Cell Cycle Mitosis and Meiosis Monohybrid and Dihybrid cross Cell and Cell Cycle DNA Replication DNA is always synthesized in the 5 to 3 direction Features of leading strand Features of lagging strand ID: 784025
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Slide1
DNA Replication and Cell Cycle Mitosis and Meiosis Monohybrid and Dihybrid cross
Slide2Cell and Cell Cycle
Slide3DNA ReplicationDNA is always synthesized in the 5' to 3' directionFeatures of leading strandFeatures of lagging strandhttp://www.mantorlab.unimi.it/mantorlab/sito/Home.htmlhttps://www.youtube.com/watch?v=dKubyIRiN84
Slide43’3’
leading strand
5’
Okazaki Fragments
5’
RNA primer
'
5'
3'
3'
5
1. Replication of DNA molecule. Draw the new synthetized
dna
with the polarity of the strands and the Okazaki fragments.
lagging strand
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Slide52. How is the structure of a chromosome before and after the replication process? Perform a scheme with the chromosome like a line and the centromere like a circle.
BEFORE
AFTER
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Homologous chromosomes and sister chromatids
Slide63. Complete the scheme of cell cycle of a diploid eukaryotic cell. Draw the chromosomes at different stages of the cycle.
Cell
Cycle
Phase M
Mitosis
Phase S
Dna Synthesis
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).
Diploid cells contain two complete sets (2n) of chromosomes.
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23 pairs of chromosomes(n)
46 chromosomes
46 pairs of chromosomes
92 chromosomes
Slide7Draw a schematic picture of chromosomes of a diploid cell with n=1 In different mitosis stage. The analysed individual is heterozygous for gene A. G1/S/G2PROPHASEMETAPHASE
ANAPHASE
Daughter cells
A
a
A
A
a
a
a
a
A
A
A
a
A
a
A
a
Phase S: 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
The chromosomes congregate at the equatorial plane (no pairing of homologous chromosomes).
The chromatids are attached to the spindle fibers at the centromeres.
The two chromatids of each chromosome separate and move to opposite poles
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Slide8Meiosis: diploid cellshaploid cellsTetrads; Synapsis; Recombination and Crossing overEquatorial alignment is random: independent assortmentIf n is 23, the possible combinations of dna in gametes are 2n over a milion
Slide9Draw a schematic picture of chromosomes of a diploid cell with n=1 in different meiosis stage. The analysed individual is heterozygous for gene A.
A
a
A
a
G
AMETOCYTE
Prophase I
Metaphase I
Anaphase I
A
a
A
a
Phase S: 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)
The homologous chromosomes congregate at the equatorial plane randomly.
Each
homologous chromosome
is attached to the spindle
fibers
.
The two homologous chromosomes separate and move to opposite poles
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Slide10½ A½ a
A
a
A
a
A
a
a
a
A
A
A
a
A
a
Anaphase I
Metaphase II
Anaphase II
Telophase I
Gametes:
4 haploid
gametes
The two chromatids of each chromosome separate and move to opposite poles
The chromosomes congregate at the equatorial plane.
The
chromatids
are attached to the spindle fibers at the centromeres.
Frequency
Slide11Which structures migrate at the opposite poles of the spindle?in mitosis SISTER CHROMATIDSb) in meiosis, division I HOMOLOGOUS CHROMOSOMESc) in meiosis, division II SISTER CHROMATIDShttp://www.mantorlab.unimi.it/mantorlab/sito
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Slide12Slide13Monohybrid and Dihybrid crosses
Slide14Gamete: a mature reproductive cell that is specialized for sexual fusion.Each gamete is haploid and fuses with a cell of the same origin but of opposite sex to produce a diploid zygote.Cross: a mating between two individuals fusion of gametesZygote: the cell produced by the fusion of male and female gametesGene: the determinant of a characteristic of an organismAlleles: alternative forms of a GeneLocus: the specific place on a chromosome where a gene is locatedGenotype is the complete heritable genetic identity (Aa)Phenotype is a description of physical characteristics (A)Dominant character (Homozygous AA Heterozygous Aa)a mendelian character that is expressed when it is transmitted by a single alleles (upper case).Recessive Character (Homozygous aa)a mendelian character that is expressed only when transmitted by both alleles (one from each parent) determining the trait (lower case).A true-breeding organism, sometimes also called a purebred, is an organism that always passes down certain phenotypic traits (i.e. physically expressed traits) to its offspring of many generations.
