A B ¾ x ¾ 916 A b ¾ x ¼ 316 a B ¼ x ¾ 316 a b ¼ x ¼ 116 GENE INTERACTIONS Consider two independent genes A and B two heterozygous individuals for both genes are crossed Complete the scheme below ID: 928074
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Slide1
GENE INTERACTIONS
Slide2¾ A
(AA, Aa)
A B: ¾ x ¾ = 9/16
A b: ¾ x ¼ = 3/16
a B: ¼ x ¾ = 3/16
a b: ¼ x ¼ = 1/16
GENE INTERACTIONS
Consider two independent genes, A and B; two heterozygous individuals for both genes are crossed. Complete the scheme below:
AaBb X AaBb
Phenotypes and frequencies for gene APhenotypes and frequencies for gene BPhenotypes and frequencies for genes Aand B
¼ a
(aa)
¾ B
(BB, Bb)
¼ b
(bb)
¾ B
(BB, Bb)
¼ b
(bb)
Slide3Because of the gene interaction phenomenon, some phenotypic classes are reduced in number, as two or more classes may display the same phenotype.
Which are the phenotypic classes of an F2 in which:
a) the dominant allele of a gene (A) masks the effect of the other gene (B)?
9/16 AB
3/16 Ab
3/16 aB
1/16 ab
9/16+3/16=
12/16
12:3:1b) the recessive homozygote (b) masks the effect of the other gene?
9/16 AB3/16 Ab
3/16 aB1/16 ab
3/16+1/16=
4/169:4:3
C)the phenotype due to the dominant allele of a gene is not observable from the phenotype due to the dominant allele of the other gene? Distinguish the two situations: when there is addictive effect and when there isn’t.
9/16 AB3/16 Ab
3/16 aB1/16 ab
3/16+3/16=
6/169:6:1
9/16 AB
3/16 Ab3/16 aB1/16 ab
9/16 + 3/16+3/16=
15/1615:1
Slide4The
A gene encodes the enzyme α, the independent gene
B
encodes the enzyme β; these enzymes are both involved in the following metabolic pathway for cyanide production. The recessive alleles,
a and b, encodes for inactive enzymes.
a) Identify the genotypes of P and F1 individuals and determine which phenotypic classes are expected in F2 and with which frequencies.
a
b
S1 S2
cyanide
P
F1
Not producing plant X not producing plant
Producing plants
X
The parental plants don’t produce cyanide but ,after the cross, F1 plants produce cyanide. In parental plants one of two genes is homozygous recessive (inactive gene), the other gene is homozygous dominant (active enzyme)
Since the F1 plants are producing plants, the gene homozygous recessive are different between the parents.
(P1) AA bb X (P2) aa BB(F1) Aa Bb
Slide5b)
b) what’s the name of genes that interact in the way showed here?
c)
which phenotypic classes are expected from the backcross of an F1 plant with the recessive homozygote?
Complementary Genes
, these genes act on the same metabolic chain.
(F1) Aa Bb x Aa Bb
(F2) 9:3:3:1
9/16 AB3/16 Ab
3/16 aB1/16 ab9/16 Producing plants
3/16 + 3/16 + 1/16= 7/16 Not producing plants Aa Bb x aa bb
1/4 AB1/4 Ab
1/4 aB1/4 ab
1/4 Producing Plants
¼ + ¼ + ¼ = ¾ Not Producing plants
Slide6The two independent genes P and C control two different consecutive steps in the same metabolic pathway.
a) PPcc
P
C
S1 S2
Yellow fruit
The dominant alleles
P
and
C
encode for active enzymes, while the recessive alleles,
p
and
c
, encode for inactive enzymes. Which phenotype is expected for individuals with the following genotype (intermediate substrates are colourless):This plant produces the enzyme encoded by gene P, but not produces the other one (gene C). Metabolic chain is blocked at step S2
. Phenotype: white melonb) PpCcThe plant has both enzymes thus the metabolic chain works until the end.
Phenotype: yellow melon. c) ppCC
The plant hasn’t the enzyme encoded by gene P. It produces the enzyme encoded by the gene C but it doesn’t convert S1 in S2. Phenotype white melon.d) Ppcc
The plant produces the first enzyme but not the second one. Then the metabolic chain is blocked at S1 step. Phenotype white melon.
Slide7The two independent genes, A e E, that originate through duplication, regulate the same step in the metabolic pathway that leads to the synthesis of a pigment.
