1

Linkage and Chromosome Mapping

in Eukaryotes

Fundamental GeneticsLecture 8

John Donnie A. Ramos, Ph.D. Dept. of Biological Sciences

College of Science University of Santo Tomas

Linked Genes

Genes located in the same chromosomes

Initiated by Thomas Morgan and Alfred Sturtevant

Will not segregate independently

Affected by crossing over

2

Linkage vs Independent Assortment

Linked Genes in DrosophilaRed eyes (bw+) dominant to mutant brown eyes (bw)

Thin wing veins (hv+) dominant to mutant heavy wing veins (hv)

3

X-Linked Genes in DrosophilaCross A

Gray body (y+) dominant to mutant yellow body (y)

Red eyes (w+) dominant to mutant white eyes (w)

Cross B

Red eyes (w+) dominant to mutant white eyes (w)

Normal wings (m+) dominant to mutant miniature wings (m)

Due to linkage and crossing over that occurred during meiosis

Distance between linked genes is related degree of crossing over

Chromosome MappingDetermining the distances between genes and the order of sequence in a chromosomeUses the frequency of crossing

The shorter the distance between linked genes, the lower the frequency of crossing-over.The longer the distance, the higher the frequency of crossing over to occur.

Frequencies of crossing over:1. Yellow, white 00.5 %2. White, Miniature 34.0 %3. Yellow, miniature 35.4 %

1% of crossing over = 1 map unit (centimorgan, cM)

4

Distance Affects Crossover

Single Crossover

50% are recombinant gametes50 % non-crossover gametesIf distance between genes is more than 50 map units, ~100 % crossing over will occur.

5

Multiple Crossover

Occurrence of more than one crossovers between non-sister chromatids.Produces double crossover (DCO) gametesIf the probability of crossover between A and B is 20% (0.20) and the probility of crossover between B and C is 30% (0.30), the frequency of DCO is 6 % (0.06) Product Law: (0.2)(0.3)=0.06

Three-Point Mapping

The percentage of crossing over could be used to map genes in a chromosomeThree criteria needed for successful mapping:

Genotypes of organisms producing the crossover gametes must be heterozygous for all gene loci Cross must be constructed so that the genotypes of gametes could be determined based on the phenotypes of the offspring.Large number of offspring must be produced

6

Traits considered:

1. Body color

Gray(y+) dominant to yellow (y)

2. Eye color

Red eyes (w+) dominant to white (w)

3. Eye shape

Normal (ec+) dominant to echinus (ec)

10,000

Determining Gene SequenceSteps:

Determine 3 possible ordersw-y-ec (y at the middle)y-ec- w (ec at the middle)y-w-ec (w at the middle)

Perform a theoretical double cross overCompare the theoretical DCO with actual DCO (least no.)Perform theoretical NCOI and NCOII and compare with data

White echinus eyes, gray body

Red normal eyes, yellow body

Yellow body, normal white eyes

Gray body, echinus red eyes

Yellow body, echinus red eyes

Gray body, white normal eyes

7

Unknown Gene Sequence

Total=1109

Unknown Gene Sequence

8

Not all crossovers can be detected

Degree of inaccuracy increases with increasing distance between linked genes

Observed vs Expected DCO

Observed DCO = double cross-over that actually occurred

Example: (44 + 42)/1109 = 0.078

Expected DCO = theoretical double crossoversProduct of all the SCOI and SCOIIExample: (82+79+44+42) / 1109 = 0.223

(200+195+44+42) / 1109 = 0.434DCO exp= (0.223)(0.434) = 0.097

9

Coefficient of Coincidence and Interference

Coefficient of Coincidence (C)The measure of actual DCOs that occurredC = Observed DCO / Expected DCO

= 0.078/0.097= 0.804 or 80.4%

Interference (I)phenomenon when a crossover event in one region of a

chromosome inhibits a second event to occur in a nearby region)

I = 1-C= 1-0.804 = 0.196 or 19.6%

Interpretation: 19.6% of expected DCO did not occur or only 80.4% of expected DCO was observed

