MATERI 7 DAN 8 SEX LINKAGE DAN CROSSING OVER. Gamet yang terbentuk dari individu dihibrid

MATERI 7 DAN 8SEX LINKAGE DAN CROSSING OVER

Gamet yang terbentuk dari individu dihibrid

Gamet yang terbentuk dari individu dihibrid

A.Gen tidak berangkai B.Kedua Gen berangkai : Couple Phases, Kedudukan Sis

Cara penulisan: (AB) ab

C. Kedua Gen berangkai : Repulsion Phases, Kedudukan TransCara penulisan: (Ab)

aB

Lalat Drosophila gen terangkai:

Cu = gen untuk sayap normalcu = gen untuk sayap keriput

Sr = gen untuk dada polos (normal)sr = gen untuk dada bergaris-garis

Cu Srcu sr

Cu srcu Sr

Gen terangkai susunan sis

Gen terangkai susunan Trans

Rangkai Sempurna Susunan sis

Lalat jantan sayap keriput dada bergaris dikawinkandengan lalat betina bersayap normal (sayap dan dadanya) homozigot

Cu SrCu Sr

cu srcu sr

Betina jantan

Cu Srcu srF1

F2 : 3 : 0 ; 0 ; 1

Rangkai Sempurna Sususan Trans

Crossing Over of John Edward’s Chromosomes

Linkage & Mapping

Eukaryotic chromosomes are long ds DNA molecules

Typical chromosome contains thousands of genes (loci)

Linkage loci located on the same chromosome linked loci tend to be transmitted as a unit

Linkage

Linkage

Because they are a group of genes linked together, chromosomes are functionally linkage groups

# of linkage groups = the # of types of chromosomes

Crossing over causes loci that are far apart on the same chromosome to sometimes independently assort known as incomplete linkage

Crossing Over Produce Recombinant Phenotypes

Crossing over (meiotic recombination) Occurs during prophase I of meiosis at the

bivalent stage zygotene - pachytene - diplotene

Non-sister chromatids of homologous chromosomes exchange DNA segments

Figure 5.1

Linkage Prevents Independent Assortment

Figure 5.1

gametes with a combination of alleles NOT found in the

original chromosomes as a result of meiotic recombination

These are termed parental gametes

These are called nonparental or recombinant gametes

Crossing Over May Produce Recombinant

Phenotypes

Example of Linkage

Bateson and Punnett conducted a cross in sweet pea involving two traits Flower color and pollen shape

Dihybrid cross expected to give 9:3:3:1 phenotypic ratio of F2 phenotypes Observed linkage Called it coupling

Purple flowers, long pollenPurple flowers, round pollenRed flowers, long pollenRed flowers, round pollen

296192785

15.61.01.44.5

240808027

9331

Observednumber

ObservedRatio

ExpectedRatio

Expectednumber

Purple flowers,long pollen (PPLL)

F2 offspring phenotypes

F1 offspring

Purple flowers,long pollen (PpLl)

Self-fertilization

x

Red flowers,round pollen (ppll)

Example of Linkage

(9:3:3:1)P

P

NPNP

Morgan Provided Evidence for the Linkage of Several X-linked Genes

The first direct evidence of linkage came from studies of Thomas Hunt Morgan

Morgan investigated several traits that followed an X-linked pattern of inheritance Body color Eye color Wing length

Linkage to a Particular

Chromosome

Tan body, red eyes, normal wingsTan body, red eyes, miniature wingsTan body, white eyes, normal wingsTan body, white eyes, miniature wingsYellow body, red eyes, normal wingsYellow body, red eyes, miniature wingsYellow body, white eyes, normal wingsYellow body, white eyes, miniature wings

439208

1570

178365

319193

01150

139335

758401

116120

317700

Females Males Total

y w m/y w m

y+w+ m+/ y w m

F2 generation

F1 generation

F1 generation contains wild-typefemales and yellow-bodied,white-eyed, miniature-wingedmales.

x

y w m / Y

x

y+ w+ m+ Y

Sex linkage of all traits places them all on X

chromosome

/

P

P

5-13Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display

Morgan observed a much higher proportion of the combinations of traits found in the parental generation

P Males

P Females

Morgan’s explanation: All three genes are located on the X chromosome Therefore, they tend to be transmitted together as a unit

However, Morgan still had to interpret two key observations

Why did the F2 generation have a significant number of nonparental combinations?

Why was there a quantitative difference between the various nonparental combinations?