The colourless phenotype can be seen only in the recessive homozygote state for both genes.
Aa Ee x Aa Ee
9/16 AE
3/16 Ae
3/16 aE
1/16 ae
1/16 white fur
9/16 + 3/16 + 3/16= 15/16 brown fur
Allele A produces enzyme
a
that converts S1 in pigment; allele a does not produce pigment.
Allele E produces enzyme e that converts S1 in pigment; allele e does not produce pigment.
Enzima a
S1 pigment Enzima e
Which phenotypic ratio is expected from the cross between two di-hybrids?
Slide8Two pure lines of pepper are crossed: first line with Red fruits and the second one with orange fruits. The plants of F1 produce only red peppers, but crossing two individuals of F1 we obtain:
192 RED FRUITS, 47 ORANGE FRUITS, 14 WHITE FRUITS
Pure lines = they pass the same alleles to all
offsprings
and so they are homozygous
We know that the colour fruit is controlled by 2 genes (A and B), which kind of interaction is present? Determine the genotypes and hypothesize a possible action on metabolic chain.
(P1) RED FRUITS X (P2) ORANGE FRUITS
(F1) RED FRUITSF2:192 red fruits 47 orange fruits
14 white fruits When the plant has an allele of A the phenotype related to gene B is not shown. Phenotypes AB and Ab produce red fruitsWhen A is homozygous recessive, the plant show the B phenotype: if it has dominant allele B the fruits will be orange, if it is homozygous bb the fruits will be white.AA bbaa BB
Aa Bb9/16 AB + 3/16 Ab
3/16 aB1/16 ab192+47+14= 253 (:16= 15.8)
The dominant allele A is epistatic on gene B
Slide9Phenotypes
Xo
H
Xa
(
Xo
–
Xa
)2: Xac2Red
19212/16189,755,06189,750,03Orange473/1647,430,1847,430,01
White141/1615,813,2715,81
0,21Total25316/16253
0,25We have to compare expected individuals (epistasis hypothesis) with observed individuals.
Degrees of freedom = 3-1 = 2c2 = 0,25P =80 - 90%Conclusion:
Hypothesis accepted
Slide10Slide11FILE 8:
POINT MUTATIONS
Slide12GENETIC CODE
Slide131.
In a mRNA sequence (wt) there is a triplet UUU. After a mutation, the triplet changes in UUA. What kind of mutation happened and which effects have the mutation on the protein encoded by the gene?
To switch from UUU to UUA is necessary a mutation in the DNA
From
T T
T to T T A
AAA A A
TThis is a Transversion: it refers to the substitution of a purine for a
pyrimidine or vice versa, in deoxyribonucleic acid (DNA).
The aminoacid Phenylalanine encodes by UUU will be substitute with Leucine (UUA). This is a Missense mutation. The effect on the protein function is not predictable.
Slide142.
This is a sequence of wt mRNA
5’- AUG AGA CCC ACC….
What kind of effects has a mutation the substitute the fifth base from G to U?
5’– AUG AUA CCC ACC…
In the second triplet the mutation changes the codon and the aminoacid encoded:
Arg
(Encoded by
AGA) to Ile (encoded by AUA).What kind of effects has a deletion of the sixth base? 5’– AUG AUA CCC ACC…
Compare the two previous mutations:The deletion causes a frameshift of the codon code. All the aminoacids after will be different.
First case: Probably the protein will be active (Missense mutation)Second case: the protein is totally different from the original protein and it will probably not be active. (FrameShift Mutation).
Slide153.
A nucleotide sequence below
Knowing that the transcription of this sequence is from left to right
(5’->3’)
, write the resulting mRNA sequence:
b) Knowing that this mRNA sequence contains the translation start codon, identify the starting codon and indicate the amino acid sequence of the resulting peptide.
5 10 15 20 25 30 35
5'
ATTCGATGGG
A
TGGCAG
T
G
C
CAAAGTGGTGATGGC
3'
3'
TAAGCTACCC
T
ACCGTC
A
C
G
GTTTCACCACTAC
CG
5'
5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC
Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly
Slide165'
ATTCGATGGG
A
TGGCAG
T
G
C
CAAAGTGGTGATGGC
3'
3'
TAAGCTACCC
T
ACCGTC
A
C
G
GTTTCACCACTAC
CG
5'
C) Find which consequences will have on the amino acid sequence:
- a transition of the T base pair in position 18;
In the DNA sequence: switch from TA to CG;
In the mRNA sequence there is a switch between U and C thus between the triplet UGC (Cys) to
C
GC (Arg) that causes a MIS-SENSE mutation.
