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Sex Chromosomes. ... ‘X’ and ‘Y’ chromosomes that determine the sex of an individual in many organisms, Females: XX Males: XY. a. a. A. hemizygous: condition where gene is present in only one dose (one allele). Differential Region. Differential Region. Paring Region. Paring Region.

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Sex chromosomes
Sex Chromosomes

...‘X’ and ‘Y’ chromosomes that determine the sex of an individual in many organisms,

Females: XX

Males: XY


a

a

A

hemizygous: condition where gene is present in only one dose (one allele).

Differential

Region

Differential Region

Paring Region

Paring Region

XX: female

XY: male


X linkage
X Linkage

…the pattern of inheritance resulting from genes located on the X chromosome.

X-Linked Genes…

…refers specifically to genes on the X-chromosome, with no homologs on the Y chromosome.


Blue is dominant.

P

x

Blue Female

Pink Male

Gametes

or


Gametes

or

F1

Blue Female

Blue Male


F1

x

Blue Female

Blue Male

Gametes

or

or


Gametes

or

or

F2

Blue Female

Blue Male

Blue Female

Pink Male


F2

Blue Female

Blue Male

Blue Female

Pink Male

3 : 1 Blue to Pink

1 : 1 Female to Male


P

x

Pink Female

Blue Male

Gametes

or


Gametes

or

F1

Blue Female

Pink Male


Gametes

or

or

F2

Pink Female

Pink Male

Blue Female

Blue Male


F2

Pink Female

Pink Male

Blue Female

Blue Male

1

1

1

1

1 : 1 Female to Male

1 : 1 Pink to Blue


Sex linkage to ponder
Sex Linkage to Ponder

  • Female is homozygous recessive X-linked gene,

    • what percentage of male offspring will express?

    • what percentage of female offspring will express if,

      • mate is hemizygous for the recessive allele?

      • mate is hemizygous for the dominant allele?

  • Repeat at home with female heterozygous X-linked gene!


Sex linked vs autosomal
Sex-Linked vs. Autosomal

  • autosomal chromosome: non-sex linked chromosome,

  • autosomal gene: a gene on an autosomal chromosome,

  • autosomes segregate identically in reciprocal crosses.


X linked recessive traits characteristics
X-Linked Recessive TraitsCharacteristics

  • Many more males than females show the phenotype,

    • female must have both parents carrying the allele,

    • male only needs a mother with the allele,

  • Very few (or none) of the offspring of affected males show the disorder,

    • all of his daughters are carriers,

      • roughly half of the sons born to these daughters are carriers.


X linked dominant
X-Linked Dominant

  • Affected males married to unaffected females pass the phenotype to their daughters, but not to their sons,

  • Heterozygous females married to unaffected males pass the phenotype to half their sons and daughters,

  • Homozygous dominant females pass the phenotype on to all their sons and daughters.


Autosomal dominant
Autosomal Dominant

  • Phenotypes appear in every generation,

  • Affected males and females pass the phenotype to equal proportions of their sons and daughters.


Pedigree for very rare trait kid with trait

(p)boy

Pedigree for Very Rare Trait? = kid with trait

1/2

1/2

Recessive?

---> Yes!

1/2 x 1/2 x ?

1/2 = 1/8

x 1/2 = 1/16

Autosomal?

X-Linked?

---> Yes!


X linked dominant examples omim
X-Linked Dominantexamples (OMIM)

  • HYPOPHOSPHATEMIA: “Vitamin-D resistant Rickett’s”,

  • LISSENCEPHALY: “smooth brain”,

  • FRAGILE SITE MENTAL RETARDATION: mild retardation,

  • RETT Syndrome: neurological disorder,

  • More on OMIM…


Linkage
Linkage

  • Genes linked on the same chromosome may segregate together.


Independent Assortment

A

a

B

b

2n = 4

A

A

a

B

b

B

a

b


Meiosis no cross over

2n = 1

MeiosisNo Cross Over

A

a

Parent Cell

B

b

A

A

a

a

B

B

b

b

Daughter Cells Have Parental Chromosomes


Meiosis with cross over

2n = 1

MeiosisWith Cross Over

A

a

Parent Cell

B

b

A

A

a

a

B

b

B

b

Daughter Cells Have Recombinant Chromosomes


Dihybrid cross
Dihybrid Cross

phenotype

genotype

gametes

genotype

yellow/round green/wrinkled

GGWW x ggww

GW gw

GgWw

P

F1


Gamate formation in f1 dihybrids p ggww x ggww independent assortment

.25

.25

.25

.25

Gamate Formation in F1 Dihybrids P: GGWW x ggww, Independent Assortment

F1 Genotype: GgWw

G g W w

alleles

gametes

GW

Gw

gW

gw

probability


How do you test for assortment of alleles

GW

Gw

gW

gw

.25

.25

.25

.25

How do you test for assortment of alleles?