Morgan Provided Evidence for the Linkage of Several X-linked Genes

Gray body, red eyes 1,159

Yellow body, white eyes 1,017

Gray body, white eyes 17

Yellow body, red eyes 12

Total 2,205

Reorganize Morgan’s data considering pairs of genes separately

Red eyes, normal wings 770

White eyes, miniature wings 716

Red eyes, miniature wings 401

White eyes, normal wings 318

Total 2,205

It was fairly common to get this nonparental combination

But this nonparental combination was rare

Figure 5.4

These parental phenotypes are the most common offspring

because the genes are far apart

These recombinant offspring are not uncommon

Figure 5.4

because the genes are very close together

These recombinant offspring are fairly uncommon

These recombinant offspring are very unlikely1 out of 2,205

it is product of single cross over probabilities

Incomplete Linkage

Linked loci that are sometimes separated by recombination are inherited in a pattern between linked and independently assorting not always together not present in normal dihybrid ratios

The percentage of offspring with the loci linked vs those with the loci separated is a measure of the physical distance separating the loci on the chromosome

An undergraduate in the laboratory of T. H. Morgan Constructed first genetic map in 1911 Sturtevant wrote:

“In conversation with Morgan … I suddenly realized that the variations in the length of linkage, already attributed by Morgan to differences in the spatial orientation of the genes, offered the possibility of determining sequences [of different genes] in the linear dimension of the chromosome. I went home and spent most of the night (to the neglect of my undergraduate homework) in producing the first chromosome map, which included the sex-linked genes, y, w, v, m, and r, in the order and approximately the relative spacing that they still appear on the standard maps.”

Alfred Sturtevant’s Analysis

Estimating the relative distances between linked genes, based on the amount of recombination occuring between them allows us to generate genetic maps If the genes are far apart many recombinant offspring If the genes are close very few recombinant offspring

Map distance = Number of recombinant offspring

Total number of offspringX 100

The units of distance are called map units (mu) They are also referred to as centiMorgans (cM)

One map unit is equivalent to 1% recombination frequency

Linkage and Genetic Maps

Genetic mapping experiments are typically accomplished by carrying out a testcross

Example of a two-point mapping cross Cross of two linked genes affecting bristle length

and body color in fruit flies e = ebony body color + = gray body color

s = short bristles + = normal bristles

One parent double recessive (homozygous recessive at both loci) – s/s ; e/e

Other parent is heterozygous at both loci (+/s ; +/e)

Linkage Analysis and Mapping

5-47Figure 5.9

Chromosomes are the product of a crossover during

meiosis in the heterozygous parent

Recombinant offspring are fewer

in number than nonrecombinant

offspring

The phenotype data are used to estimate the distance between the two loci

Map distance =

Number of recombinant offspring

Total number of offspringX 100

Therefore, the s and e genes are 12.3 map units apart from each other along the same chromosome

76 + 75

542 + 537 + 76 + 75X 100=

= 12.3 map units

Linkage Analysis and Mapping

Data from trihybrid crosses can also yield information about map distance and gene order

The following experiment outlines a common strategy for using trihybrid crosses to map genes In this example, we will consider fruit flies that differ in

body color, eye color and wing shape


Trihybrid Crosses

b = black body color b+ = grey body color

pr = purple eye color pr+ = red eye color

vg = vestigial wings vg+ = normal wings

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-60

Step 1: Cross two true-breeding strains that differ at three loci.

Female is mutant for all three traits

Male is homozygous wildtype for all three

traits

The goal in this step is to obtain F1 individuals that are heterozygous for all three genes


Step 2: Perform a testcross by mating F1 female heterozygotes to homozygous recessive, male flies

During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles


Step 3: Collect data for the F2 generation

Phenotype Number of Observed Offspring (males and females)

Gray body, red eyes, normal wings 411

Gray body, red eyes, vestigial wings 61

Gray body, purple eyes, normal wings 2

Gray body, purple eyes, vestigial wings 30

Black body, red eyes, normal wings 28

Black body, red eyes, vestigial wings 1

Black body, purple eyes, normal wings 60

Black body, purple eyes, vestigial wings 412


Analysis of the F2 generation flies will allow us to map the three genes The three genes exist as two alleles each Therefore, there are 23 = 8 possible combinations of

offspring If the genes assorted independently, all eight combinations

would occur in equal proportions It is obvious that they are far from equal

In the offspring of crosses involving linked genes, Parental phenotypes occur most frequently Double crossover phenotypes occur least frequently Single crossover phenotypes occur with “intermediate”

frequency


The combination of traits in the double crossover tells us which gene is in the middle A double crossover separates the gene in the middle from

the other two genes at either end

In the double crossover categories, the recessive purple eye color is separated from the other two recessive alleles Thus, the gene for eye color lies between the genes for

body color and wing shape


Step 4: Calculate the map distance between pairs of genes To do this, one strategy is to group the data

according to pairs of phenotypes resulting from non-crossovers, single crossovers, & double crossovers