Slide175'
ATTCGATGGG
A
TGGCAG
T
G
C
CAAAGTGGTGATGGC
3'
3'
TAAGCTACCC
T
ACCGTC
A
C
G
GTTTCACCACTAC
CG
5'
- a transversion of the CG base pair in position 20;
Case 1: from CG to GC
Case 2: from CG to AT
mRNA: switch between C to G
From UGC (
Cys
) to UG
G
(
Trp
)
MISSENSE MUTATION: protein with one different
aminoacid
mRNA: switch between C to A
From UGC (Cys) to UG
A
(STOP)
NON SENSE MUTATION
thus a truncated protein
Slide185'
ATTCGATGGG
A
TGGCAG
T
G
C
CAAAGTGGTGATGGC
3'
3'
TAAGCTACCC
T
ACCGTC
A
C
G
GTTTCACCACTAC
CG
5'
- an insertion of a base pair after pair 11 (AT)
+1
AUG GGA
N
UG GCA GUG CCA AAG UGG UGA UGG C
Met-Gly-
aaX-Ala-Val-Pro-Lys-Trp-
Stop
After the insertion all the
aminoacids
will be different: frame-shift mutation.
The frame-shift creates a stop codon.
Slide19+1
Met-Gly-
aaX-Glu
-Cys-Gln-Ser-Gly-Asp-Gly
d) How can we abolish the insertion mutation in position 11?
AUG-GGA-
N
UG-GAG-UGC-CAA-AGU-GGU-GAU-GGC
The second mutation restore the correct reading frame.
Only the aminoacids located between the two mutations will be different.
If in a position near the first insertion a deletion happens, we can restore the correct frame of aminoacids. For example in position 15 (intragenic suppression-suppressor mutation)
DNA with insertion and subsequent suppressor mutation:5’ ATTCGATGGGANTGGAGTGCCAAAGTGGTGATGGC 3’
3’ TAAGCTACCCTNACCTCACGGTTTCACCACTACCG 5’
Slide20This is a
polypeptidic
sequence of a protein. Wild type and mutant sequences are compared. What type of mutations did happen?
WT :
Met-
Arg-Phe-Thr
…Mutant 1: Met-Ile-
Phe-
Thr…Mutant 2: Met-Ser-Ile-TyrWe compare mutant 1 with the wt: the second amino acid is different Arg could be encoded by CGU, CGC, CGA, CGG, AGA, AGG
Ile could be encoded by AUU, AUC, AUA Phe could be encoded by UUU/UUC and Thr by ACU/ACC/ACA/ACG, The possible DNA sequence will be:Sequence of wt: AUG – AGA – UUPy – ACN Py=C o TSequence of mutant 1: AUG – AUA – UUPy – ACN -
It is probable that the codon encoding Arg could be AGA, and a mutation occurred originating AUA
Slide21WT :
Met-
Arg
-
Phe-Thr
…
Mutant 2: Met-Ser-Ile-Tyr
We compare mutant 2 with wt: all the amicoacids after the Methionine are different.
A frameshift mutation caused by an insertion. AUG – AGA – UUPy – ACN –Ser is encoded by UCN or AGU-AGC , We can hypotesize the sequence:AUG –AGN–AUU-PyAC-With N
=U/CThe third codon AUU = IleThe fourth codon will be UAC = Tyr
Slide225-
An Escherichia coli mutant auxotroph for tryptophan (Trp-) has an amino acid substitution in tryptophan-synthetase: Glycine at position 210 is replaced by an Arginine.
On the basis of the genetic code, find which kind of mutation (on the DNA) you think has caused the amino acid replacement
Gly Codons : GGU
GGC
GGA GGG
Arg Codons : CGU CGC CGA
CGG AGA AGG
We can hypothesize a single base substitutionFirst base G could switch to C or AGGU->CGU; GGC->CGC; GGA->CGA;
GGG->CGG;GGA->AGA; GGG->AGG
Slide23In the Escherichia coli
metA
gene a base substitution occurred. Because of this mutation, in the mRNA a UAA codon is present inside the gene.
- Which consequence will this mutation have on protein synthesis?