Problem 1

A stock of corn homozygous for the recessive linked genes colorless (c), shrunken (sh), and waxy (wx) was crossed to a stock of homozygous for the dominant wild type alleles of the genes (+ + +). The F1 plants were then backcrossed to the homozygous recessive stock. The F2 results were as follows:

Phenotype Number Phenotype Number

+ + + 17,959 + + wx 4,455

c sh wx 17,699 c sh + 4,654

+ sh wx 509 + sh + 20

c + + 524 c + wx 12

a. Determine the distance between the c and sh

b. Determine the distance between the sh and wx

c. Determine the distance between c and wx

d. Give the coefficient of coincidence

e. Compute for the interference

10

Problem 1: Solutionc sh wxc sh wx

+ + ++ + +x

x c sh wxc sh wx

c sh wx+ + +

NCO + + + = 17,959c sh wx = 17,699

SCOI + sh wx = 509c + + = 524

SCOII + + wx = 4,455c sh + = 4,654

DCO + sh + = 20c + wx = 12Total = 45,832

35,658

1,033

9,109

32

=

=

=

= 00.07 %

19.87 %

02.25 %

77.80 %

Problem 1: Solution

Distance between c and sh = (509 + 524 + 20 +12) / 45,832

= 0.0232 or 2.32 %

Distance between sh and wx = (4466 + 4654 + 20 + 12) / 45832

= 0.1994 or 19.94 %

Distance between c and wx = 2.32 + 19.94

= 22.26

11

Problem 1: Solution

c sh wx

2.32 mu 19.94 mu

22.26 mu

C = (0.0007) / (0.0232)(0.1984) = 0.1521 or 15.21%

I = 1-C= 1-0.1521= 0.8479 or 84.79 %

Problem 2

In a variety of tomato plant, the mutant genes o (oblate fruit), h (hairy fruit), and c (compound inflorescence) are all located in chromosome 2. These genes are recessive to their wild type alleles round fruit, hairless and single inflorescence, respectively. A testcross mating of an F1 heterozygote for all three genes resulted in the following phenotypes:

Phenotypes Number Phenotypes Number

+ + + 73 + + c 348

+ h + 2 + h c 96

o + + 110 o + c 2

o h + 306 o h c 63

a. Determine the sequence of the 3 genes in chromosome 2

b. Give genotypes of the homozygous parents (P1) used in making the F1 heterozygote.

c. Compute for the map distances between the genes

d. Give the coefficient of coincidence and interference

12

Problem 2: Solution+ + +

o h c

o h c

o h cx

NCO: o h + = 306

+ + c = 348

DCO: + h + = 2

o + c = 2

Inference from given data:

Sequence of genes is not correct

One chromosome contains 2 wild type alleles while the homolog contains the 3rd wild type allele

Three possible orders of the genes involved:

o h c

o c h

h o c

Find a sequence that will satisfy both NCO and DCO

o h ++ + c Satisfies NCO but not DCO

+ c ho + + Satisfies DCO but not NCO

+ o hc + +

Satisfies both NCO and DCO

Problem 2: Solution

+ o h

c + +

o h c

o h cx

NCO: o h + (same as + o h) = 306+ + c (same as c + +) = 348

DCO: + h + (same as + + h) = 2o + c (same as c o +) = 2

Try if the sequence can satisfy the SCOs

SCO I: + + + = 73o h c (same as c o h) = 63

SCOII: o + + (same as + o +) = 110+ h c (same as c + h) = 96

Total = 1,000

654 = 0.654 or 65.4%

4 = 0.004 or 0.4%

136 = 0.136 or 13.6%

206 = 0.206 or 20.6%

13

Problem 1: Solution

Distance between c and o = (73 + 63 + 2 + 2) / 1,000

= 0.140 or 14 % / cM

Distance between o and h = (110 + 96 + 20 + 12) / 1000

= 0.210 or 21 % / cM

Distance between c and wx = 14 + 21

= 35 cM

Problem 1: Solution

c o h

14 cM 21.0 cM

35 cM

C = (0.004) / (0.14)(0.21) = 0.1361 or 13.61%

I = 1-C= 1-0.1361= 0.8639 or 86.39 %