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DNA Replication and Cell Cycle
 Mitosis and Meiosis
 Monohybrid and Dihybrid cross
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DNA Replication
DNA is always synthesized in the 5' to 3' direction
Features of leading strand
Features of lagging strand
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1. Replication of DNA molecule. Draw the new synthetized dna
with the polarity of the strands and the Okazaki fragments.
lagging strand
5'
3’
5’
Okazaki Fragments
5’
3’
3'
RNA
primer
leading strand
3'
5'
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2. How is the structure of a chromosome before and after the
replication process?
BEFORE
AFTER
Performe a scheme with the chromosome like a line and the
centromere like a circle.
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G1 phase. Metabolic changes prepare the cell for
division. At a certain point - the restriction point - the
cell is committed to division and moves into the S
phase.
S phase. DNA synthesis replicates the genetic material.
Each chromosome now consists of two sister
chromatids.
G2 phase. Metabolic changes assemble the
cytoplasmic materials necessary for mitosis and
cytokinesis.
M phase. A nuclear division (mitosis) followed by a cell
division (cytokinesis).
Phase M
Mitosis
Cell
Cycle
Phase S
Dna Synthesis
Diploid cells contain
two complete sets
(2n) of chromosomes.
3. Complete the scheme of cell cycle of a diploid eukaryotic cell.
Draw the chromosomes at different stages of the cycle.
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Draw a schematic picture of chromosomes of a diploid cell with n=1
In different mitosis stage. The analysed individual is heterozygous for
gene A.
A
G1
a
PROPHASE
A a
a
a
METAPHASE
ANAPHASE
The chromosomes congregate at the equatorial plane (no
pairing of homologous chromosomes).
The chromatids are attached to the spindle fibers at the
centromeres.
A
A
a
a
The two chromatids of each chromosome separate and
move to opposite poles
A
Daughter cells
a
A
The chromosomes replicate themselves to form
pairs of identical sister chromosomes, or
chromatids. The two chromatids remain attached
to one another at a region called the centromere
A
a
A
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Draw a schematic picture of chromosomes of a diploid cell with n=1 in
different meiosis stage. The analysed individual is heterozygous for gene
A.
GAMETOCYTE
A
a
Prophase I
A a
The chromosomes replicate themselves to form
pairs of identical sister chromosomes, or
chromatids. The two chromatids remain attached
to one another at a region called the centromere.
Pairing of homologous chromosomes.
(Recombination events)
Metaphase I
A a
The homologous chromosomes congregate at the
equatorial plane.
Each homologous chromosome is attached to the
spindle fibers.
Anaphase I
The two homologous chromosomes
separate and move to opposite poles
A a
Anaphase I
A a
Telophase I
A
a
Metaphase II
The chromosomes congregate at the equatorial
plane.
The chromatids are attached to the spindle fibers at
the centromeres.
A
a
Anaphase II
AA
Gametes
4 gametes
haploid
A
A
½A
The two chromatids of each
chromosome separate and
move to opposite poles
a a
a
a
½a
Frequency
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Which structures migrate at the opposite poles of the spindle?
a) in mitosis
SISTER CHROMATIDS
b) in meiosis, division I
HOMOLOGOUS CHROMOSOMES
c) in meiosis, division II
SISTER CHROMATIDS
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Which types of gametes and in which proportions are produced by individuals that
have the following genotypes?
a) genotype AA;
b) genotype Aa;
c) genotype aa
ONLY GAMETES A
½ GAMETES A
½ GAMETES a
ONLY GAMETES a
Genotype is the complete heritable genetic identity
Phenotype is a description of physical characteristics
Dominant character
a mendelian character that is expressed when it is transmitted by a single gene (upper case).
Recessive Character
a mendelian character that is expressed only when transmitted by both genes (one from each
parent) determining the trait (lower case).
For each cross determine genotypic and phenotypic classes expected in the progeny and relative
frequencies.