The triplet UAA is a stop codon,
Thus the protein synthesis will be interrupted.
Slide24Exon
INTRON
mRNA
The primary transcript of chicken ovalbumin RNA is composed by 7 introns (white) and 8 exons (black):
If the Ovoalbumin DNA is isolated, denaturated and hybridized to its cytoplasmic mRNA, which kind of structure do we expect?
Structure with loops corresponding to introns.
Slide25if a deletion of a base pair occurs in the middle of the second intron, which will be the likely effects on the resulting polypeptide?
if a deletion of a base pair occurs in the middle of the first exon, which will be the likely effects on the resulting polypeptide?
We don’t have any effect if the mutation is not in the splicing site.
We have a frame-shift effect or a truncated protein.
Slide26FILE 8
Slide27A man has the chromosome 21 translocated on the 14. Draw the karyotype (only the chromosomes involved in the mutation).
What kind of gametes will be produced by this person?
Which will be the consequences on the progeny, if this man has a child together with a normal woman?
14
21
14-21
14
21
14-21
14
21
14-21
14
21
14-21
Pairing of homologous chromosomes during meiosis
Karyotype
GAMETES
Slide2814
21
14-21
2
chromosomes
14, 1
chromosomes
21
14
21
14-21
3
chromosomes
14, 2
chromosomes
21
1
chromosome
14, 2
chromosomes
21
14
21
14-21
2 chromosomes
14, 2 chromosomes
21
2 chromosomes
14, 2 chromosomes
21
2 chromosomes 14, DOWN 3 chromosomes 21 SYNDROME
Parent with translocation
Normal Parent
Normal Gamete
Gametes (First parent)
21 Trisomy : Down’s Syndrome
21 Monosomy : not compatible with life
14 Trisomy : not compatible with life
Monosomy 14 n:ot compatible with life
Zigote healthy carrier: normal phenotype
Normal Zigote: normal phenotype
NOT VITAL
NOT VITAL
NOT VITAL
CARRIER
TRASLOCATION
NORMAL
Slide291/6+1/6+1/6: ZIGOTEs are not compatible with life
1/6 1/6 1/6
Which will be the consequences on the progeny, if this man has a child together with a normal woman?
Slide302-
In humans trisomy of chromosome 21 is responsible of the Down syndrome.
- which gametes originated an affected person?
draw a scheme of the meiotic stages that can give rise to the mutated gamete and indicate the name of this process.
Chromosomes are distributed to gametes incorrectly
The gametes either are missing or have an extra chromosome 21It is caused by the NONDISJUNCTION of chromosome 21 during meiosis (homologous chromosomes or sister chromatids).
Slide313.
A man carries a heterozygous paracentric inversion.
A B C D
E F G H
a b c d
g f e h
Draw a scheme of homologous chromosomal pairing during meiosis
Which gametes are produced (in particular which gametes are missing)? Explain why.
A B C D
a b c d
g
f
e
G
F
E
h
H
Draw only one cromatide for each chromosome.
A B C D
E F G H
a b c d
g f e h
Parental gametes
Slide32Does the presence of the mutation change the fertility of this man?
Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC FRAGMENT CHROMOSOME (lost).
A B C D
a b c d
g
f
e
G
F
E
h
H
No, if inversion is not extended and is limited to a small part of chromosome.
RECOMBINATION
(CROSSING-OVER)
A B C D
E f g
h
e F G H
d c b a
RECOMBINANT GAMETES
We will have PARENTAL GAMETES but we don’t have vital recombinant gametes produced by a crossing over happens in the inverted region
Slide33AA (1/2)
A (1/2)
BB (1/2)
B (1/2)
BB (1/2)
B (1/2)
CC (1/2)C (1/2)
CC (1/2)C (1/2)CC (1/2)C (1/2)CC(1/2)C (1/2)
AABBCC (1/8)AABBC (1/8)AABCC (1/8)AABC (1/8)ABBCC (1/8)ABBC (1/8)ABCC (1/8)ABC (1/8)
In a triploid cell which gametes are produced?
2/8 are gametes that could generate
an individual 6/8 are not compatible with life
6 - how can a triploid organism originate?
Draw a scheme of meiosis process in a triploid cell with n = 3.
From the cross between gamete n + gamete 2n.
