Recombination & Genetic Analysis

Recombination & Genetic Analysis

-The maximum recombination frequency

-Quiz section 5 revisited

-Genetic vs. Physical maps

Test your understanding

Dihybrid female

X

Predict the number of progeny for each type of offspring that result from the following cross. Assume 100 total offspring.

b+ sp+

b sp

Testcross male

tan no speck

black speck

tan speck

black no speck

21

21

29

29

58.2 cM

P

R

25

25

25

25

Recombinants never exceed 50%

vg+

vg

Are b and sp linked?

• b and sp appear to be unlinked even though they are on the same chromosome!

• These genes are so far apart, that they assort independently from one another.

How do we know that b and sp are on the same chromosome? b is linked to vg and vg is linked to sp.

No!

b sp+

b+ sp+

b sp

b+ sp

All four gamete

types are

equally frequen

t

sp +b+

sp +b+

spb

spb

1

23

4

In-class experiment…Mark two crossovers anywhere between the homologues:

Write down the parental types with respect to A/a and B/b:

Write down the parental types with respect to A/a and D/d:

After you are given the locations of loci A, B, and D…

Why do we observe IA?

In-class experiment…Mark two crossovers anywhere between the homologues:

A B and a b

A D and a d

Write down the parental types with respect to A/a and B/b:

Write down the parental types with respect to A/a and D/d:

After you are given the locations of loci A, B, and D…

A

Aa

a

B

Bb

b

D

Dd

d

1

2

3

4


In-class experiment (cont’d)Looking first at just loci A/a and B/b…

What are the genotypes of the products from your meiosis?

1.

2.

3.

4.

Then look at just loci A/a and D/d…What are the genotypes of the products from your meiosis?

1.

2.

3.

4.

Are these gametes all parental? All recombinant? 2 of each?

Are these gametes all parental? All recombinant? 2 of each?

Class aggregate data:

2 P, 2 R

AB, ab, Ab, aB

4 RAb, aB

4 PAB, ab

Number R gametesNumber P gametesNumber?P or R?Genotype

A-B

# Recombinants% Rec?

2

43

X

81

0

86

324

8

86

0

Total # Gametes

X

X=

What must have happened to create these

gametes?

In-class experiment (cont’d)

Class aggregate data:

2 P, 2 R

AD, ad, Ad, aD

4 RAd, aD

4 PAD, ad

Number R gametesNumber P gametesNumber?P or R?Genotype

A-D

# Recombinants% Rec?

19

61

X

40

0

122

160

76

122

0

Total # Gametes

76+122

76+122 +160 +122=

In-class experiment (cont’d)

Everyone in the class drew crossovers somewhere between A/a and D/d, yet the overall % recombinants for the class was only ~50%. If we look at a large enough sample, even genes that are very far apart on the same chromosome cannot show more than 50% recombinant products.


A

Aa

a

B

Bb

b

D

Dd

d

Need to look closer at meiosis itself to see why.

What is the maximum recombination frequency in any interval?

Parental

Parental

RecombinantRecombinant

Reminder… one crossover gives 2 parental, 2 recombinant gametes

Consider 100 cells undergoing meiosis…

if one cell has a crossover

2 recombinant out of 400

0.5%

if 10 cells have crossovers


5.0%

if all cells have crossovers


50%

The range of possibilities: tightly linked independent assort.

Maximum recombination frequency = 50% for single recombination events

The effect of multiple crossovers:

# xovers resulting gametes

4 parental

2(2 strands)

2 parental2 non-par.

2(3 strands)

4 non-par.

2(4 strands)

6 parental6 non-par.

= 50% recombinant and 50% parental.Also true for triple Xover, quadruple Xover, etc.

What is the maximum recombination frequency in any interval?

Human X-chromosome map…

how could we get 180 cM?

Now consider independent assortment

…the “ultimate” in non-linkageRefer to one of Mendel’s F1 x F1 dihybrid cross (round yellow X round yellow):

What were the parental types for the F1?

What were the parental and recombinant gametes made by the F1 plants?

What was the % recombinant gametes?

RY and ry

1/4 RY1/4 ry1/4 Ry1/4 rY

1/4 + 1/4 = 50%!

So, even for independently assorting genes, the % recombinant products is only 50%

R Y

r y

R Y

r yX

Thus, the maximum recombinant frequency = 50%Loci can appear to be unlinked because:

• They are on separate chromosomes

• They are so far apart on the same chromosome that recombination always occurs

Conclusion

1) An odd number of crossovers gives, on average, an equal number of parental and recombinant types. 2) An even number of crossovers gives, on average, an equal number of parental and recombinant types.3) Alleles on two different chromosomes line up on the metaphase plate independently, giving on average equal numbers of parental and recombinant types.

For widely separated genes

Practice question

In a certain plant species…

flower fragrance (F) is dominant over unscented (f)blue flower color (B) is dominant over white (b)rounded leaves (R) is dominant over pointy (r); andthorny stems (T) is dominant over smooth stems (t).