homozygote
GENETIC TERMINOLOGY
Slide15Which 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 http://www.mantorlab.unimi.it/
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½ GAMETES A ½ GAMETES a
ONLY GAMETES a
only GAMETES A
Slide16For 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
Aa x aa
Aa x Aa
A (1)
a (1)
½ A
½ a
a (1)
½ A
½ a
½ A
½ a
Aa (1x1=1)
Aa (½ x 1= ½)
aa (½ x 1= ½)
AA (½ x ½= ¼ )
Aa ( ¼ + ¼ = 2/4)
aa (½ x ½= ¼ )
A (1)
A (½)
a (½)
A
(¼+2/4=3/4)
a (¼)
Monohybrid
Cross
Typical ratio
3:1
Slide17A purple-flower pea plant is crossed with a white-flowered pea plant. All the F1 plants produced purple flowers. When the F1 plants are allowed to self-pollinate, 401 of the F2 have purple flowers and 131 have white flowers. What are the genotypes of the parental and F1 generation plants?Ratio in F2 is very close to 3:1 monohybrid crossF1xF1 cross in which both are identically heterozygousParents differed in phenotypes F1 only one phenotype parents were true breedingF1 resembled one of the parental phenotypes purple is dominant over white
Slide18In 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
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 observe the phenotypes of the progeny: all individuals have short hair. Then the first parent will be homozygous dominant
PP
.
X
Slide19PARENTAL PHENOTYPE
n° OF INDIVIDUAL OF THE PROGENY
PARENTAL GENOTYPE
SHORT HAIR
LONG HAIR
a) short x long
100
0
b) short x long
50
50
PP x pp
X
b) We have already 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).
Pp x 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.
Slide20PARENTAL PHENOTYPE
n° OF INDIVIDUAL OF THE PROGENY
PARENTAL GENOTYPE
SHORT HAIR
LONG HAIR
a) short x long
100
0
b) short x long
50
50
c) short x short
150
50
PP x pp
Pp x pp
X
c) The parents have the same phenotype (short hair) but in the progeny we have an alternative phenotype (long hair): both individuals will be heterozygous to produce homozygous recessive in the progeny pp (with frequency of ¼).
Ratio 3:1
monohybrid cross
Pp x 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.
Slide21Wild-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.
Slide22Parental phenotypesn° of individuals in the progeny
Parental genotypes
Red
purple
total
a) red x red
125
35
160
b) purple x purple
c) red x red
d) purple x red
pr+ pr
x
pr+ pr
Crossing two individuals with Red phenotypes we obtain individual with purple phenotype.
The parents are heterozygous and Red is the dominant character (3/4 Red, ¼
Purple
RATIO
3:1
).
This phenotype is controlled by the
pr
gene, which has two allelic states
pr+
and
pr
. The following crosses have been done:
Slide23Parental phenotypes
n° of
individuals
in the
progeny
Parental
genotypes
Red
purple
total
a)
red
x
red
125
35
160
b)
purple
x
purple
0
45
45
c)
red
x
red
d)
purple
x
red
pr+ pr x pr+ pr
pr
pr
x
pr
pr
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:
Slide24Parental phenotypesn° of individuals in the progeny
Parental genotypes
Red
purple
total
a) red x red
125
35
160
b) purple x purple
0
45
45
c) red x red
177
63
240
d) purple x red
pr+ pr x pr+ pr
pr
pr
x pr
pr
pr+ pr
x
pr+ pr
c) In the progeny we observe purple individuals.
The progeny is distributed: 3/4 red ¼ purple
3:1
The parents are heterozygous.
This phenotype is controlled by the
pr
gene, which has two allelic states
pr
+
and
pr
. The following crosses have been done:
Slide25Parental phenotypesn° of individuals in the progeny
Parental genotypes
Red
purple
total
a) red x red
125
35
160
b) purple x purple
0
45
45
c) red x red
177
63
240
d) purple x red
45
55
100
pr+ pr x pr+ pr
pr
pr
x pr
pr
pr+ pr x pr+ pr
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.
pr
pr
x
pr+ pr
This phenotype is controlled by the
pr
gene, which has two allelic states
pr
+
and
pr
. The following crosses have been done:
Slide26The major part of genes possesses more than 2 forms although all diploid individual could carry up to 2 alleles . In rabbit the colour of fur is encoded by gene C. The wild type allele C is dominant over the other alleles and correspond to a dark grey colour. The allele ch is recessive respect the wild type and if homozygous shows a Himalayan phenotype (white fur with some dark parts). The c allele is recessive respect all the previous alleles and show a phenotype totally white. We can describe the relation between alleles:C > ch > ca) White rabbit x dark gray rabbit progeny: dark gray rabbitsb) Himalayan rabbit x Himalayan rabbit
progeny: ¾
himalayan
¼ white
c) Dark
gray
rabbit x white rabbit
progeny: ½ dark
gray
½ white
d) Dark
gray
rabbit x Dark
gray
rabbit
progeny: ¾ dark
gray
¼ white
e) Dark
gray
rabbit x white rabbit
½ dark gray ½
himalayan
a) cc x CCb) c
hc x chcc) Cc x cc
d) Cc x Cce) Cch x cc
Slide27INHERITANCE OF TWO OR MORE INDEPENDENT GENES(Principle of Independent Assortment)
Slide28GametocyteProphase IMetaphase IDraw 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
A
B
b
a
A
B
b
FASE S (DNA
replication
)
Homologous
chromosomes
will
be
separated
a
A
B
b
a
A
B
b
Slide29Metaphase I
a
A
B
b
a
A
B
b
A
B
a
b
A
b
a
B
Gametes
A
B
a
b
A
b
a
B
A
B
a
b
A
b
a
B
¼ AB
¼ ab
¼ Ab
¼
aB
Metaphase
II
Slide30Now use the branch diagram to determine type and frequency of the gametes produced by this cell. AaBbGene A (frequency) Gene B (frequency) Gametes
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………...(…….)
