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LECTURE 7: TETRAD ANALYSIS; GENE CONVERSION ..., Exercises of Genetics

Recombination frequency, RF = [(NPD + 1/2T) / total number of tetrads] x 100. In our example, this is ([3 + 1/2(70)]/200) x 100 = 19 map units.

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LECTURE 7: TETRAD ANALYSIS; GENE CONVERSION, continued
Reading: Ch. 5, p 142-151; Ch. 6, p 192-194
Problems: Ch. 5, solved problem III, also 28a, 29ab, 30, 32, 34, 35; Ch. 6, #30, 31
Announcements:
*You must be enrolled in the section that you are attending. If you are not enrolled, you must contact both GSIs
involved and we will work with the MCB Undergraduate Affairs office to get you into a section. You may not get
your first choice. Friday sections tend to be over-enrolled.
*You must take the quiz in the section in which you are enrolled. If you have a conflict, you must arrange to take the
quiz in a different section before you regular section meets.
First, let’s review how to calculate map distance in yeast:
Consider and example of two linked genes, ARG3 and URA2.
P: arg3 ura2 x ARG3 URA2
Diploid: arg3 ura2 / ARG3 URA2
Meiotic products:
PD (arg3 ura2; arg3 ura2; ARG3 URA2; ARG3 URA2) = 127
NPD (arg3 URA2; arg3 URA2; ARG3 ura2; ARG3 ura2) = 3
T: (arg3 ura2; arg3 URA2; ARG3 ura2; ARG3 URA2) = 70
Recombination frequency, RF = [(NPD + 1/2T) / total number of tetrads] x 100. In our example,
this is ([3 + 1/2(70)]/200) x 100 = 19 map units. We will modify the equation to obtain a better
estimate of map distance. If you draw out the possible crossover events between two linked
genes, you can see the different tetrads that result; see Fig. 5.17 (p. 147).
No crossovers --------> PD
Single crossover --------> T
Double crossover (2-strand) --------> PD
Double crossover (3-strand) --------> T
Double crossover (3-strand) --------> T
Double crossover (4-strand) --------> NPD
You can see how we can modify the equation to make it more accurate. Remember that half
(2/4) the strands recombine if there is a single crossover event and that 4 strands recombine if
there is a double crossover event (even if all of the strands don’t participate, some participate
more than once.)
Map distance = (total rec. events / total tetrads) x 100 = [(1/2[SCO] + DCO) / total tetrads] x 100
Map distance = ([1/2 (T – 2 NPD) + 4 NPD]/ total tetrads) x 100
Map distance = (1/2 T + 3 NPD) / total tetrads x 100
For our example above, map distance = ([1/2 (70) + 3 (3)] / 200) x 100 = 22 map units
This modified equation makes 2 assumptions: (1) there are no more than two crossovers in the
interval and (2) there is no chromosomal interference (all types of DCOs occur with equal
frequency.
pf3

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LECTURE 7: TETRAD ANALYSIS; GENE CONVERSION, continued

Reading: Ch. 5, p 142-151; Ch. 6, p 192-

Problems: Ch. 5, solved problem III, also 28a, 29ab, 30, 32, 34, 35; Ch. 6, #30, 31

Announcements:

*You must be enrolled in the section that you are attending. If you are not enrolled, you must contact both GSIs

involved and we will work with the MCB Undergraduate Affairs office to get you into a section. You may not get

your first choice. Friday sections tend to be over-enrolled.

*You must take the quiz in the section in which you are enrolled. If you have a conflict, you must arrange to take the

quiz in a different section before you regular section meets.

First, let’s review how to calculate map distance in yeast:

Consider and example of two linked genes, ARG3 and URA.

P: arg3 ura2 x ARG3 URA

Diploid: arg3 ura2 / ARG3 URA

Meiotic products:

PD ( arg3 ura2 ; arg3 ura2 ; ARG3 URA2 ; ARG3 URA2 ) = 127

NPD ( arg3 URA2 ; arg3 URA2 ; ARG3 ura2 ; ARG3 ura2 ) = 3

T: ( arg3 ura2 ; arg3 URA2 ; ARG3 ura2 ; ARG3 URA2 ) = 70

Recombination frequency, RF = [(NPD + 1/2T) / total number of tetrads] x 100. In our example,

this is ([3 + 1/2(70)]/200) x 100 = 19 map units. We will modify the equation to obtain a better

estimate of map distance. If you draw out the possible crossover events between two linked

genes, you can see the different tetrads that result; see Fig. 5.17 (p. 147).

No crossovers --------> PD

Single crossover --------> T

Double crossover (2-strand) --------> PD

Double crossover (3-strand) --------> T

Double crossover (3-strand) --------> T

Double crossover (4-strand) --------> NPD

You can see how we can modify the equation to make it more accurate. Remember that half

(2/4) the strands recombine if there is a single crossover event and that 4 strands recombine if

there is a double crossover event (even if all of the strands don’t participate, some participate

more than once.)

Map distance = (total rec. events / total tetrads) x 100 = [(1/2[SCO] + DCO) / total tetrads] x 100

Map distance = ([1/2 (T – 2 NPD) + 4 NPD]/ total tetrads) x 100

Map distance = (1/2 T + 3 NPD) / total tetrads x 100

For our example above, map distance = ([1/2 (70) + 3 (3)] / 200) x 100 = 22 map units

This modified equation makes 2 assumptions: (1) there are no more than two crossovers in the

interval and (2) there is no chromosomal interference (all types of DCOs occur with equal

frequency.

ORDERED TETRADS AND GENE-CENTROMERE DISTANCE

In Neurospora crassa , meiosis occurs within the tight confines of a narrow ascus, resulting in the

formation of ordered tetrads. Because of the precise positioning of each meiotic product within

the ascus, one can infer the arrangement (and segregation) of each chromatid of homologous

chromosomes during Meiosis I and II. This gives information about the distance between the

gene and its centromere. (Meiosis II is followed by mitosis; each pair of genetically identical

daugthers sits adjacent to one another. Each ascus is thus made of up 8 haploid ascospores.)

Consider a gene required for ascospore color (ws

gives black spores and ws gives white

spores):

P: ws

x ws

Diploid: ws

/ ws (immediately undergoes meiosis)

If no recombination between ws gene and the centromere occurs, then the resulting ascospores

are arranged in a neat array with black and white spores clearly segregated from one another

(they separated during meiosis I), each type cleanly segregated to either side of the imaginary

line separating the 4

th

and 5

th

ascospores. This is called a first division segregation pattern.

Since the daughters of the mitotic division lie right next to one another, we can simplify the two

possible configurations to:

(ws

ws

ws ws)

(ws ws ws

ws

If recombination occurs between ws and the centromere, then a second division segregation

pattern is observed. Now, both types of spores are found on either side of the imaginary line

between the 4

th

and 5

th

ascospores. Now there are four possible configurations:

(ws

ws ws

ws)

(ws ws

ws

ws)

(ws

ws ws ws

(ws ws

ws ws

When an ascus shows a second division segregation pattern, we know that half of the chromatids

are recombinant and the other half have not participated in crossovers. Thus, we can calculate

the distance of a gene from its centromere simply by dividing the percentage of second division

octads by 2.

Gene-centromere distance = ([# of second division octads / total octads] x 100) / 2

To examine linkage of two genes in Neurospora, we can use the same formulas as we did for

Baker’s yeast.