Phenotype Number of Observed Offspring

Gray body, purple eyes, vestigial wings

30

Black body, red eyes, normal wings

28

Gray body, red eyes, vestigial wings

61

Black body, purple eyes, normal wings

60

Gray body, purple eyes, normal wings

2

Black body, red eyes, vestigial wings

1

30 + 28

1,005= 0.058

Single crossover between b and pr

61 + 60

1,005= 0.120

Single crossover between pr and vg

1 + 2

1,005= 0.003

Double crossover, between b and pr, and between pr and vg


To determine the map distance between the genes, we need to consider both single and double crossovers

To calculate the distance between b and pr Map distance = (0.058 + 0.003) X 100 = 6.1 mu

To calculate the distance between pr and vg Map distance = (0.120 + 0.003) X 100 = 12.3 mu

To calculate the distance between b and vg The double crossover frequency needs to be multiplied by two

Because both crossovers are occurring between b and vg Map distance = (0.058 + 0.120 + 2[0.003]) X 100

= 18.4 mu


Step 5: Construct the map Based on the map unit calculation the body color and

wing shape genes are farthest apart The eye color gene is in the middle

The data is also consistent with the map being drawn as vg – pr – b (from left to right)

In detailed genetic maps, the locations of genes are mapped relative to the centromere


Alternatively, the distance between b and vg can be obtained by simply adding the map distances between b and pr, and between pr and vg Map distance = 6.1 + 12.3 = 18.4 mu

The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover

Interference

P (double crossover) = P (single crossover between b

and pr)

P (single crossover between

pr and vg)

X

= 0.061 X 0.123 = 0.0075

Based on a total of 1,005 offspring the expected number of double crossover offspring is

= 1,005 X 0.0075 = 7.5


Therefore, we would expect seven or eight offspring to be produced as a result of a double crossover

However, the observed number was only three! Two with gray bodies, purple eyes, and normal sings One with black body, red eyes, and vestigial wings

This lower-than-expected value is due to a common genetic phenomenon, termed positive interference The first crossover decreases the probability that a

second crossover will occur nearby

Interference

Interference (I) is expressed as I = 1 – C

where C is the coefficient of coincidence

C = Observed number of double crossovers

Expected number of double crossovers

3

7.5 C =

I = 1 – C = 1 – 0.4

= 0.6 or 60% This means that 60% of the expected number of

crossovers did not occur

= 0.40


Since I is positive, this interference is positive interference

Rarely, the outcome of a testcross yields a negative value for interference This suggests that a first crossover enhances the rate of

a second crossover

The molecular mechanisms that cause interference are not completely understood However, most organisms regulate the number of

crossovers so that very few occur per chromosome

Somatic Cell Hybridization Fusion of human cell and rodent cell Eventually, most human sequences lost

entire chromosomes, portions (arms) of chromosomes Lines carrying sub-sets of human chromosomes are

compared to determine which express the given gene product

have to have a detectible DNA sequence or protein Process of elimination determines on which

chromosome (region of chromosome) a gene is located Hybrid Panel

Other Mapping Techniques

Cytogenetic Mapping Have a mutant phenotype Perform karyotype Observe altered chromosomes

aneuploidies translocations deletions

Aberrant chromosome probablylocation of gene

normal

translocation t(14;17)

Patau Syndrome - Trisomy 13 Karyotype: 47, 13+

Partial Chromosomal Deletions

cri-du-chat - 46, 5p- Fragile X mental retardation – 46,

Xp-

Monosomy loss of an entire chromosome lethal in all cases for autosomes XO – Turner’s syndrome is only viable

monosomy

Mapping Strategies

Deletion Mapping Cross recessive heterozygote to

deletion heterozygote lines appearance of recessive phenotype

localizes recessive gene within deletion interval

Complementation

Determines if two phenotypes are caused by same mutant gene

Complementation Analysis


Complementation Analysis of Eye Color Mutations

Complementation groups white, cherry, coral, apricot, buff

W1, wch , wc , wa , wb Alleles of white gene garnet ruby vermillion carnation


Mapping with Unknown Phase of LinkageWhen you have heterozygotes, but do not know what the

phenotypes of the parents were.

12 possible diploid arrangements for 3 linked locibm pr

bm pr

+ +

bm

+

bm

+

pr

+

brown midrib, virescent seedling, purple aleurone

Use Results of Cross to Determine Phase of Linkage

NCOs = parental chromosomes = phase

DCOs = fewestDo not directly detect middle locus

NCO Gives 3 Possibilities for Phase

12 possible diploid arrangements for 3 linked loci

bm pr

bm pr

+ +

bm

+

bm

+

pr

+

DCO Gives Loci Order

Documents

MATERI 7 DAN 8 SEX LINKAGE DAN CROSSING OVER. Gamet yang terbentuk dari individu dihibrid