………...(…….)
A ½
a ½
B ½
b ½
B ½
b ½
AB ¼
Ab ¼
aB
¼
ab ¼
Slide31Which type of gametes and in which proportions are produced by individuals that have the following genotype (use the branch diagram)?a) aa bbb) Aa bbc) Aa Bb ab (1)
a (1/2)
B (1/2)
b (1/2)
B (1/2)
b (1/2)
A (1/2)
ab (1/4)
AB (1/4)
Ab (1/4)
aB (1/4)
a (1/2)
b (1)
b (1)
A (1/2)
Ab (1/2)
ab (1/2)
Slide32For 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 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 bb x aa BB
Aa bb x aa Bb
Aa Bb x aa bb
Ab (1)
aB (1)
Aa Bb (1)
A B (1)
Ab (1/2)
ab (1/2)
aB (1/2)
ab (1/2)
¼ AaBb
¼ Aabb
¼ aaBb
¼ aabb
¼ AB
¼ Ab
¼ aB
¼ ab
¼ AB
¼ Ab
¼ aB
¼ ab
ab (1)
¼ AaBb
¼ Aabb
¼ aaBb
¼ aabb
¼ AB
¼ Ab
¼ aB
¼ ab
Slide33Now use the branch diagram to calculate the phenotypic classes. b) Aa bb X aa BbPhenotypes for A gene Phenotypes for B gene Phenotypic classes (cross Aa X aa) (cross bb X Bb)
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………….(……..) ………….(……..)
………...(…….)
………...(…….)
A ½
a ½
B ½
b ½
B ½
b ½
AB ¼
Ab ¼
aB ¼
ab ¼
Slide34In Drosophila melanogaster, body colour is determined by the e gene: the recessive allele is responsible for the black colour of the body, the dominant allele e+ is responsible for the grey body. Vestigial wings are determined by the recessive allele vg, normal wings are determined by the dominant allele vg+. These two genes are independent. If dihybrid flies for these two genes are crossed and resulting progeny is composed by 368 individuals, how many individuals are present in every phenotypic class? e+ e+ e e
vg
+
vg
+
vg
vg
Parental Genotypes:
e
+
e vg
+
vg
X
e
+
e vg
+
vg
Gametes
e (1/2)
vg
+
(1/2)
vg
(1/2)
vg
+
(1/2)
vg
(1/2)
e
+
(1/2)
e vg
(1/4)
e+ vg+ (1/4)
e+ vg (1/4) e vg+ (1/4)
Slide35Phenotypes for gene e Phenotypes for gene vg Phenotipical classes (cross e+e X e+e) (cross vg+vg X vg+vg) vg+ (3/4) e (1/4)
e
+
(3/4)
vg
(1/4)
e
+
vg
+
(¾ X ¾= 9/16)
e
+
vg
(¾ X ¼ = 3/16)
vg
+
(3/4)
vg
(1/4)
e vg
+
(¾ X ¼ = 3/16)
e vg
(¼ X ¼=1/16)
9/16 di 368 = 207 individuals
e
+
vg
+
grey body and normal wings
3/16 di 368 = 69 individuals
e+ vg grey body and vestigial wings 3/16 di 368 = 69 individuals e vg+ black body and normal wings 1/16 di 368 = 23 individuals e vg black body and vestigial wings
Slide36e+ vg+ 1/4
Genotypes? - Punnet square
e
+
vg
1/4
e
vg
+
1/4
e
vg
1/4
e
+
vg
+
1/4
e
+
vg
1/4
e
vg
+
1/4
e vg 1/4
Slide37e+ e+ vg+ vg+ 1/16
e
+
vg
+
1/4
Genotypes? - Punnet square
e
+
vg
1/4
e
vg
+
1/4
e
vg
1/4
e
+
vg
+
1/4
e
+
vg
1/4
e
vg+ 1/4e vg 1/4
e+ e+ vg+ vg
1/16e+ e vg+ vg+ 1/16
e+ e vg+ vg 1/16e
+ e+ vg+ vg 1/16e+ e+ vg vg
1/16e+ e vg+ vg 1/16
e+ e vg vg 1/16e+ e
vg+ vg+ 1/16e+ e vg+ vg 1/16
e e vg+ vg+ 1/16e e
vg+ vg 1/16e+ e vg+
vg 1/16e+ e vg vg 1/16
e e vg+ vg 1/16e
e vg vg 1/16
Slide38Genotypes1/16 of 368 = 23 individuals e+ e+ vg+ vg+ 2/16 of 368 = 46 individuals e+ e+ vg+
vg
2/16 of 368 = 46
individuals
e
+
e
vg
+
vg
+
1/16 of 368 = 23
individuals
e
+
e+
vg
vg 4/16 of 368 = 92 individuals e
+
e
vg+
vg
2/16 of 368 = 46 individuals e+
e vg vg 1/16 of 368 = 23
individuals e e vg+
vg+ 2/16 of 368 = 46 individuals e e
vg+ vg 1/16 of 368 = 23 individuals e e
vg vg 9/16 of 368 = 207 individuals
e+ vg+ grey body and normal wings3/16 of 368 = 69 individuals
e+ vg grey body and vestigial wings
3/16 of 368 = 69 individuals e vg+ black body and normal wings 1/16 of 368 = 23 individuals
e vg black body and vestigial wingsRATIO 9:3:3:1
