Genes,Chromosomes,and the
mechanism of Mendelian
Inheritance
Cells reproduce via a cell cycle
Meiosis is the key to sexual reproduction and Mendel’s rules
Sex determination,sex linkage and chromosome
nondisjunction supported the chromosome theory,
Cells reproduce by a cell cycle
Cells in the body
reproduce by
mitosis.
Mitosis produces
cells with exact
copies of the
parental cell
chromosomes and
genes.
Fig,3.1
Sexual reproduction,Meiosis
Germ cells undergo two special nuclear divisions with
only one chromosome replication.
These meiotic divisions reduce the number of
chromosomes in the gametes.
Chromosome behavior in meiosis is the first evidence
that genes reside on chromosomes.
Meiosis I
During
prophase I,
homologous
chromosomes
condense,
pair
(becoming
bivalents),and
recombine,
The bivalents
attach to
spindle fibers.
Meiosis II
The
chromosomes
split at the
centromere
3.11
Meiosis,The Bottom Line
A diploid cell (2N) gives
rise to four haploid cells
(N).
Recombination and
segregation of
chromosomes generates
diversity.
2 N
N
N
N
N
Mendel and Meiosis
Chromosome segregation parallels allele segregation,
F1 cell Gametes
A
A
a
a
Meiosis
A
A
a
a
SEX DETERMINATION
Sex determination:
In Drosophila sex is determined by the ratio of
autosomes to X-chromosomes,An X to one set
of autosomes = male,two Xs = female
In mammals (humans) only the presence of a Y
chromosome is necessary to determine
maleness,no Y = female
The Y chromosome is nearly completely inactive,
Genes on the X in a male are therefore
hemizygous.
Sex Linkage
As the Y chromosome is nearly inactive,genes
are not passed from the Y chromosome to
offspring,
In males,a recessive mutation is phenotypically
visible (it is hemizygous),but not in females.
Consequently,a sex-linked trait in a male has a
different phenotypic inheritance pattern than in a
female,In other words,reciprocal crosses give
different results.
This is demonstated by inheritance of the white
allele in Drosophila (next slide),
T.H,Morgan’s Observation,Inheritance of
white
X
X
1/2
1/2
P
F1
F2
The white mutation
is inherited
differently in
different sexes,
sex-linked
inheritance.
Morgan’s Experiment,Reciprocal Crosses
P
F1
F2
X
X
w
w
w
+
w
+
w
w
+
w
w
+
1/4
1/4
1/4
1/4
1/4
1/4
X
+ X
+
+
w
+
+
+
w
+
w
+
1/4
1/4
w
C,B,Bridges’ Observation,Unusual Inheritance
of white
X
1/2000 1/2000
P
F1
The white phenotype is
occasionally inherited
differently,
Hypothesis,X
chromosome
nondisjunction.
Nondisjunction,Chromosomes segregation abnormally
during meiosis giving rise to gametes with an incorrect
number of chromosomes.
Nondisjunction,
Failure of
segregation in
meiosis.
3.25
Bridges’ Hypothesis
P
F1
X chromosome
nondisjunction
creates XXY
females and XO
males.
X
w
w
w
+
+
w
Bridges’ Experiment
P? Crossing XXY females should
produce progeny
with unique
karyotypes.
X
+
w
w
w
+
w
w
w
w
+
w
w
+
ww
+
+
w w
Conclusions
Studies of nondisjunction provided the final proof of the
chromosome theory.
Sex linked genes segregate with the X chromosome
even in the rare cases when the chromosome
segregates abnormally.
Problem
Attached-X,When two X chromosomes fused at their ends
they are called attached-X,Attached-X cannot separate
during meiosis,This property is often used in the
genetic manipulation of Drosophila.
1) Are attached-X female flies conveniently maintained
and propagated as XX or XXY stocks?
2) If an attached-X white-eye female fly is crossed to an
red eye male,what is the eye color of the progenies?
Cross F1 F2
Dominant Recessive Ratio
tall X short tall 787 277 2.84:1
axial X terminal flowers axial 651 207 3.15:1
Coin flip Face side Back side Ratio
50 times total 30 20 3:2
Do these results fit expected ratio?
Tests of results,Chi Square (i)
The chi square test is used to test how well the
observed data fits the expectations.
2 =(observed-expected)2/expected
Example,Mendel’s test cross of the F1 from his
dihybrid cross gives these expected results:
Expected phenotypes Expected ratio
Round Yellow 1/4
Round green 1/4
wrinkled Yellow 1/4
wrinkled green 1/4
Chi Square Test
Hypothesis,the progeny actually are in a 1:1:1:1 ratio.
2 test:
genotype obs exp (0bs-exp)2/exp
Rr Yy 55 51.75 0.204
Rr yy 51 51.75 0.011
rr Yy 49 51.75 0.146
rr yy 52 51.75 0.001
Total 207 207 0.362
Degrees of freedom = 4 -1 = 3 (one less than the number
of progeny classes)
Chi Square Test
The probability of this?2 value,given the degrees of
freedom,is determined from a Chi square table
For this example,probability > 0.90
This much deviation from a correct hypothesis will occur
more than 9 of 10 times by random chance.
The hypothesis is accepted,We conclude these genes
really do assort independently.
Hypotheses that give P values > 0.05 are accepted.
If P is < 0.05,the hypothesis is rejected.
Problem
In a cross between ebony flies and wild type flies,
the F1 displayed wild type body color,F2 counting
derived the following numbers,
Normal body color male 152 female 168
Ebony body color male 29 female 27
1) Does male:female ratio fit 1:1?
