sex chromosomes
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
Sex Chromosomes
... ‘X’ and ‘Y’ chromosomes that determine the sex of an individual in many organisms,
Females: XX
Males: XY
Differential
Region
Differential Region
Paring RegionParing Region
XY: male XX: female
a aA
hemizygous: condition where gene is present in only one dose (one allele).
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 Female
Pink Male
xP
Gametesor
Blue is dominant.
Gametesor
F1
Blue Female Blue Male
F1
Blue Female Blue Male
x
Gametesoror
Gametesoror
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
Pink Female
Blue Male
xP
Gametesor
F1
Blue Female Pink Male
Gametesor
F2
Pink Female Pink Male Blue Female Blue Male
Gametesoror
F2
Pink Female Pink Male Blue Female Blue Male
1 1 1 1
1 : 1 Female to Male1 : 1 Pink to Blue
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
• autosomal chromosome: non-sex linked chromosome,
• autosomal gene: a gene on an autosomal chromosome,
• autosomes segregate identically in reciprocal crosses.
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
• 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
• Phenotypes appear in every generation,
• Affected males and females pass the phenotype to equal proportions of their sons and daughters.
Recessive? ---> Yes!
Pedigree for Very Rare Trait? = kid with trait
Autosomal?X-Linked? ---> Yes!
1/2
1/2
1/2 x 1/2 x ? 1/2 = 1/8
x 1/2 = 1/16
(p)boy
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
• Genes linked on the same chromosome may segregate together.
A
b2n = 4
Independent Assortment
a
A A
B
B b a B a b
MeiosisNo Cross Over
Parent Cell
Daughter Cells Have Parental Chromosomes
A a
B b
A
B
a
b
a
b
A
B
2n = 1
MeiosisWith Cross Over
Parent Cell
Daughter Cells Have Recombinant Chromosomes
A a
B b
A
B
A
b
a
b
a
B
2n = 1
Dihybrid Cross
yellow/round green/wrinkled
GGWW x ggww
GW gw
GgWw
phenotype
genotype
gametes
genotype
P
F1
Gamate Formation in F1 Dihybrids P: GGWW x ggww, Independent Assortment
G g W w
GW Gw gW gw
alleles
gametes
probability.25 .25 .25 .25
F1 Genotype: GgWw
How do you test for assortment of alleles?
GW Gw gW gw
.25 .25 .25 .25
F1: GgWw
Test Cross: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.
Test CrossGgWw x ggww
GW (.25)
Gw (.25)
gW (.25)
gw (.25)
G gww (.25)
GgWw (.25)
ggWw (.25)
ggww (.25)
gw (1)
gw (1)
gw (1)
gw (1)
x
x
x
x
Test CrossGgWw x ggww
GW (.25)
Gw (.25)
gW (.25)
gw (.25)
gw (1)
gw (1)
gw (1)
gw (1)
Ggww (.25)
GgWw (.25)
ggWw (.25)
ggww (.25)
P
P
F1 parental types GgWw and gwgw
R
R
recombinant types Ggww and ggWw
x
x
x
x
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.
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
• 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.
A B
In this example, there is a 2/10 chance of recombination.
a b
A C
In this example, there is a 4/10 chance of recombination.
a c
homologs
Linkage Ratio
P GGWW x ggww
Testcross F1: GgWw x ggww
# recombinant
# total progeny
GW Gw gW gw
? ? ? ?
x 100 = Linkage Ratio
Units: % = mu (map units) - or - % = cm (centimorgan)
determine
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.
Study Figs 4.2, 4.3, and 4.5
Simple Mapping
• white eyes x miniature = 36.9 mu,
• white eyes x yellow body = 0.5 mu,
• miniature x yellow body = 38 mu,
my
38 mu
36.9 mu
w
0.5 mu
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
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.
target
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
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
• 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 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.
target
What recombination frequency do you expect between the target and HY2?
What recombination frequency do you expect between the target and TT2?
Classical mapping in humans requires pedigrees…
Three Point Testcross
Triple Heterozygous
(AaBbCc )
x
Triple Homozygous Recessive
(aabbcc)
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.
Representing linked genes...
W G D w g d
x
w g dw g d
w g d
P
Testcross
= WwGgDd
= wwggdd
Representing linked genes...
+ + + w g d
x
w g dw g d
w g d
P
Testcross
= WwGgDd
= wwggdd
Phenotypic Classes
W-
ww
G-
gg
G-
gg
D-
dd
D-
dd
D-
dd
D-
dd
W-G-D-
W-G-dd
W-gg-D
W-gg-dd
wwG-D-
wwG-dd
wwggD-
wwggdd
W-G-D-
W-G-dd
W-gg-D
W-gg-dd
wwG-D-
wwG-dd
wwggD-
#
179
52
46
4
22
22
2
wwggdd 173Parentals
Recombinants,double crossover
Recombinants 1 crossover, Region I
Recombinants 1 crossover, Region II
W G D
w g d
III
Arbitrarily name regions between genes…
W-G-D-
W-G-dd
W-gg-D
W-gg-dd
wwG-D-
wwG-dd
wwggD-
#
179
52
46
4
22
22
2
wwggdd 173Parentals
Recombinants,double crossover
Recombinants 1 crossover, Region I
Recombinants 1 crossover, Region II
W G D
w g d
I
Total = 500
Region I:
46 + 52 + 2 + 4
500x 100
= 20.8 mu
W-G-D-
W-G-dd
W-gg-D
W-gg-dd
wwG-D-
wwG-dd
wwggD-
#
179
52
46
4
22
22
2
wwggdd 173Parentals
Recombinants,double crossover
Recombinants 1 crossover, Region I
Recombinants 1 crossover, Region II
W G D
w g d
II
Total = 500
Region II:
22 + 22 + 2 + 4
500x 100
= 10.0 mu
20.8 mu
W G D
w g d
W-gg-D
wwG-dd 4
2Recombinants,double crossover
Total = 500
10.0 mu 20.8 mu
0.1 x 0.208 = 0.0208
6/500 = 0.012
NO GOOD!
Coefficient of Coincidence = ObservedExpected
Interference = 1 - Coefficient of Coincidence
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
• 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...
–
#
2001
52
46
589
990
887
600
1786
Which one is the “odd” one?
A C B
a c b
II I
A-B-C-
A-B-cc
A-bb-C-
A-bb cc
aaB-C-
aaB-cc
aabbC-
aabbcc
A-B-C-
A-B-cc
A-bb-C-
A-bb cc
aaB-C-
aaB-cc
aabbC-
#
2001
52
46
589
990
887
600
aabbcc 1786
Region I
A C B
a c b
I
990 + 887 + 52 + 46
6951x 100
= 28.4 mu
A-B-C-
A-B-cc
A-bb-C-
A-bb cc
aaB-C-
aaB-cc
aabbC-
#
2001
52
46
589
990
887
600
aabbcc 1786
Region II
A C B
a c b
28.4 mu
600 + 589 + 52 + 46
6951x 100
= 18.5 mu
II18.5 mu
Fig. 4.18. DNA molecule containing three
EcoRI cleavage sites
Fig. 4.19
Fig. 4.20a
Molecular Mapping Markers
Fig. 4.20b
p. 143. Fluorescent dyes are often used to label DNA so that the
positions of DAN fragments in a gel can be identified.
Assignments
• 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.