Phenotypes
Slide39Phenotype of individuals used for the cross BlackRoughBlack SmoothBrownRoughBrownSmootha) Black Rough X Brown Smooth5000
0
b) Black Rough
X Black Rough
c) Black Rough
X Brown Smooth
d) Brown Rough
X Black Rough
a)The parents have different phenotypes and the progeny is all Black and Rough:
Black and Rough are dominant characters (B, black; R, rough)
The parents are homozygous: BB RR X bb
rr
We do not perform statistic test because we do not have differences between
expected
progeny and
observed
progeny.
n° of individuals in the progeny (observed)
In guinea pigs, the
R
gene determinates fur rough or smooth
, while the
B
gene controls fur colour
. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.
Slide40Phenotypes of individuals used for the cross BlackRoughBlack SmoothBrownRoughBrownSmootha) Black Rough X Brown Smooth50
0
0
0
b) Black Rough
X Black Rough
185
60
57
18
c) Black Rough
X Brown Smooth
d) Brown Rough
X Black Rough
b) We know that Black and Rough are dominant traits.
The crossed individuals have the same phenotype but in the progeny we have the recessive character, thus both parents are double heterozygous:
Bb Rr X Bb Rr
If the two genes assort independently, in the progeny we
expect
:
9/16 BR
(Black rough):
3/16 Br
(black smooth):
3/16
bR
(brown rough): 1/16
br (brown smooth)
n° of individuals in the progeny (observed)In guinea pigs, the
R gene determinates fur rough or smooth, while the B gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.
Slide41PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –
Xe
)
2
Xe
Black
Rough
185
9/16
Black Smooth
60
3/16
Brown Rough
57
3/16
Brown Smooth
18
1/16
Total
320
16/16
C
2
(chi-square) test is used to help in making the decision to hold onto or reject the hypothesis.
We want to compare experimentally observed numbers of items in several different categories with numbers that are predicted on the basis of our hypothesis.
It is important to understand that usually the observed phenotypic ratios among progeny rarely match exactly the expected ratios.
If the difference between observed and predicted results is too large
reject hypothesis
Slide42PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
Black
Rough
185
9/16
180
(185-180)
2
=
25
25/180= 0,14
Black Smooth
60
3/16
60
0
0/60= 0
Brown Rough
57
3/16
60
(57-60)
2
=
9
9/60=0,15
Brown Smooth
18
1/16
20(18-20) 2=
44/20= 0,20
Total
32016/16320
S
=0,49
c2 = 0,49Degrees of freedom = 4 – 1 = 3 (number of phenotypes -1)C2 (chi-square) test is used to help in making the decision to hold onto or reject the hypothesis.
We want to compare experimentally observed numbers of items in several different categories with numbers that are predicted on the basis of our hypothesis.
Slide43Accepted
Rejected
c
2
= 0,49
P >90%, we accept the hypotesis
5% is the conventional decision line
Slide44Phenotype of individuals used for the cross BlackRoughBlack SmoothBrownRoughBrownSmootha) Black Rough X Brown Smooth5000
0
b) Black Rough
X Black Rough
185
60
57
18
c) Black Rough
X Brown Smooth
105
100
98
97
d) Brown Rough
X Black Rough
c) The parents have different phenotypes. The second parent is homozygous recessive for both genes (bb
rr
). The first parent has the dominant phenotypes for both characters B - R -.
In the progeny we observe classes of individuals with recessive phenotypes thus the first individual is double heterozygous: Bb Rr.