Genotype of the
individuals used
for the cross
Gametes of the
first individual
(Frequency)
Gametes of the
second individual
(Frequency)
Genotypes and
frequency of the
progeny
Phenotypes and
frequency of the
progeny
AA x aa
A (1)
a (1)
Aa (1x1=1)
A (1)
Aa x aa
½ A
½ a
a (1)
Aa (½ x 1= ½)
aa (½ x 1= ½)
A (½)
a (½)
½ A
½ a
A
AA (½ x ½= ¼ )
(¼+2/4=3/4)
Aa ( ¼ + ¼ = 2/4)
aa (½ x ½= ¼ )
a (¼)
Aa x Aa
½ A
½ a
In dogs hair length is determined by a gene, P, that can be present in
two alternative alleles, P and p. Accordingly to the crosses reported
below, determine the genotype of the individuals used for the crosses.
PARENTAL
PHENOTYPE
a) SHORT x LONG
n° OF INDIVIDUAL OF THE PROGENY
SHORT HAIR
LONG HAIR
100
0
PARENTAL GENOTYPE
PP x pp
X
a) The parents have different phenotypes then different genotypes. The progeny is
homogeneous (short hair) then short hair (P) is dominant over long hair (p). The
parent with long hair will be homozygous recessive (pp) while the parent with
short hair could be PP o Pp. In order to determine the genotype of the first
parent I observed the phenotypes of the progeny: all individuals have short hair.
Then the first parent will be homozygous dominant PP.
In dogs hair length is determined by a gene, P, that can be present in
two alternative alleles, P and p. Accordingly to the crosses reported
below, determine the genotype of the individuals used for the crosses.
PARENTAL
PHENOTYPE
n° OF INDIVIDUAL OF THE PROGENY
PARENTAL GENOTYPE
SHORT HAIR
LONG HAIR
a) short x long
100
0
PP x pp
b) short x long
50
50
Pp x pp
X
b) We have established that short is dominant over long: the parent with long
hair is homozygous recessive pp while the parent with short hair could be PP
o Pp.
In the progeny we have long and short hair individuals in the same
proportion. The parent with the short hair will be heterozygous (Pp).
In dogs hair length is determined by a gene, P, that can be present in
two alternative alleles, P and p. Accordingly to the crosses reported
below, determine the genotype of the individuals used for the crosses.
PARENTAL
PHENOTYPE
n° OF INDIVIDUAL OF THE PROGENY
PARENTAL GENOTYPE
SHORT HAIR
LONG HAIR
a) short x long
100
0
PP x pp
b) short x long
50
50
Pp x pp
c) short x short
150
50
Pp x Pp
X
c) The parents have the same genotypes (short hair) but in the progeny we
have an alternative phenotype (long hair): both individuals will be
heterozygous to produce homozygous recessive (with frequency of ¼).
Wild-type Drosophila melanogaster has red eyes. Mutants with
purple eyes exist. This phenotype is controlled by the pr gene, which
has two allelic states pr+ and pr.
This phenotype is controlled by the pr gene, which has two
allelic states pr+ and pr. The following crosses have been done:
Parental phenotypes n° of individuals in the progeny
a) red x red
Red
purple
total
125
35
160
Parental genotypes
pr+ pr x pr+ pr
b) purple x purple
c) red x red
d) purple x red
a) Crossing two individuals with Red phenotypes we obtain individual with
purple phenotype. The parent is heterozygous and Red is the dominant
character (3/4 Red, ¼ Purple).
This phenotype is controlled by the pr gene, which has two
allelic states pr+ and pr. The following crosses have been done:
Parental phenotypes n° of individuals in the progeny
a) red x red
b) purple x purple
Red
purple
total
Parental
genotypes
125
35
160
pr+ pr x pr+ pr
0
45
45
pr pr x pr pr
c) red x red
d) purple x red
b) In the progeny we have only purple individuals (recessive). The parents are
homozygous recessive.