From the following crosses, can you determine whether the fragrance gene is linked to any of the other genes; if so, at what map distance?

Bb Ff x bb ff

270 blue, fragrant281 blue, non-fragrant268 white, fragrant275 white, non-fragrant

Rr ff x rr Ff

219 rounded, fragrant222 rounded, non-fragrant209 pointy, fragrant216 pointy, non-fragrant

Tt Ff x tt ff

333 thorny, fragrant36 thorny, non-fragrant39 smooth, fragrant342 smooth, non-fragrant

F not linked to B

Can’t tell!

F and T linked at 10 cM

R f

r f

r F

r f

X

The parental and recombinant types are the same! Need to be heterozygous at both loci

QS7 revisited

What were the main points of QS5?

etc.

…were designed to help set up specific predictions

-To give you an opportunity to see actual data from a meiosis and to draw conclusions from the data based on your knowledge of this process.-To show what can be learned from looking at all four products of a single meiosis.

The diagrams used in quiz section…

Setting up predictions for meiosis outcomes…

2 genes are linked

they are independently assorting but each close to a centromere

they are unlinked and at least one is distant from a centromere

If… then we expect…a parental ditype (PD)

PD > T >> NPD

either PD or non-parental ditype (NPD), with a 50:50 chance of eachPD = NPD > T

But if we can’t see all of the products from a single meiosis we expect…mostly parental types

an equal proportion of parental and recombinant types

an equal proportion of parental and recombinant types

Can’t distinguishCan distinguish!

mostly tetratypes (T), but also some PD, and NPD

What did the data from quiz section tell you?

Mat haploid parent = ade his

Mata haploid parent = ADE HIS

Spore phenotype

# of spores

ADE HIS 9

ADE his 11

ade HIS 11

ade his 9

Total = 40

Mat haploid parent = his LEU

Mata haploid parent = HIS leu

Spore phenotype

# of spores

HIS LEU 3

HIS leu 17

his LEU 17

his leu 3

Total = 40

Mat haploid parent = LEU TS

Mata haploid parent = leu ts

Spore phenotype

# of spores

LEU TS 11

Leu ts 9

leu TS 9

leu ts 11

Total = 40

Conclusion? Conclusion? Conclusion?

Probably not linked

Probably linked

Probably not linked

Looking at the 10 tetrads in terms of LEU & TS

How many PD?

How many NPD?

How many T?

4

5

1

So what do you conclude about the LEU and TS genes?they are independently assorting but each close to a centromere!

Our completed map

Diploid genotype:

MATa ADE HIS leu ts URA1 ura2 MAT ade his LEU TS ura1 URA2

leu HIS

LEU his ts

TS

ADE

ade

ADE2

HIS4

LEU2

CDC7 (TS)

How well did we do?

Let’s look in SacchDB…

The actual gene names…

ADE2 CDC7 (TS)HIS4 LEU2

Not bad!

What about URA?

Know the parental types

Look at spore phenotypes

Parental types?U1 u2

&u1 U2

What spore genotypes would you expect in a PD tetrad?

U1 u2

U1 u2

u1 U2

u1 U2

Phenotype on -ura plate?

no growth

no growth

no growth

no growth

So, given these parental types… 0/4 spores growing is diagnostic of PD

What spore genotypes would you expect in a NPD tetrad?

Growth phenotype on -ura?Genotype?

What about URA?


&u1 U2

U1 U2

U1 U2

u1 u2

u1 u2

GROWTH

GROWTH

no growth

no growth

What spore genotypes would you expect in a T tetrad?

Growth phenotype on -ura?Genotype?

What about URA?


&u1 U2

U1 u2

U1 U2

u1 U2

u1 u2

no growth

GROWTH

no growth

no growth

Looking at the 10 tetrads…

How many PD?

How many NPD?

How many T?

So what do you conclude about the ura genes?

Brown seed pods (B) in a plant species is dominant to green (b), and elongated pods (E) is dominant over squished (e).

(a) A fully heterozygous plant has the dominant alleles linked in trans (i.e., dominant alleles not on the same homologue) at a map distance of 20 cM. What will be the genotypes of gametes produced by this plant, and in what frequencies (or percentages)?

(b) If this plant is self-pollinated, what progeny phenotypes will you expect to see, and in what frequencies? Use a Punnett square to illustrate your answer.