In the progeny we expect 4 phenotypic classes with ratio:
1:1:1:1
n° of individuals in the progeny
In guinea pigs, the
R
gene determinates fur rough or smooth, while the
B
gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.
Slide45PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
Black
Rough
105
¼
Black Smooth
100
¼
Brown Rough
98
¼
Brown Smooth
97
¼
Total
400
4/4
c
2
Slide46PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
Black
Rough
105
¼
100
(105-100)
2
=
25
25/100= 0,25
Black Smooth
100
¼
100
0
0/60= 0
Brown Rough
98
¼
100
(98-100)
2
=
4
4/100=0,04
Brown Smooth
97
¼
100(97-100) 2=
99/100= 0,09
Total
4004/4400
S
=0,38
c2 = 0,38Degree of freedom = 4 – 1 = 3c2
P =>90%, we accept the hypothesis
Slide47Phenotypes of individuals used for the cross BlackRoughBlack SmoothBrownRoughBrownSmootha) Black Rough X Brown Smooth50
0
0
0
b) Black Rough
X Black Rough
185
60
57
18
c) Black Rough
X Brown Smooth
105
100
98
97
d) Brown Rough
X Black Rough
63
17
58
22
d) The first individual is homozygous recessive for colour (bb) and he has the dominant phenotype for Rough gene (R -). The second individual is dominant for both characters (B - R -).
In the progeny we observe the classes with recessive phenotypes for brown and smooth hair, thus the original individuals have a recessive allele to donate to the progeny: b b R r X B b R r
n° of individuals in the progeny
In guinea pigs, the
R
gene determinates fur rough or smooth, while the B
gene controls fur color. Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test.
Slide48Gene B Gene R phenotypic Classesbb X Bb Rr x Rr R (3/4) B R (1/2 x 3/4 = 3/8) black roughB (1/2) r (1/4) B r (1/2 x 1/4 = 1/8) black smooth R (3/4) b R (1/2 x 3/4 = 3/8) brown roughb (1/2) r (1/4) b r (1/2 x 1/4 = 1/8) brown smooth
With branch diagram determine expected phenotypic classes:
Slide49PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –
Xe
)
2
Xe
Black
Rough
63
3/8
Black
smooth
17
1/8
Brown
Rough
58
3/8
Brown smooth
22
1/8
Totale
160
8/8
c
2
Slide50PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
Black
Rough
63
3/8
60
(63-60)
2
=
9
9/60= 0,15
Black
smooth
17
1/8
20
(17-20)
2
=
9
9/20= 0,45
Brown
Rough
58
3/8
60
(58-60)
2=
44/60=0,07Brown smooth
221/8
20(22-20) 2=
44/20= 0,2
Totale1608/8
160
S =0,87c2 = 0,87Freedom of degree= 4 – 1 = 3
c2 P =80-90%, we accept the hypothesis
Slide51Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red color, gene D determines the plant heightPhenotype of individuals used for the crossRed tall
Red
short
White
tall
White short
a) red tall x red tall
120
0
45
0
b) red tall x white short
c) red tall x white short
d) white tall x red tall
e) red tall x red tall
Colour: the crossed individuals have the same phenotype but in the progeny we have also recessive phenotype individuals: both parents are heterozygous (
Ww
) and Red (W) is dominant over white (w).
Height: we cannot establish which is the dominant allele since the character does not segregate. All the progeny is tall. Both parents could be recessive (dd) or D D x D - o or D - x D D
Numero di
individui
della
progenie
Slide52PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –
Xe
)
2
Xe
Red tall
120
3/4
123,75
(120-123,75)
2
=
14
14/123,75= 0,11
White tall
45
1/4
41,25
(45-41,25)
2
=
14
14/41,25= 0,33
Total
165
4/4
165
S
=
0,44
c2 = 0,44
Freedom of degree = 2 – 1 = 1
c2
P =50-70%, we can accept the hypothesis
Slide53Phenotypes of individuals used for the crossRed tallRed shortWhite
tall
White short
a) red tall x red tall
120
0
45
0
b) red tall x white short
100
0
105
0
c) red tall x white short
d) white tall x red tall
e) red tall x red tall
b) The crossing plants have two different phenotypes.