F1: GgWw

Test Cross: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.


Test cross ggww x ggww

GW (.25)

gw (1)

GgWw (.25)

x

Gw (.25)

x

gw (1)

G gww (.25)

gW (.25)

x

gw (1)

ggWw (.25)

gw (.25)

gw (1)

ggww (.25)

x

Test CrossGgWw x ggww


Test cross ggww x ggww1

P

R

R

P

F1 parental types GgWw and gwgw

recombinant types Ggww and ggWw

Test CrossGgWw x ggww

GW (.25)

x

gw (1)

GgWw (.25)

Gw (.25)

x

gw (1)

Ggww (.25)

gW (.25)

gw (1)

ggWw (.25)

x

gw (.25)

gw (1)

ggww (.25)

x


Recombination frequency
Recombination Frequency

…or Linkage Ratio: the percentage of recombinant types,

  • if 50%, then the genes are not linked,

  • if less than 50%, then linkage is observed.


Linkage1
Linkage

  • Genes closely located on the same chromosome do not recombine,

    • unless crossing over occurs,

  • The recombination frequency gives an estimate of the distance between the genes.


Recombination frequencies
Recombination Frequencies

  • Genes that are adjacent have a recombination frequency near 0%,

  • Genes that are very far apart on a chromosome have a linkage ratio of 50%,

  • The relative distance between linked genes influences the amount of recombination observed.


homologs

A

B

In this example, there is a 2/10 chance of recombination.

A

C

a

c

In this example, there is a 4/10 chance of recombination.

a

b


Linkage ratio p ggww x ggww testcross f1 ggww x ggww

GW

Gw

gW

gw

determine

?

?

?

?

Linkage RatioP GGWW x ggwwTestcross F1: GgWw x ggww

# recombinant

# total progeny

x 100 = Linkage Ratio

Units: % = mu (map units) - or - % = cm (centimorgan)


Fly crosses simple 3 point mapping white eyes minature yellow body

Study Figs 4.2, 4.3, and 4.5

Fly Crosses (simple 3-point mapping)(white eyes, minature, yellow body)

  • In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu,

  • In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu,

  • When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu.


Simple mapping

0.5 mu

36.9 mu

y

w

m

38 mu

Simple Mapping

  • white eyes x miniature = 36.9 mu,

  • white eyes x yellow body = 0.5 mu,

  • miniature x yellow body = 38 mu,


Do we have to learn more mapping techniques
Do We have to Learn More Mapping Techniques?

  • Yes,

    • three point mapping,

  • Why,

    • Certainty of Gene Order,

    • Double crossovers,

    • To answer Cyril Napp’s questions,

    • and, for example: over 4000 known human diseases have a genetic component,

      • knowing the protein produced at specific loci facilitates the treatment and testing.


cis

“coupling”


trans

“repulsion”


Classical mapping

target

Classical Mapping

Cross an organism with a trait of interest to homozygous mutants of known mapped genes.

  • Then, determine if segregation is random in the F2 generation,

    • if not, then your gene is linked (close) to the known mapped gene.

What recombination frequency do you expect between the target and HY2?

What recombination frequency do you expect between the target and TT2?


Gene order
Gene Order

  • It is often difficult to assign the order of genes based on two-point crosses due to uncertainty derived from sampling error.

    A x B = 37.8 mu,

    A x C = 0.5 mu,

    B x C = 37.6 mu,


Double crossovers
Double Crossovers

  • More than one crossover event can occur in a single tetrad between non-sister chromatids,

    • if recombination occurs between genes A and B 30% of the time (p = 0.3), then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or nearly one map unit.

  • If there is a double cross over, does recombination occur?

    • how does it affect our estimation of distance between genes?


Genetics:

…in the News


Classical mapping1

target

Classical Mapping

Cross an organism with a trait of interest to homozygous mutants of known mapped genes.

  • Then, determine if segregation is random in the F2 generation,

    • if not, then your gene is linked (close) to the known mapped gene.

What recombination frequency do you expect between the target and HY2?

What recombination frequency do you expect between the target and TT2?