Heterozygote genotype =

B e

b E

Practice question

3-point testcrosses

The problem with using two markers (like a and d below)…double crossovers can go undetected

underestimation of recombinant frequency

Solution: include a third marker between the other two…

more DCOs revealed

Plus… gene order revealed (more later)

3-point testcross—predicting progeny from a known map

Predict the progeny phenotypes and numbers from this cross:

Parent 1:

Parent 2:

+ + a

b c +

b c a

b c a

+ = wild type, dominant

Map:b c a3 cM 7 cM

Step 1.Determine the number of DCO products

Probability of recombinant product in (b-c) =

Probability of recombinant product in (c-a) =

Probability of recombinant product in both =

3% = 0.03

Count 10,000 progeny

7% = 0.07

0.03 x 0.07 = 0.0021

Predicting progeny from a known map (cont’d)Heterozygous parent:

only one chromatid of each homologue shown on next slide

DCO: both together =

10000 x 0.0021 = 21

SCO in b-c interval:Both together =(10000 x 0.03)-21 = 279

SCO in c-a interval:Both together =(10000 x 0.07)-21 = 679

NCO (non-crossover):Both together =10000-(SCO + DCO) = 9021

Predicting progeny from a known map (cont’d)

SCO

SCO

SCO

DCO

DCO

NCO

NCO

3-point testcross—constructing a linkage mapConstruct a linkage map (gene order and map distance) for the following genes in Drosophila:

Genespr/+ (purple or red eyes)v/+ (vestigial or long wings)b/+ (black or tan body)

ParentsFemale: pr/+ v/+ b/+Male: pr/pr v/v b/b

Progeny phenotypes+ + + 564b 32v 4125pr 266v b 272pr b 4137pr v 30pr v b 574

Step 1. Expand the shorthand

pr+ v+

b+ pr+

b+ v+

pr+

v+

b+

Step 2. Identify the NCO and DCO classes

Step 3. Which gene is in the middle?

Total = 10000

ASK: Which order allows us to go from the NCO genotype to the DCO genotype.

Constructing a linkage map… Step 3 (cont’d)We know: (b+ v pr+)

(b v+ pr)DCO (b v+ pr+)

(b+ v pr)

The process:

• Try out the parental genotypes in the 3 possible orders

• Do a “virtual double crossover” to see which one would give the correct DCO genotype.

order unknown

b+ v pr+

b v+ prX X

b+ v+ pr+

b v princorre

ct

b+ pr+ v

b pr v+X X

b+ pr v

b pr+ v+corre

ct!

Constructing a linkage map (cont’d)

Step 4.Calculate % recombinant products

b+ pr v+

b pr+ v

NCO:4125+4137 = 8262

SCOb-pr:266+272 = 538

b+ pr+ v+

b pr v

SCOpr-v:564+574 = 1138

b+ pr v

b pr v+

DCO:30+32 = 62

% recombinants in b-pr interval=(538+62)/10000

=600/10000

=6%% recombinants in pr-v interval=(1138+62)/10000

=1200/10000

=12%

Step 5.Draw the map

b pr v6 cM 12 cM

bprv6 cM12

cM

or

Constructing a linkage map (cont’d)

Interference and coefficient of coincidence (COC)Interference: Lower-than-expected frequency of DCO products- Chiasma at one one location blocks

other chiasmata from forming nearby

COC =

observed DCOexpected DCO

Interference = 1 - COC

In our example…expected DCO = 0.06 x 0.12 x 10000 = 72Observed DCO = 62COC = 62/72 = 0.86Interference = 1 - 0.86 = 0.14

alleles!alleles!

Genetic vs. physical maps

Genetic maps…

based on recombinant frequencies between markers

variation at location #1

variation at location #2pr+ vg

vg+pr

Alleles are detected as associated phenotypes

New combination of phenotypes new combination of alleles recombination

recombination:

how frequent?

Physical maps…

based on DNA sequence or landmarks in sequence

The number of chromosomal bands separating the known locations of genes.

For example:A B

The number of bp of DNA.

The pattern of restriction sites in a DNA sequence.

site1 site2 site4 site5site3

Genetic vs. physical maps (cont’d)

Which chromosome?

Relation to centromere and telomere?pr+ vg+

vgpr

How do we know where to place these genes relative to the centromere and telomeres?

Number of bp?

Questions for Thought

How do we know which physical entity (chromosome) our linkage group describes?

What is the relationship between crossover frequency and gene order/distance (in bp of DNA) along the chromosomes?

“Linkage group” = chromosome

white is X-linked

Linking the Physical and Genetic Maps

Color is on chromosome IX

knob

extra DNA

C

Wx

Novel Strain

Cri du chat is on chromosome V

From lecture 6

From lecture 8

Metaphase chromosom

es

* * *Fluorescently labeled single-stranded DNA

-Fluorescence in situ hybridization (FISH)

Partially denature DNA

***

Linking the Physical and Genetic Maps

Double-stranded DNA segment from a particular location in the human genome

Fragile XR-G colorblindness 6 cM

genetic map (recombination)units = cM

Haemophilia10 cM

What does the genetic map position tell us?

physical map (DNA)~6Xbp ~10Xbp

units = bp

Fragile X HaemophiliaR-G colorblindness

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Physicalmap (FISH)

units = chrom. bands

-Order of genes is conserved in genetic and physical maps.-Distance separating markers in genetic and physical maps is ~proportional (but X varies in different organisms).

telomere centromere

Recombination & Genetic Analysis

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