BIO341F2014

Figures for:

Topic Two“Tetrad Analysis “

Several figures are derived from Chapters 3 & 4, Griffiths et al. (8th Edition).

It is best to view this this file in Power Point “presentation mode” so that the animations will work!

James B. Anderson

A tetrad:

• contains all four meiotic products, which can be physically separated from one another and genotyped.

What questions can be answered by examining the genotypes of all four

products of each of several meioses?

• Is allele segregation 1:1 with each meiosis?

• Is each recombination event reciprocal? (In the cross AB X ab, is the aB recombinant ALWAYS accompanied by the Ab recombinant?)

• Which parts of which chromatid strands were involved in crossing over?

There are two kinds of tetrads:

• 1.Unordered

AND:• 2. Ordered

For ordered, let's consider the case of

no crossover between a gene

locus and it's centromere.

In this case, segregation of alleles occurs at meiosis I. We will call this pattern "MI segregation".

Let's look at another case in which there is a crossover between a heterozygous

gene locus and its centromere.

In this case, segregation of alleles occurs at meiosis II. We will call this pattern "MII segregation".

A - MII segregation

B - MI segregation

MII segregation pattern extends distal from the point of exchange.

Tetrad analysis

• Step one: Determine individual genotypes within tetrads.

• Step two: Pool tetrad types. Lump reversals of halves and, within halves, between spore pairs.

a+ +b++ = abab +++b a+

a+ ++++ a+

=ab +b+b ab

• Step three: List according to abundance.• Step four: Set up a table.

– Register each pair of markers as N, P, or T. – Score each allele pair as 1st or 2nd division

segregation.

P N Tab a+ abab a+ a+++ +b +b++ +b ++all all allparental recomb- four possible geno-

inant types present

1. How many genotypes are present?

2. Are these genotypes parental or recombinant?

Consider the following cross of Neurospora crassa:

me + a X + ad A

P

T 2

1P P

P

P

P

P

T

T T

T

T

T

T TT

T

1 1

11

1 1

1

1

1 22

2 2

2

2

2

Step five: Calculate recombinationfrequencies.

gene-to-gene:RF = N + 0.5T/Total

gene-to-centromere:RF = 0.5 MII/Total

ad Ameme/ad

me/A

ad/A

Step six: Construct the map

• me/ad, RF = 0.5 X 9/61 = 0.07• me/A, RF = 0.5 X 13/61 = 0.11• ad/A, RF = 0.5 X 9/61 = 0.07• me/cent, RF =0.5 X11/61 = 0.09• ad/cent, RF = 0.5 X2/61 = 0.02• A/cent, RF = 0.5 X7/61 = 0.06

me A

0.11

0.07 0.07

ad

0.09

0.060.02

Note that recombination frequencies are not additive!!

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

me + a

+ ad A

+ ad A

Parental (solved map order) Tetrad no. 1

Step seven: Locate the crossovers (Note: ALWAYS used the solved map order)

NO CROSSOVERS

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

+ + a

me ad A

+ ad A

Parental (solved map order) Tetrad no. 2

Step seven (cont.) Locate the crossovers.

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

me + A

+ ad a

+ ad A

Parental (solved map order) Tetrad no. 3

Step seven (cont.) Locate the crossovers.

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

+ ad a

me + A

+ ad A

Parental (solved map order) Tetrad no. 4

Step seven (cont.) Locate the crossovers.

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

+ + A

me ad a

+ ad A

Parental (solved map order) Tetrad no. 5

Step seven (cont.) Locate the crossovers.

me + 1 a

me + a 2+ ad 3 A

+ ad A 4

me + a

+ + A

me ad A

+ ad a

Parental (solved map order) Tetrad no. 6

Step seven (cont.) Locate the crossovers.

What is the relationship between the frequency of crossing over and the

frequency of recombination?

• Distances based on recombination frequencies are not additive. Why?

• Some meioses with more than one crossover produce only parental genotypes.

• “True map distance” is the mean number of crossovers per meiotic product within an interval X 100. This is NOT the same as recombination frequency, especially when the interval is large.

• True map distance is expressed in centimorgans (cM)

Mapping: relating recombination frequency to

crossover frequency

Meioses with one or more crossovers yield, on average 50% recombination

P

T

T

N

The Poisson distribution:

f(i) = (e-m mi )/i!

i=0 i=1 i=2 i=n

e-m e-mm (e-mm2)/2 (e-m mn)/n!

0% recombination

Ave. 50% recombination

Ave. 50% recombination

Ave. 50% recombination, whenever n is one or

more

wherei = the number crossovers per meiosisandm = the average number of crossovers per meiosis

1

Let’s first worry about f(i) where i=0

• This is e-m

• f(i) where i > 1 is: 1- e-m

• Half of the meiotic products in these meioses are recombinant.

• Therefore, RF = 0.5 (1- e-m )

• This assumes that crossovers are distributed randomly - i.e., no interference

RF = 0.10 = (aB +Ab)/ total

A B C

Crossover interference (cross abc X ABC)

What frequency of double crossovers would be expected?

RF = 0.30 = (Bc +bC)/ total

Answer 0.10 X 0.30 = 0.03 = (AbC+aBc)/total

Example of NO interference: in 1000 random meiotic progeny, there were about 100 recombinants for AB, about 300 for BC. Of those, about 30 were double crossovers (AbC+aBc)

0.5

1.0RECOMBINATION

MEAN NO. CROSSOVERS/MEIOSIS (m)2.01.0

R = 0.5 (1 - e )-m

MEAN NO. CROSSOVERS/MEIOTIC PRODUCT (x)

"TRUE" MAP DISTANCE (cM)

1.00.5

50 100

Remember, this assumes NO INTERFERENCE.

(What if there were TOTAL interference?)

Correction for double crossovers• Map distance = 0.5(SCO +

2DCO)

• SCO = T-2N

• DCO = 4N

• Map distance = 0.5(T-2N+8N)

• Map distance = 0.5 (T+6N)

Summary of Tetrad Analysis and Mapping

• Alleles segregate 1:1 and recombination is reciprocal within individual meioses.

• We can use tetrad types for mapping.

• For individual tetrads, we can locate the crossovers that occurred in a meiosis.

• We can use tetrad analysis to relate crossover frequency to recombination frequency.