Download Document

DNA Replication and Cell Cycle
 Mitosis and Meiosis
 Monohybrid cross
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'
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.
Phase M
Mitosis
Cell
Cycle
Phase S
Dna Synthesis
3. Complete the scheme of cell cycle of a diploid eukaryotic cell.
Draw the chromosomes at different stages of the cycle.
Draw a schematic picture of chromosomes of a diploid cell with n=1
Indifferent mitosis stage. The analysed individual is heterozygous for
gene A.
A
G1
a
PROPHASE
A a
a
a
A
A
a
a
A
A
METAPHASE
ANAPHASE
Daughter cells
a
A
a
A
Draw a schematic picture of chromosomes of a diploid cell with n=1
indifferent meiosis stage. The analysed individual is heterozygous for
gene A.
GAMETOCITE
A
a
Prophase I
A a
Metaphase I
A a
Anaphase I
A a
Anaphase I
A a
Telophase I
A
a
A
a
Metaphase II
Anaphase II
AA
a a
Gametes
A
A
½A
a
a
½a
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
Which types of gametes and in which proportions are produced by individuals that
have the following genotypes?
a) genotype AA;
ONLY GAMETES A
b) b) genotype Aa;
c)
½ GAMETES A
d) c) genotype aa
ONLY GAMETES A
½ GAMETES a
For each cross determine genotypic and phenotypic classes
expected in the progeny and relative frequencies.
Genotype of
the
individuals
used for the
cross
AA x aa
Aa x aa
Aa x Aa
Gametes
of the first
individual
(Frequenc
y)
Gametes of
the second
individual
(Frequency)
Genotypes and Phenotypes
frequency of
and frequency
the progeny
of the progeny
A (1)
a (1)
Aa (1x1=1)
A (1)
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)
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 coul be PP o Pp. In order to determine the genotype of the first parent
I observed the phenotypes of the progeny: all individuals with 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. 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
Gametocite
FASE S (DNA replication)
A
B
a
b
Prophase I
Homologous chromosome will be
separated
Metaphase I
A
B
A
b
a
b
a
B
Metaphase I
A
A
B
A
b
a
b
a
B
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.
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