Chromosomes A, B and C
gamete 2n= AABBCC
gamete n = ABCindividual 3n=AAABBBCCC
Slide34MUTANTS in MICROORGANISMS
Slide35The reproduction of the bacterium
Escherichia coli
is achieved by binary fission, after his genome replication. Complete the following scheme representing the chromosome in various stages.
Bacteria has one chromosome and don’t have nucleus.
Mitosis is performed in order to add cells to a population
Slide36A prototroph bacterium can grow on a minimum media, composed by inorganic minerals and containing an
organic source of carbon
. Glucose is the most simple source of carbon.
Alternative carbon sources can be used by wild bacteria. Some mutants loose this ability
.The ability of bacteria to grow on different media is reported in the following table; identify their phenotypes.
STRAIN
Minimum media containing
PHENOTYPE
Glucose
Galactose
Lactose
Arabinosen. 1
++++
n. 2++
-+n.
3+--+
n. 4
+++-
wt
Lac-
Gal- Lac-Ara-
Slide37A prototroph bacterium can grow on a minimum media, because it can synthesize what it needs. Mutants that aren’t able to grow on a minimum medium (auxotrophs) can be analysed on media enriched with different kinds of molecules, to determine their phenotypes.
The ability of bacteria to grow on different media is reported in the following table; identify their phenotype
STRAIN
Complete Media
Minimum media+ glucose and…
Phenotype
- -
arginine
pyrimidine
adenine
biotin
n. 1
+
-
+---
n. 2+-
-+--
n. 3+-
--+-
n. 4+
----+
Arg
-Pyr-
Ade-Bio-
Slide38A mutant bacterium isn’t able to synthesize the prolin amino acid and to metabolize lactose. Indicate which molecules must be added to a minimum media to grow this mutant.
Minimum media + glucose and proline
30°C it grows
37°C it grows (optimal)
42°C it does not grow
An Escherichia coli mutant carries a temperature-sensitive mutation in the polymerase III gene that is necessary for DNA replication. At which temperature can this bacterium grow?
Slide39Medium with streptomycin
Synthesize methionine
The wild-type
Escherichia coli
can’t grow on streptomycin-enriched media. If in a culture of 10 millions cells, there are about 100 streptomycin-resistant mutant cells, how can I select them (i.e. grow only the resistant cells)?
Two
Escherichia coli mutants display both the Met- phenotype, i.e. they aren’t able to....
The two mutations are located on different genes. How can you explain this situation?
S1 S2 methionine a b
Genes: met1 met2
Slide40No because the pathways of galactose degradation is composed by different genes. If only one of the genes is inactive the galactose is not metabolized.
Two Escherichia coli mutants can’t grow on galactose medium if galactose is the unique source of Carbon.
Assign the phenotypes at the mutants
……………..
Are you sure that the mutations regard the same gene?
Gal
-
Slide41FILE 10
Slide421.
In Escherichia coli, the lac (lactose) operon, is made of the following genes and sites. Specify what is the function of the ones indicated below:
- promoting site
- operator site
- repressor gene - structural genes.
PROMOTER
OPERATOR
STRUCTURAL GENESREPRESSOR
DNA region where RNA polymerase sits to start transcription. DNA locus where the repressor could bind to stop transcription.gene that encodes for a protein that negatively regulates transcription.
Genes that are usefull for a cellular function; for example metabolism of lactose.
Slide432.
What would be the result of a base substitution that inactivates the following genes:
Since the gene encodes for
b
-galattosidase enzyme, a mutation in this gene probably inactivates the function of the gene.
LacZ
-
LacI-
This gene encodes for the repressor of lactose operon. The mutation will have different effects depending on the protein domain where it occours:a) If the mutation inactivates the protein (frame-shift, stop codon, mis-sense), we have the absence of the repressor and constitutive transcription of the structural genes (recessive mutation LacI-)
b) If the mutation alters the allosteric domain of the protein where the inducer binds, we have constitutive repression of the structural genes because the repressor is bound to its site and is not influenced by the presence of lactose. (dominant mutation, LacIs)
Slide443.
What would happen if a base deletion occurs in the operator region?
OPERATOR is the DNA locus bound by the repressor to stop transcription.
After the mutation in the operator, the repressor could be unable to recognize the locus. In fact the repressor (protein) is able to recognize the Operator (sequence).
Thus, we have the constitutive expression of the genes. The mutant in operator constitutive lacO
c
(cis DOMINANT)
Cis indicates that the mutation acts on the nearest genes.
