V+ Cv+ Ct+ 94v Cv+ ct 3V+ cv Ct+ 5

Parent

DCO

24

Solving three-point crosses

1. Identify the two parental combinations of alleles

2. The two most rare classes represent the product of double crossover.

Parent v Cv+ Ct+ & V+ cv ct

DCO v Cv+ ct & V+ cv Ct+

3. With this knowledge, you can establish a gene order in which a double cross produces the allelic combination observed in the most rare class.

There are three possible relative order of the three genes in the parent:

25

There are three possible gene orders for the parental combination

**basically we want to know which of the three is in the middle**

Parent v Cv+ Ct+ & V+ cv ctvermillion rednormal vein crossveinlessnormal wing cut wing

DCO v Cv+ ct & V+ cv Ct+vermillion rednormal vein crossveinlesscut wing normal wing

You are driving along Rte1 and you are told that there are three towns along this route- San Francisco, Half moon bay and Santa Cruz.You have no idea which town you will encounter first, second and last.How many possible orders are there?

San Francisco----Santa Cruz----Half moon bay

San Francisco----Half moon bay----Santa Cruz

Half moon bay----San Francisco----Santa Cruz

26

Parent v Cv+ Ct+ & V+ cv ctvermillion rednormal vein crossveinlessnormal wing cut wing

Observed DCO v Cv+ ct & V+ cv Ct+

There are three possible gene orders for the parental combination

**basically we want to know which of the three is in the middle**

Each relative order in the parent gives a different combination of the rarest class (DCO)

v----Cv+----Ct+V+---cv-----ct

OR

v----Ct+----Cv+V+---ct-----cv

OR

Ct+----v----Cv+ct-----V+---cv

predicted DCO

v----cv----Ct+V+---Cv+---ct

v----ct----Cv+ *V+---Ct+---cv

Ct+----V+---Cv+ct-----v----cv

27

Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified

v cv ct

v Cv+ Ct+ 580

V+ cv ct 592

v cv Ct+ 45

V+ Cv+ ct 40

v cv ct 89

V+ Cv+ Ct+ 94

v Cv+ ct 3

V+ cv Ct+ 5

Gene Order v----ct----cv

REWRITE THE COMBINATION IN THE PARENTS v---Ct+---Cv+ and V+---ct---cv

28

3. Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified

v..Ct+..Cv+ V+..ct..cv

v..Ct+..cvV+..ct..Cv+

v..ct..cv V+..Ct+..Cv+

v..ct..Cv+ V+..Ct+..cv

Gene Order

v----ct----cv

REWRITE THE COMBINATION IN THE PARENTS v---Ct+---Cv+ and V+---ct---cv

v Ct+ Cv+

v Ct+ Cv+

V+ ct cv

V+ ct cv

P

SCO-II

SCO-I

DCO

v cv ct

v Cv+ Ct+ 580 V+ cv ct 592

v cv Ct+ 45 V+ Cv+ ct 40

v cv ct 89 V+ Cv+ Ct+ 94

v Cv+ ct 3 V+ cv Ct+ 5

CO1 CO2

29

Now the non-recombinants, single recombinants, and double recombinants are readily identified

Recombination freq in region I = 89+94 + 3+5

SCOI DCO

Recombination freq in region II = 45+40 + 3+5

SCOII DCO

Now the DCO are not ignored.With this information one can easily determine the map distance between any of the three genes

v--------13.2 m.u.--------ct--------6.4m.u.-------cv

30

Now the non-recombinants, single recombinants, and double recombinants are readily identified

Parental input:

(As a check that you have not made a mistake, reciprocal classes should be equally frequent)

With this information one can easily determine the map distance between any of the three genes:

31

Interference

Interference: this is a phenomenon in which the occurrence of one crossover in a region influences the probability of another crossover occurring in that region.

Interference is readily detected genetically. For example, we determined the following map for the genes v ct and cv.

v--------13.2 m.u.--------ct--------6.4m.u.-------cv

Expected double crossovers = product of single crossovers

The expected frequency of a double crossover is the product of the two frequencies of single crossovers:

DCO= 0.132 x 0.064= 0.0084

Total progeny = 1448

Expected number of DCO is 0.0084 x 1448 = 12

Observed number of DCO = 8

The coefficient of coincidence is calculated by dividing the actual frequency of double recombinants by this expected frequency:c.o.c. = actual double recombinant frequency / expected double recombinant frequency

Reduction is because of interference

32

Interference is often quantified by the following formula:

I= 1- observed frequency of doubles/ expected frequency of

Doubles

I= 1- 8/12 = 4/12 = 33%

If actual frequency is the same as expected frequency then Interference is 1-1=0

33

Example

P sc ec vg x Sc+ Ec+ Vg+sc ec vg Sc+ Ec+ Vg+

F1 Sc+ Ec+ Vg+ x sc ec vg

sc ec vg sc ec vg

F2

If these genes were on separate chromosomes, they should be assorting independently.