2) Does normal:ebony fit 3:1 ratio? Can you give a
likely explanation for this deviation?
mechanism of Mendelian
Inheritance
Cells reproduce via a cell cycle
Meiosis is the key to sexual reproduction and Mendel’s rules
Sex determination,sex linkage and chromosome
nondisjunction supported the chromosome theory,
Cells reproduce by a cell cycle
Cells in the body
reproduce by
mitosis.
Mitosis produces
cells with exact
copies of the
parental cell
chromosomes and
genes.
Fig,3.1
Sexual reproduction,Meiosis
Germ cells undergo two special nuclear divisions with
only one chromosome replication.
These meiotic divisions reduce the number of
chromosomes in the gametes.
Chromosome behavior in meiosis is the first evidence
that genes reside on chromosomes.
Meiosis I
During
prophase I,
homologous
chromosomes
condense,
pair
(becoming
bivalents),and
recombine,
The bivalents
attach to
spindle fibers.
Meiosis II
The
chromosomes
split at the
centromere
3.11
Meiosis,The Bottom Line
A diploid cell (2N) gives
rise to four haploid cells
(N).
Recombination and
segregation of
chromosomes generates
diversity.
2 N
N
N
N
N
Mendel and Meiosis
Chromosome segregation parallels allele segregation,
F1 cell Gametes
A
A
a
a
Meiosis
A
A
a
a
SEX DETERMINATION
Sex determination:
In Drosophila sex is determined by the ratio of
autosomes to X-chromosomes,An X to one set
of autosomes = male,two Xs = female
In mammals (humans) only the presence of a Y
chromosome is necessary to determine
maleness,no Y = female
The Y chromosome is nearly completely inactive,
Genes on the X in a male are therefore
hemizygous.
Sex Linkage
As the Y chromosome is nearly inactive,genes
are not passed from the Y chromosome to
offspring,
In males,a recessive mutation is phenotypically
visible (it is hemizygous),but not in females.
Consequently,a sex-linked trait in a male has a
different phenotypic inheritance pattern than in a
female,In other words,reciprocal crosses give
different results.
This is demonstated by inheritance of the white
allele in Drosophila (next slide),
T.H,Morgan’s Observation,Inheritance of
white
X
X
1/2
1/2
P
F1
F2
The white mutation
is inherited
differently in
different sexes,
sex-linked
inheritance.
Morgan’s Experiment,Reciprocal Crosses
P
F1
F2
X
X
w
w
w
+
w
+
w
w
+
w
w
+
1/4
1/4
1/4
1/4
1/4
1/4
X
+ X
+
+
w
+
+
+
w
+
w
+
1/4
1/4
w
C,B,Bridges’ Observation,Unusual Inheritance
of white
X
1/2000 1/2000
P
F1
The white phenotype is
occasionally inherited
differently,
Hypothesis,X
chromosome
nondisjunction.
Nondisjunction,Chromosomes segregation abnormally
during meiosis giving rise to gametes with an incorrect
number of chromosomes.
Nondisjunction,
Failure of
segregation in
meiosis.
3.25
Bridges’ Hypothesis
P
F1
X chromosome
nondisjunction
creates XXY
females and XO
males.
X
w
w
w
+
+
w
Bridges’ Experiment
P? Crossing XXY females should
produce progeny
with unique
karyotypes.
X
+
w
w
w
+
w
w
w
w
+
w
w
+
ww
+
+
w w
Conclusions
Studies of nondisjunction provided the final proof of the
chromosome theory.
Sex linked genes segregate with the X chromosome
even in the rare cases when the chromosome
segregates abnormally.
Problem
Attached-X,When two X chromosomes fused at their ends
they are called attached-X,Attached-X cannot separate
during meiosis,This property is often used in the
genetic manipulation of Drosophila.
1) Are attached-X female flies conveniently maintained
and propagated as XX or XXY stocks?
2) If an attached-X white-eye female fly is crossed to an
red eye male,what is the eye color of the progenies?
Cross F1 F2
Dominant Recessive Ratio
tall X short tall 787 277 2.84:1
axial X terminal flowers axial 651 207 3.15:1
Coin flip Face side Back side Ratio
50 times total 30 20 3:2
Do these results fit expected ratio?
Tests of results,Chi Square (i)
The chi square test is used to test how well the
observed data fits the expectations.
2 =(observed-expected)2/expected
Example,Mendel’s test cross of the F1 from his
dihybrid cross gives these expected results:
Expected phenotypes Expected ratio
Round Yellow 1/4
Round green 1/4
wrinkled Yellow 1/4
wrinkled green 1/4
Chi Square Test
Hypothesis,the progeny actually are in a 1:1:1:1 ratio.
2 test:
genotype obs exp (0bs-exp)2/exp
Rr Yy 55 51.75 0.204
Rr yy 51 51.75 0.011
rr Yy 49 51.75 0.146
rr yy 52 51.75 0.001
Total 207 207 0.362
Degrees of freedom = 4 -1 = 3 (one less than the number
of progeny classes)
Chi Square Test
The probability of this?2 value,given the degrees of
freedom,is determined from a Chi square table
For this example,probability > 0.90
This much deviation from a correct hypothesis will occur
more than 9 of 10 times by random chance.
The hypothesis is accepted,We conclude these genes
really do assort independently.
Hypotheses that give P values > 0.05 are accepted.
If P is < 0.05,the hypothesis is rejected.
Problem
In a cross between ebony flies and wild type flies,
the F1 displayed wild type body color,F2 counting
derived the following numbers,
Normal body color male 152 female 168
Ebony body color male 29 female 27
1) Does male:female ratio fit 1:1?
2) Does normal:ebony fit 3:1 ratio? Can you give a
likely explanation for this deviation?