This phenotype is controlled by the pr gene, which has two
allelic states pr+ and pr. The following crosses have been done:
Parental phenotypes n° of individuals in the progeny
a) red x red
b) purple x purple
c) red x red
Red
purple
total
Parental
genotypes
125
35
160
pr+ pr x pr+ pr
0
45
45
pr pr x pr pr
177
63
240
pr+ pr x pr+ pr
d) purple x red
c) In the progeny we observe purple individuals.. The parents are heterozygous
and the progeny is distributed: 3/4 red ¼ purple.
This phenotype is controlled by the pr gene, which has two
allelic states pr+ and pr. The following crosses have been done:
Parental phenotypes n° of individuals in the progeny
Red
purple
total
Parental
genotypes
125
35
160
pr+ pr x pr+ pr
0
45
45
pr pr x pr pr
c) red x red
177
63
240
pr+ pr x pr+ pr
d) purple x red
45
55
100
pr pr x pr+ pr
a) red x red
b) purple x purple
d) The first parent is purple thus homozygous recessive pr pr. The second parent
is Red and its genotype could be pr+ pr+ o pr+ pr.
In the progeny we observe individuals homozygous recessive, thus the second
parent is heterozygous pr+ pr.
Draw a scheme of meiosis process of a diploid cell with n=2. One chromosome carries
gene A, the other carries gene B. The analyzed individual is heterozygous for both
genes. Represent the two possible relative positions of the chromosomes in
metaphase I.
A
B
a
b
A
B
a
b
Gametocyte
FASE S (DNA replication)
Prophase I
Homologous chromosomes will be
separated
Metaphase I
A
B
A
b
a
b
a
B
Metaphase I
A
B
A
b
a
b
a
B
Metaphase II
A
B
A
B
a
a
b
b
Gametes
A
A
B
a
A
b
a
A
B
a
¼ AB
b
B
¼ ab
B
b
b
¼ Ab
a
¼ aB
Now use the branch diagram to determine type and frequency of the
gametes produced by the same cell.
AaBb
Gene A (frequency)
A ½
………...(…….)
Gene B (frequency)
Gametes
………….(……..)
B ½
………….(……..)
AB ¼
………….(……..)
b ½
………….(……..)
………….(……..)
B ½
………….(……..)
b ½
………….(……..)
Ab ¼
aB ¼
a ½
………...(…….)
………….(……..)
ab ¼
Which type of gametes and in which proportions are produced by
individuals that have the following genotype (use the branch diagram)?
a) aa bb
b) Aa bb
ab (1)
A (1/2)
b (1)
Ab (1/2)
a (1/2)
b (1)
ab (1/2)
B (1/2)
AB (1/4)
b (1/2)
Ab (1/4)
B (1/2)
aB (1/4)
b (1/2)
ab (1/4)
c) Aa Bb
A (1/2)
a (1/2)
For each cross determine genotypic and phenotypic classes expected in the progeny
and relative frequencies (A and B genes are independent)
genotype of the
individuals used
for the cross
AA bb x aa BB
Aa bb x aa Bb
Aa Bb x aa bb
Gametes of the
Gametes of the
Genotypes and
second
first individual
frequency of the
individual
(frequency)
progeny
(frequency)
Ab (1)
aB (1)
Ab (1/2)
ab (1/2)
aB (1/2)
ab (1/2)
¼ AB
¼ Ab
¼ aB
¼ ab
ab (1)
Aa Bb (1)
phenotypes
and
frequency of
the progeny
A B (1)
¼
¼
¼
¼
AaBb
Aabb
aaBb
aabb
¼
¼
¼
¼
AB
Ab
aB
ab
¼
¼
¼
¼
AaBb
Aabb
aaBb
aabb
¼
¼
¼
¼
AB
Ab
aB
ab
Now use the branch diagram to calculate the phenotypic classes.
b) Aa bb X aa Bb
Phenotype for A gene
(cross Aa X aa)
A ½
………...(…….)
Phenotypes for B gene
(cross bb X BB)
Phenotypical classes
………….(……..)
B ½
………….(……..)
AB ¼
………….(……..)
b ½
………….(……..)
………….(……..)
B ½
………….(……..)
b ½
………….(……..)
Ab ¼
aB ¼
a ½
………...(…….)
………….(……..)
ab ¼