Are all three assorting independently, are two assorting independently or are none assorting independently

sc ec vg sc ec vgsc ec vg

sc ec vg

Sc+ Ec+ Vg+

sc ec Vg+

Sc+ Ec+ vg

sc Ec+ vg

Sc+ ec Vg+

sc Ec+ Vg+

Sc+ ec vg

Sc= scutellar bristleEc= echinus rough eyeVg= vestigial wing

235 89 130

241 94 145

243 45 132

233 40 149

12 3 145

14 5 132

14 580 133

16 590 145

34

Meiosis consists of two divisions, meiosis I and II, by which a diploid cell produces four haploid daughters. Reduction in ploidy occurs at meiosis I, when homologous chromosomes (homologs) disjoin. This event is prepared during meiotic prophase, when homologs recognize each other and form stable pairs (bivalents) that can line up in the metaphase I spindle. In most eukaryotes, including mouse and yeast, both the recognition of homologs and the formation of stable bivalents depend on recombinational interactions between homologs (reviewed in ref. 1). For this process, the meiotic prophase cell actively induces DNA double-strand breaks (DSBs) and repairs them by homologous recombination, using preferably a nonsister chromatid of the homolog as template (2). In species such as yeast and mouse, most interhomolog recombinational interactions are not resolved as reciprocal exchanges [crossovers (COs)] and probably serve homolog recognition and alignment (3, 4). A small proportion, however, yields COs, which become cytologically visible as chiasmata and are essential for the stable connection of homologs. COs are not randomly distributed among and along bivalents; every bivalent forms at least one CO (obligate CO), and, if multiple COs occur, they are more evenly spaced along the bivalent than would be expected if they were randomly placed. This phenomenon was originally detected genetically by the finding that the frequency of double recombinants involving a pair of adjacent or nearby intervals was lower than the frequency expected from recombinant frequencies for each of those intervals (reviewed in refs. 5 and 6). Interference has also been analyzed cytologically, from spatial distributions of chiasmata (7, 8) or recombination complexes along chromosomes during meiotic prophase, when recombination is in progress (9). How interference is imposed is not known.

Concomitantly with meiotic recombination, the sister chromatids of each chromosome form a common axis, the axial element (AE), and the AEs of homologs align. Then, numerous transverse filaments connect the AEs of homologs, and a zipper-like structure, the synaptonemal complex (SC), is formed between the homologs (1). Protein complexes that mediate, and mark the sites of, recombination have been localized to AEs or SCs by both EM and immunocytology (reviewed in refs. 10 and 11). These studies (9, 12), together with molecular genetic analyses (13, 14), have elicited several specific questions regarding the imposition of interference: At which step in meiotic recombination is interference first detectable? Is the level of interference the same among recombination complexes representing early and late steps in meiotic recombination? Does the SC contribute to interference? We have analyzed these questions in the mouse by examining how protein complexes that are thought to mark intermediate and late events in meiotic recombination are distributed along SCs in two stages of meiotic prophase.

In mouse, many recombination-related proteins have been identified, and the meiotic time courses of immunofluorescent foci containing these proteins have been described (15, 16). The mouse transverse filament protein SYCP1 is also known (17, 18), and SYCP1-deficient mice have been constructed (19). We have analyzed the distributions of four types of foci along mouse SCs or AEs in wild-type and/or Sycp1–/– strains: (i) MLH1 foci, which occur during pachytene and specifically mark the sites of COs (9, 20); (ii and iii) MSH4 and replication protein A (RPA) foci, which appear earlier, during zygotene, and were analyzed here at late zygotene. In mouse, these foci outnumber the prospective COs. However, a subset of them likely matures into MLH1 foci and then into COs, because early MLH1 foci colocalize with MSH4 (16, 21) but then lose MSH4 at later stages; (iv) because Sycp1–/– strains do not form MLH1 foci (19), we analyzed {gamma}H2AX signals in Sycp1–/– pachytene spermatocytes. In wild-type meiosis, {gamma}H2AX signals occur from leptotene until pachytene (22). Based on their timing and other evidence (reviewed in refs. 13 and 23), MSH4 and RPA foci likely mark early intermediate stages of recombination involving strand exchange, whereas MLH1 foci likely mark the latest stages, e.g., conversion of double Holliday junctions to COs. {gamma}H2AX signals mark various DNA lesions, including DSBs (24); in Sycp1–/– pachytene, they probably represent (perhaps diverse) unresolved recombination intermediates (19).

For the detection of genetic interference, the coefficient of coincidence (CC) is often used. However, CC is problematic as a measure for the level of interference because it is not based on the precise positions of genetic exchanges but instead is based on the frequencies of recombinants for genetic markers that delimit two adjacent or nearby chromosomal intervals.