In the progeny we observe plants red and white: we know that red is dominant over white thus the genotypes are W w X w w
Height: since we have only tall plants we establish that tall is dominant over short. In the progeny there isn’t segregation of character then the tall plant is homozygous. The genotypes are: D D X d d
n° of individuals in the progeny
Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red colour, gene D determines the plant height
Slide54c2 = 0,12Degrees of freedom = 2 – 1 = 1P = 70-80%, we accept the hypothesisWw DD X ww ddWe expect: ½ WD (red and tall) and ½ wD (white and tall).c2
Phenotypes
Xo
(observed individuals)
H
(hypothesis)
Xa
Expected individuals
(Xo –Xa)
2
(Xo –Xa)
2
Xa
Red tall
100
½
102,5
(100-102,5)
2
=
6,25
6,25/102,5= 0,06
White tall
105
½
102,5
(105-102,5)
2
=
6,25
6,25/102,5= 0,06Total
2052/2205
S
=0,12
Slide55Phenotypes of individuals used for the crossRed tallRed shortWhite
tall
White short
a) red tall x red tall
120
0
45
0
b) red tall x white short
100
0
105
0
c) red tall x white short
45
43
48
44
d) white tall x red tall
e) red tall x red tall
c) The crossed plants have different phenotypes. In the progeny we observed segregation of the characters: we know that red is dominant over white and we assess that the cross is between a heterozygote and a homozygote recessive: W w X w w
Height: Tall is dominant over short, thus also for height the cross is between a heterozygote and a homozygote recessive D d X d d
n° of
individuals
in the
progeny
Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red
color
, gene D determines the plant height
Slide56PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –
Xe
)
2
Xe
Red tall
45
¼
Red short
43
¼
White tall
48
¼
White short
44
¼
Total
180
4/4
test
c
2
W w D d X w w d d
We expect 4 phenotypic classes with ratio 1 : 1 : 1 : 1.
Slide57PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
Red tall
45
¼
45
(45-45)
2
=
0
0/45= 0
Red short
43
¼
45
(43-45)
2
=
4
4/45= 0,09
White tall
48
¼
45
(48-45)
2
=
9
9/45=0,2
White short44¼
45
(44-45) 2=11/45= 0,02
Total180
4/4180
S =
0,31c2 = 0,31Degrees of freedom = 4 – 1 = 3
test c2 P =>90%, we can accept the hypothesisW w D d X w w d dWe expect 4 phenotypic classes with ratio 1 : 1 : 1 : 1.
Slide58Phenotype of individuals used for the crossRed tallRed shortWhite
tall
White short
a) red tall x red tall
120
0
45
0
b) red tall x white short
100
0
105
0
c) red tall x white short
45
43
48
44
d) white tall x red tall
175
67
182
58
e) red tall x red tall
d) In the progeny recessive phenotypic classes are present.
We can suppose that the cross is between: w w D d X W w D d
n° of
individuals
in the
progeny
Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red colour, gene D determines the plant height
Slide59We expect 4 phenotypic classes: 3/8 : 1/8 : 3/8 : 1/8Gene W Gene D Phenotypic classesww X Ww Dd x Dd D (3/4) W D (1/2 x 3/4 = 3/8) Red TallW (1/2) d(1/4) W d (1/2 x 1/4 = 1/8) Red short D (3/4) w D (1/2 x 3/4 = 3/8) White Tallw (1/2) d (1/4) w d (1/2 x 1/4 = 1/8) White short
What
are the
expected
phenotypic
classes
?
Slide60PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –Xe)
2
Xe
Red
tall
175
3/8
Red
short
67
1/8
White
tall
182
3/8
White short
58
1/8
Total
482
8/8
Slide61PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2(Xo –Xe)2Xe
Red tall
175
3/8
180,75
(175-180,75)
2
=
33,06
33,06
/180,75= 0,18
Red short
67
1/8
60,25
(67-60,25)
2
=
45,56
45,56
/60,25= 0,75
White tall
182
3/8
180,75
(182-180,75)
2
=
1,56
1,56/180,75=0,1
White short
581/860,25(58-60,25)
2=5,065,06/60,25= 0,08
Total4828/8
482S =1,11
c2 = 0,31Degrees of freedom = 4 – 1 = 3P =70-80%, Hypothesis accepted
Slide62Phenotype of individuals used for the crossRed tallRed shortWhite
tall
White short
a) red tall x red tall
120
0
45
0
b) red tall x white short
100
0
105
0
c) red tall x white short
45
43
48
44
d) white tall x red tall
175
67
182
58
e) red tall x red tall
265
92
93
28
e)In the progeny we observe recessive phenotypes. The individuals must be double heterozygous: W w D d X W w D d
From a dihybrid cross we expect 4 phenotypic classes:
9/16 WD (Red tall) : 3/16
Wd
(Red short) : 3/16
wD
(white tall) : 1/16
wd
(white short).n° of
individuals in the progeny Determine the genotype of individuals used for the crosses and verify your hypothesis with χ2 test. For each cross create a table, according to the scheme reported below. Gene W determines Red colour, gene D determines the plant height
Slide63PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(
Xo
–Xe)
2
Xe
Red
tall
265
9/16
Red
short
92
3/16
White
tall
93
3/16
White short
28
1/16
Total
478
16/16
Slide64PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)
2
Xa
Red
tall
265
9/16
268,87
(265-268,87)
2
=
14,97
14,97
/268,87= 0,05
Red
short
92
3/16
89,62
(92-89,62)
2
=
5,66
5,66
/89,62= 0,06
White
tall
93
3/16
89,62
(93-89,62) 2=
11,4211,42/89,62=0,12
White short281/16
29,87(28-29,87) 2=3,493,49/29,87= 0,11Total47816/16478
S =0,34c2
= 0,34Degrees of freedom = 4 – 1 = 3P >90%, hypothesis accepted
Slide65Crossing a pure line (true breeding organism) of melons whit white (G gene) and spherical fruits (R gene)with another pure line of melons with yellow and flat fruits the f1 progeny is melons with white and spherical fruits. Crossing two plants of F1 we have:Determine the genotypes of individuals of P1, P2 e F1, make the hypothesis of traits segregation and verify the results with X2
P1
P2
X
F1
148
Plants with white and spherical fruits
52
Plants with yellow and spherical fruits
49
Plants with white and flat fruits
23
Plants with yellow and flat fruits
F2
From first cross we know that spherical is dominant over flat and white is dominant over yellow. WE decide to name R the gene that control the shape melon and G the gene that control the
color
melon: the first parent is homozygous dominant and the second parent is homozygous recessive. The progeny will be heterozygous for both genes.
