Chapter 5 Linkage, recombination, and the mapping of genes on chromosomes.

Chapter 5

Linkage, recombination, and the

mapping of genes on chromosomes

Fig. 5.1

linkage

recombination

wY+

W+y

wY+

W+y

wY+

wY+

wY+

W+y

W+Y+wy

When genes are linked, parental

combinations outnumber

recombination types.

W+Y+

wy

W+Y+

W+Y+

wy

W+Y+

Two parental types

Two recombination types

W+Y+

wy

W+Y+

W+Y+

wy

W+Y+

Fig. 5.5

Autosomal genes can also exhibit linkage

bb: black cc: curved

bc+

bc+

b+c

bc+

b+c

b+c

The Chi square test pinpoints the probability that experimental results are

evident for linkage

Chi test measures the “goodness of fit”:

How often an experimentally observed deviation from the prediction of a particular hypothesis will occur solely by chance.

Fig. 5.6

Assume A and B genes are not linked.

F1

Chi Square – Experiment 1 & 2

2 = (observed – expected)2

number expected

2 = (31 – 25)2 + (19 – 25)2 25 25

= 2.88

2 = (62 – 50)2 + (38 – 50)2 50 50

= 5.76

Experiment 1

Experiment 2

Table 5.1

A and B are not linked A and B are linked

difference is significantdifference is non-significant

Recombination results when crossing-over during meiosisseparates linked genes

Fig. 5.7

Evidence that recombination results from reciprocal exchangebetween homologous chromosomes

X chromosome

Fig. 5.8

Recombination through the light microscope

(synaptonemal complex)

anaphase

Fig. 5.9

Recombination frequencies are the basis of genetic map

RF: recombination frequency;

1% RF= 1 Centimorgan (cM)=1 map unit (m.u.)

Fig. 5.10

Unlinked genes show a recombination frequency of 50%

ry ry+

tkv tkv+

ry

tkv

ry

tkv+

ry+

tkv

ry+

tkv+

Fig. 5.10

Unlinked genes show a recombination frequency of 50%

Locus: chromosomal position of a gene

Mapping: the process of determining that locus

Fig. 5.11

Mapping genes by comparison of two-point crosses

The limitation of two-point cross

1. Gene order is difficult to determine if they are very close.

2. Actually distance do not always add up.

Fig. 5.12

Vestigial wingsBlack bodyPurple eye color

Vg to b: (252+ 241+131+118)/4197=0.177, 17.7%Vg to pr:(252+241+13+9)/4197=0.123, 12.3%B to pr:(131+118+13+9)/4197=0.064, 6.4%

Fig. 5.13

Three point-crosses allow correction for double cross-over

Vg to b (three-point cross):(252+241+131+118+13+13+9+9)/4197=0.187, 18.7%

Vg to b (two-point cross): (252+ 241+131+118)/4197=0.177, 17.7%

For greatest accuracy, it is always best to constructa map using many genes separated by relative short

distance.

Mapping genes at X chromosome by two-point cross

Fig. 5.14

RF between Y and W: 49+41+1+2/6823 X100=1.3 m.u.

RF between m and W: 1203+1092+2+1/6823 X100=33.7 m.u

RF between m and y: 1203+1029+49+41+2+2+1+1/6823 X100=35 m.u.

y w m

The actual physical distance between genes does notalways show a direct correspondence to genetic map

distance

1. Recombination is not uniform over the length of a single chromosome,

Hot spot.

• The existence of double, triple, or even more cross-overs.

Rates of recombination differ from species to species

In human, 1 m.u. is = 1 million baseIn yeast, 1 m.u. is 1500 base pairs

In Drosophila, meiotic recombination only occurs in female.

Fig. 5.15

Linkage groups:

Fig. 5.16a

The life cycle of the yeast Saccharomyces cerevisiae

stress

Fig. 5.16b

The life cycle of the bread mold Neurospora crassa

Bread mold

Fig. 5.17ab

How meiosis can generate three kinds of tetrads

Fig. 5.17cde

When PD=NPD, two genes are unlinked

Four types of gametes when genes on different chromosome

H

HHtt hhTT

Ht hT

H

h

h

T

T

t

t

(h)

(H)

Fig. 5.18

When genes are linked, PDs exceed NPDs

Fig. 5.19abc

How crossovers between linked genesgenerate different tetrads

Fig. 5.19def

Rare!

How to calculate the recombination frequency between

two linked genes in the tetrad analyses?

RF=(NPD+1/2 T)/total tetrads x100

RF= 3+(1/2)(70)/200 x100=19 m.u. (tetrads) = (4X3)+ (2X70)/800 x100=19 m.u. (spores)

Fig. 5.20

Tetrad analyses confirms that recombination occurs at the four-strand stage

(A mistake model!) (a lot)

Fig. 5.22

How ordered tetrads form

Arrangement of the four chromatids of each homologous chromosome pair

Fig. 5.23

(Cross-over between gene and centromere)

Ordered Tetrads help locate genes in relation to the centromere

Fig. 5.24

Genetic mapping by ordered-tetrad analysis

Thr-centromere: (1/2) (16+2+2+1)/105 x100=10 m.u.

Arg-centromere:(1/2) (11+2+2+1)/105 x100=7.6 m.u.

Thr-Arg linkage: 3+(1/2)(16+11+2)/105 X100=16.7

PD PDT T T NPD NPDDouble

cross-over

NPD Tetrad

Double cross-over

Double cross-over

Double cross-over

Fig. 5.25

Mitotic recombinationWild-type tissue: y sn+/Y+ sn: wild-type color and bristle

y sn+/y sn+ Y+ sn /Y+ sny sn+/y sn+

Y+ sn /Y+ sn

Fig. 5.26

Fig. 5.27

Mitotic recombination during growth of diploid yeastcolonies can create sectors

ADE2/ade2

ade2/ade2

Recombinatioinbetweem ade2and centromere

AA

aa

A

Aa

aMitotic recombination

Fig. 5.28

Fig. 5a.p131

Fig. 5a.p143

Fig. 5.21

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