Slide454.
Which of the following genotypes will be able to produce β-galactosidase and/or permease
in the presence of lactose
?
b
gal perm.
…… …..
…… …..
…… ..….
…… ……
Lac I
+
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
-
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
-
Lac A
+
Lac P
+
Lac I
-
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
-
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
c
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
c
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Genotype 1
Genotype 2
Genotype 3
Genotype 4
+ +
+ -
+ +
+ +
Genotypes 3 and 4 -> constitutive transcription of Lac operon (no repression)
Slide464.
If the
lactose is not present
, in which mutants the expression of genes change?
b
gal perm.
…… …..
…… …..
…… ..….
…… ……
Lac I
+
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
-
Lac A
+
Lac P
+
Lac I
+
Lac O
+
Lac Z
+
Lac Y
-
Lac A
+
Lac P
+
Lac I
-
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
-
Lac O
+
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
c
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Lac I
+
Lac O
c
Lac Z
+
Lac Y
+
Lac A
+
Lac P
+
Genotype 1
Genotype 2
Genotype 3
Genotype 4
- -
- -
+ +
+ +
Genotypes 1 and 2 will not express the genes (REPRESSION)
Genotypes 3 and 4 will express constitutively the enzymes
Slide47CIS dominant mutation: it expresses the dominant phenotype but it affects only the expression of genes on the same DNA molecule where the mutation occurs.
LacO
c
, affects only neighbouring genes (for example a plasmid)
To demonstrate it we construct an heterozygote (diploid) with:
The mutation lacZ- (
Bgal enzyme) located in cis to lacOc
no production of b-galattosidase
The gene lacZ+ in trans (on plasmid) with wt lacOPhenotype in absence of induction:mutation is cis-dominant -> no b-gal activitymutation is trans-dominant ->
b-gal activity5. What does it mean that the Oc mutation is dominant in cis? How can I demonstrate it? lacOc
lacZ-
lacO+lacZ+
lacIlacI
No lactose
Slide486.
What kind of phenotype will have a bacteria
LacI
+
O+ Z+ Y+
A+ carrying on a plasmid lacIs O
+ Z+ Y+ A
+ ? Which conditions can we use to analyse its phenotype?
The bacteria will be: LacI+ O+ Z+ Y+ A+ / LacIs O+ Z+ Y+
A+ heterozygote for gene LacI+ /LacISLacI+ encodes for the repressor that is able to bind lactose (induction and repression) and is able to recognise sequence of Operator (repression) LacIS encodes for a repressor with a mutation that unables the protein to interact with lactose (constitutive repression).LacIS
repressor is always bound to the Operator repressing the transcription of Operon Lac : the operon Lac genes are not transcribed and expressedIn order to analyse the phenotype I’ll grow the mutant in a condition where I can detect the expression of enzyme…with the presence of lactose (induction).We expect that the bacteria with the plasmid does not express the structural gene
it does not grow if the unique source of carbon is Lactose. (trans dominant mutation)The wt bacteria (control) express the structural genes and grow on Lactose.
lacI
lacZ
lacI
SlacZ+
lacO
lacYlacA
lacO
lacYlacA
lactose
Slide49I complement them: I produce bacteria partially diploid carrying both mutations: one on a plasmid with all the operon
Trp
, the other on the chromosome.
7.
Two bacteria have a
Trp- phenotype, then?
The Trp- bacteria are unable to synthesize Tryptophan. It needs Tryptophan to grow.I have to add Tryptophan in the medium.How can I verify whether the two mutations are in the same gene?
To analyse the phenotype I plate them on a medium
without TryptophanCASE 1: If bacteria grow, the two mutations complement each other, because they affect two different genes.CASE 2: If bacteria don’t grow, the two mutations do not complement each other, because they affect the same gene.
Slide50POPULATION GENETICS
The study of the allele frequencies in a population
The Hardy Weinberg equilibrium
“Allele and genotype frequencies
in a population tend to remain constant in the absence of disturbing influences”
-non-random mating-mutations-selection-limited population size-random genetic driftConditions for HW equilibrium are never met in nature.
The equationsALLELES GENOTYPESp + q = 1 p
2 + 2pq + q2 =1
p=frequency of allele A p2=frequency of AA q=frequency of allele a q2=frequency of aa 2pq=frequency of Aa
Slide51An Example
Assume a population in which 36% of the population are homozygous for a certain recessive allele a. Assume the population is at equilibrium.