Three point testcross
Three Point Testcross

Triple Heterozygous

(AaBbCc )

x

Triple Homozygous Recessive

(aabbcc)


Three point mapping requirements
Three Point Mapping Requirements

  • The genotype of the organism producing the gametes must be heterozygous at all three loci,

  • You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring,

  • You must look at enough offspring so that all crossover classes are represented.


W g d
w g d

Representing linked genes...

W G D

w g d

x

w g d

w g d

P

= WwGgDd

Testcross

= wwggdd


W g d1
w g d

Representing linked genes...

+ + +

w g d

x

w g d

w g d

P

= WwGgDd

Testcross

= wwggdd


Phenotypic classes

D-

D-

D-

D-

dd

dd

dd

dd

G-

G-

gg

gg

Phenotypic Classes

W-G-D-

W-G-dd

W-

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

ww

wwggD-

wwggdd


#

Arbitrarily name regions between genes…

W-G-D-

179

Parentals

wwggdd

173

W G D

w g d

W-G-dd

46

Recombinants

1 crossover,

Region I

II

I

wwggD-

52

22

Recombinants

1 crossover,

Region II

wwG-D-

W-gg-dd

22

W-gg-D

2

Recombinants,

double crossover

wwG-dd

4


#

I

W-G-D-

179

Parentals

wwggdd

173

W G D

w g d

W-G-dd

46

Recombinants

1 crossover,

Region I

wwggD-

52

Region I:

22

Recombinants

1 crossover,

Region II

wwG-D-

W-gg-dd

22

46 + 52 + 2 + 4

500

x 100

W-gg-D

2

Recombinants,

double crossover

wwG-dd

4

= 20.8 mu

Total = 500


#

II

20.8 mu

W-G-D-

179

Parentals

wwggdd

173

W G D

w g d

W-G-dd

46

Recombinants

1 crossover,

Region I

wwggD-

52

Region II:

22

Recombinants

1 crossover,

Region II

wwG-D-

W-gg-dd

22

22 + 22 + 2 + 4

500

x 100

W-gg-D

2

Recombinants,

double crossover

wwG-dd

4

= 10.0 mu

Total = 500


W-gg-D

2

Recombinants,

double crossover

wwG-dd

4

Total = 500

Observed

Expected

Coefficient of Coincidence =

10.0 mu

20.8 mu

0.1 x 0.208 = 0.0208

W G D

w g d

NO GOOD!

6/500 = 0.012

Interference = 1 - Coefficient of Coincidence


Interference
Interference

…the effect a crossing over event has on a second crossing over event in an adjacent region of the chromatid,

  • (positive) interference: decreases the probability of a second crossing over,

    • most common in eukaryotes,

  • negative interference: increases the probability of a second crossing over.


Gene order in three point crosses
Gene Order in Three Point Crosses

  • Find - either - double cross-over phenotype…based on the recombination frequencies,

  • Two parental alleles, and one cross over allele will be present,

  • The cross over allele fits in the middle...


II

I

A C B

a c b

#

A-B-C-

2001

aabbcc

1786

A-B-cc

46

Which one is the “odd” one?

aabbC-

52

aaB-cc

990

A-bb-C-

887

A-bb cc

600

aaB-C-

589


990

+ 887

+ 46

+ 52

x 100

6951

= 28.4 mu

#

Region I

A-B-C-

2001

aabbcc

1786

A-B-cc

46

aabbC-

52

aaB-cc

990

I

A-bb-C-

887

A C B

a c b

A-bb cc

600

aaB-C-

589


600

+ 589

+ 46

+ 52

x 100

6951

= 18.5 mu

18.5 mu

#

Region II

A-B-C-

2001

aabbcc

1786

A-B-cc

46

aabbC-

52

aaB-cc

990

II

28.4 mu

A-bb-C-

887

A C B

a c b

A-bb cc

600

aaB-C-

589


Fig 4 18 dna molecule containing three eco ri cleavage sites
Fig. 4.18. DNA molecule containing three EcoRI cleavage sites


Fig 4 19

Molecular Mapping Markers

Fig. 4.19

Fig. 4.20a



p. 143. Fluorescent dyes are often used to label DNA so that the positions of DAN fragments in a gel can be identified.


Assignments
Assignments the positions of DAN fragments in a gel can be identified.

  • Read from Chapter 3, 3.6 (pp. 100-106),

  • Master Problems…3.12, 3.15, 3.20,

  • Chapter 4, Problems 1, 2,

  • Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 - 4.20 a,b,c,d.

  • Exam Wednesday.

    • One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.


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