RR GG
rr
gg
Rr
Gg
Slide66PhenotypesXo(observed individuals)H(hypothesis)XeExpected individuals(Xo –Xe)2
(Xo –
Xe
)
2
Xe
White
Spherical
148
9/16
Yellow
Spherical
52
3/16
White
Flat
49
3/16
Yellow
Flat
23
1/16
Total
272
16/16
In F2 we have 4 phenotypic classes. This is a dihybrid cross and we expect ratio 9 : 3 : 3 : 1.
test
c
2
.
Slide67PhenotypesXo(observed individuals)H(hypothesis)XaExpected individuals(Xo –Xa)2(Xo –Xa)2Xa
White
Spherical
148
9/16
153
(148-153)
2
=
25
25
/153= 0,16
Yellow
Spherical
52
3/16
51
(52-51)
2
=
1
1
/51= 0,02
White
Flat
49
3/16
51
(49-51)
2
=
4
4/51=0,08
YellowFlat231/16
17(23-17) 2=3636/17= 2,12Total27216/16272
S =2,38
c2 = 2,38Degrees fo freedom = 4 – 1 = 3P=30-50%, hypothesis acceptedIn F2 we have 4 phenotypic classes. This is a dihybrid cross and we expect frequencies 9 : 3 : 3 : 1. test c
2.
Slide68Gene AAaA (1/2)a (1/2)Gene BBbB (1/2)b (1/2)B (1/2)b (1/2)
Gene C
CC
C (1)
C (1)
C (1)
C (1)
Gametes
ABC (
½
X ½ X 1= 1/4)
AbC
(½ X ½ X 1= 1/4)
aBC
(½ X ½ X 1= 1/4)
abC
(½ X ½ X 1= 1/4)
Using the branch diagram, determine the gametes produced by individuals with the following genotype (A, B and C genes are independent):
a) Aa Bb CC
Slide69b) Aa bb Cc DdGene AAaA (1/2)a (1/2)Gene Bbbb (1)
b (1)
Gene C
Cc
C (1/2)
c (1/2)
C (1/2)
c (1/2)
Gene D
Dd
D (1/2)
d (1/2)
D (1/2)
d (1/2)
D (1/2)
d (1/2)
D (1/2)
d (1/2)
Gametes
AbCD (½ X 1 X ½ X ½ = 1/8)
AbCd (½ X 1 X ½ X ½ = 1/8)
AbcD (½ X 1 X ½ X ½ = 1/8)
Abcd (½ X 1 X ½ X ½ = 1/8)
abCD (½ X 1 X ½ X ½ = 1/8)
abCd (½ X 1 X ½ X ½ = 1/8)
abcD (½ X 1 X ½ X ½ = 1/8)
abcd (½ X 1 X ½ X ½ = 1/8)
Slide70Gene AAaA (1/2)a (1/2)Gene BBbB (1/2)b (1/2)B (1/2)b (1/2)
Gene C
Cc
C (1/2)
c (1/2)
C (1/2)
c (1/2)
C (1/2)
c (1/2)
C (1/2)
c (1/2)
c) Aa
Bb
Cc
Gametes
ABC (½ X ½ X ½ = 1/8)
ABc (½ X ½ X ½ = 1/8)
AbC (½ X ½ X ½ = 1/8)
Abc (½ X ½ X ½ = 1/8)
aBC (½ X ½ X ½ = 1/8)
aBc (½ X ½ X ½ = 1/8)
abC (½ X ½ X ½ = 1/8)
abc (½ X ½ X ½ = 1/8)
Slide71Phenotypes forGene AAa X aaA (1/2)a (1/2)Phenotypes for Gene BBb X bbB (1/2)b (1/2)B (1/2)b (1/2)Phenotypes for Gene CCc X cc
C (1/2)
c (1/2)
C (1/2)
c (1/2)
C (1/2)
c (1/2)
C (1/2)
c (1/2)
Final
Phenotypes
ABC (½ X ½ X ½ = 1/8)
ABc (½ X ½ X ½ = 1/8)
AbC (½ X ½ X ½ = 1/8)
Abc (½ X ½ X ½ = 1/8)
aBC (½ X ½ X ½ = 1/8)
aBc (½ X ½ X ½ = 1/8)
abC
(½ X ½ X ½ = 1/8)
abc
(½ X ½ X ½ = 1/8)
For the following cross, determine the phenotypic classes expected in the progeny and their frequencies (A, B and C genes are independent)
AaBbCc
x
aabbcc
Which are the gametes produced by the first individual and their frequencies?