What is the frequency of the recessive allele in this population?
36%
aaq
2=0.36q=f(a)= √ 0.36=0.6
What is the frequency of the dominant allele in this population?q =0.6p = f(A) =1-0.6=0.4What percentage of the population are homozygous for the dominant allele A?
p2=f(AA)=0.4 x 0.4=0.16 16%
What percentage of the population are heterozygous for the trait?2pq=f(Aa)=2 x 0.6 x 0.4 =0.48 48%Why do we have to start the problem with the percentage or numeber of the homozygous recessive in the population?Because only the recessive phenotype indicates only the homozygous recessive genotype…the dominant phenotype indicates heterozygous and homozygous
Slide520,1 0,4 0,5
In this case we know the number of heterozygous
We think about the number of
alleles
p =f (A) =
(2 x 100) + 400 = 0,3 2 x 1000q = f(a)=1 – 0,3 = 0,7
0,3 0,7
1 - Determine the gene and genotypic frequencies of the following populations:
Slide53p =
(2 x 300) + 200
= 0,4
2 x 1000q = 1 – 0,4 = 0,6
0,3 0,2 0,5
0,4 0,6
1 - Determine the gene and genotypic frequencies of the following populations:
Slide54p =
(2 x 700) + 200
= 0,8
2 x 1000q = 1 – 0,8 = 0,2
0,7 0,2 0,1
0,8 0,2
1 - Determine the gene and genotypic frequencies of the following populations:
(A(AA)+A(AA))+A(Aa)/(total alleles)
Slide552 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Population 1
p = 0,3 q = 0,7
f(AA) = p
2 = 0,09
f(Aa) = 2pq = 0,42f(aa) = q2 = 0,49
Genotipi
Xo
Xa(Xo – Xa)2 : Xa
AA100901,11Aa4004200,95aa500490
0,20
c2 = 2,26
Degree of Freedom = 2(p,q)-1= 1P = 0,10-0,25Population is in equilibrium
Slide56Population II
p = 0,4 q = 0,6
f(AA) = p
2 = 0,16
f(Aa) = 2pq = 0,48f(aa) = q2 = 0,36
Genotipi
Xo
Xa
(Xo – Xa)2 : XaAA300
160Aa200480aa500360
Expected and observed values are too different.
The population is not in equilibrium2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Slide57Population
III
p = 0,8 q = 0,2
f(AA) = p2 = 0,64
f(Aa) = 2pq = 0,32f(aa) = q2 = 0,04
Genotipi
Xo
Xa
(Xo – Xa)2 : XaAA700640
5,6Aa20032045aa1004090
c2 =
140,6GL = 2– 1 = 1P <0,01The population is
not at equilibrium2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it.
Slide583 -
The frequencies of
L
M and
LN alleles in a group of 200 black Americans were 0,8 e 0,2, respectively. Calculate the expected frequencies for individuals with M, N and MN blood group.
If
the population is at
equilibrium:Group M = LM LM
= p2 = 0,8 x 0,8 = 0,64 x 200 = 128Group MN = LM LN = 2pq = 2 x 0,8 x 0,2 = 0,32 x 200 = 64Group N = LN LN = q2 = 0,2 x 0,2 = 0,04 x 200 = 8
Slide59Population
I
f(aa) = 810/1000 = 0,81
Since
f(aa) = q2 f(a) = √0,81 = 0,9
f(A) 1 – 0,9 = 0,1
Population IIf(aa) = 360/1000 = 0,36f(a) = √0,36 = 0,6 f(A) 1 – 0,6 = 0,4
Population III
f(aa) = 490/1000 = 0,49f(a) = √0,49 = 0,7 f(A) 1 – 0,7 = 0,3 The individuals with phenotype A can be AA or
Aa.We are sure that individuals with phenotype a are: aaWe start from this information (aa) to figure out f(A)
Slide605 -
What is the expected frequency for dominant homozygous and heterozygous genotypes in a population at equilibrium in which the homozygous recessive genotype frequency is 0,09?
f(aa) = 0,09
f(a) = √0,09 = 0,3
f(A) = 1 – 0,3 = 0,7
f(AA) = 0,7 x 0,7 = 0,49f(Aa) = 2 x 0,7 x 0,3 = 0,42
Slide616 -
if in a population at equilibrium the frequency of Rh- phenotype is 0,0025, which is the expected frequency of heterozygous individuals?
f(Rh-) = 0,0025 = f(
dd
)
f(d) = √0,0025 = 0,05f(D) = 1 – 0,05 = 0,95f(Dd) = 2 x 0,95 x 0,05 = 0,095
Slide627
In Drosophila melanogaster the w recessive sex-linked allele, is responsible of the white colour of the eyes. In a population the frequency of this allele is 0,3. Which frequencies of male and female with white-eyes is expected if the population is at equilibrium?