Which are the Genotypes of the progeny and their frequencies?
Slide72Consider three gene pairs Aa, Bb, and Cc each of which affects a different character. In each case the upper case letter signifies the dominant allele and the lowercase letter the recessive allele. These three gene pairs assort independently of each other. Calculate the probability of obtaining:an AaBB cc zygote from a cross of individuals that are aaBBcc x AAbbCC1 x impossible x impossiblean ABC phenotype from a cross of individuals that are AaBbCC x AaBbcc3/4 x 3/4 x 1 = 9/16an abc phenotype from a cross of individual that are AaBbCc x aaBbcc
½ x ¼ x ½ =1/16
an
AaBBCc
zygote from a cross of individuals that are
AaBbCc
x
AaBbCc
½ x ¼ x ½ = 1/16
Slide73In chickens the white plumage of the leghorn breed is dominant over coloured plumage, feathered shanks are dominant over clean shanks, and pea comb is dominant over single comb. Each of the gene pairs segregates independently.If a homozygous white, feathered, pea-combed chicken, is crossed with a homozygous coloured, clean, single-comb chicken, and the F1 are allowed to interbreed, what proportion of the birds in the F2 will produce only white, feathered, pea-comb progeny if mated to coloured, clean shanked, single combed birds? W white w colouredF feathered shanks f clean shanksP pea-comb p single-combP generation: WWFFPP x ww ff pp
F1 generation:
Ww
Ff Pp
The question is: the proportion of the birds in the F2 that will produce only white, feathered, pea-combed progeny if mated to coloured, clean-shanked, single-combed birds(x).
X=
ww
ff pp
Slide74W white w colouredF feathered f cleanP pea-comb p single-combP generation: WWFFPP x ww ff ppF1 generation: Ww Ff Pp The question is: the proportion of the birds in the F2 that will produce only white, feathered, pea-combed progeny if mated to coloured, clean-shanked, single-combed birds(x).i= ww ff ppF3 phenotype = W F P
F1 generation:
Ww
Ff Pp
F2 generation: ……?????????…. X
i
=
ww
ff pp
F3 phenotype = W F P 100%
Slide75W white w colouredF feathered f cleanP pea-comb p single-combThe question is: the proportion of the birds in the F2 that will produce only white, feathered, pea-combed progeny if mated to coloured, clean-shanked, single-combed birds(x).F1 generation: Ww Ff Pp F2 generation:
WW FF PP
X
i
=
ww
ff pp
F3 phenotype = W F P 100%
F1
W
w
W
WW
Ww
w
Ww
ww
¼ probability
F1
F
f
F
FF
Ff
f
Ff
ff
¼ probability
F1
P
p
P
PP
Pp
p
Pp
pp
¼ probability
¼ x ¼ x ¼ = 1/64
Slide76PhenotypesGenotypes
Slide77Two homozygous strains of corn are hybridized. They are distinguished by six different pairs of genes, all of which assort independently and produce an independent phenotypic effect. The F1 hybrid is selfed to give an F2.What is the number of possible genotypes in the F2?b. How many of these genotypes will be homozygous at all six gene loci?c. If all gene pairs act in a dominant-recessive fashion, what proportion of the F2 will be homozygous for all dominants?d. What proportion of the F2 will show all dominant phenotypes?3n n= number of genes considered36 = 729b. How many of these genotypes will be homozygous at all six gene loci?
2 AA BB CC DD EE FF or aa bb cc dd
ee
ff
a. What is the number of possible genotypes in the F2?
Slide78c. If all gene pairs act in a dominant-recessive fashion, what proportion of the F2 will be homozygous for all dominants?AA BB CC DD EE FF ¼ x ¼ x ¼ x ¼ x ¼ x ¼ = 1/4096number of possible gametes 26 = 64 individuals in the progeny =64x64= 4096d. What proportion of the F2 will show all dominant phenotypes?¾ x ¾ x ¾ x ¾ x ¾ x ¾ = 729/4096