The
frequency
of these two
alleles in the entire population are
𝑝 and 𝑞. Let's assume the locus is
present on the X chromosome. We
will assume that the allele frequency do not differ between males and females.
Male white eyes = w YFemale white eyes = w wFrequency:Males white eyes = f(w)=
q = 0,3Females white eyes = f(ww) = q2 = 0,3 x 0,3 = 0,09
If alleles are X-linked, females
may be homozygous or heterozygous, but males
carry only a single allele for each X-linked
locus. For X-linked alleles in females, the H-W
frequencies are the same as those for
autosomal loci. p2+2pq+q2=1In males,
however, the frequencies of the genotypes will be
p and q, the same as the
frequencies of the alleles in the population. p+q=1
Slide638 -
Daltonism is due to a sex-linked recessive allele. In a population the
daltonic
male frequency is 0,1; which is the one of
daltonic females?
Daltonic MALE = d Y (XY) hemizygote freq
dY = freq allele d
Daltonic FEMALE = d d (XX)
Frequency Daltonic Males = 0,1Frequency allele d in MALES = 0,1Daltonic FEMALES = f(d d) = 0,1 x 0,1 = 0,01Healthy carriers= 2*0.1*0.9=0.18
Slide64relative Fitness (W) 2/2 = 1 2/2 = 1 1/2= 0,5
The selection is against the phenotype homozygous recessive
Fitness involves the ability of organisms to survive and reproduce in the environment in which they find themselves . The consequence of this survival and reproduction is that organisms contribute genes to the next generation.
Fitness= Its lifetime reproductive success
absolute Fitness 200/100 = 2 400/200 = 2 100/100 = 1
Slide6510 -
A population is composed of: AA 250, Aa
500,
aa 250 individuals, when a selection factor acts against the
homozygote recessive and its fitness decreases to 0. The variation of genic frequency (selection effect) could be figured out taking into consideration the contribution of the gametes of different genotypes in the next generation.
Consider the previous population, try to figure out the variation of genetic frequency of a after a generation of selection.
1
1
0
250/1000
= 0,25
500
/1000 = 0,5
250
/1000 = 0,25
1 x 0,25 = 0,25
1 x
0,5 =
0,5
0,25 x 0 = 0
[2(250
) + 500]/2(1000) = 0,5
[2(0
) + 0,5] /2(0,25 + 0,5 + 0) = 0,33
q1 – q0 = 0,33 – 0,5 = -0,17
1000x2 total of alleles2x250 aa=250a+250a2(0.25+0.5) total of freq2x0 aa=0a+0a
As expected the frequency of a allele decreases
Slide6611 – In a population there is a selection against the
homozygous recessive
with fitness = 0 but the genotypic frequencies are:
AA 0,64;
Aa
0,32;
aa 0,04,
Try to figure out the variation of genetic frequency of a after a generation of selection
1 1 0 0,64 0,32 0,04
0,64 x 1 = 0,64 0,32 x 1 = 0,32 0,04 x 0 = 0 [0,32 + 2(0,04)]/2(0,64 + 0,32 + 0,04) = 0,20 [0,32 + 2(0)]/2(0,64 + 0,32 + 0) = 0,16
q1 – q0 = 0,16 – 0,20 = -0,04Previous population Dq = -0,17 (> of q0 (0,5))
This population q0 = 0,2 With the same selection coefficient, the effect of selection against homozygous recessive change on varying of genic frequency
Slide6712 – In a population the gene frequency of A is 0,3 and the frequency of a is 0,7, and there is a selection against the
heterozygous
(s =coefficient of selection=1-w= 0,4).
Try to figure out the variation of genetic frequency of a after a generation of selection
genotypes
AA
Aa
aa
fitness
Genotypic frequency
Gamete contribution
Frequency
of
q0
Frequency of
q1
D
q
1 1 – 0,4 = 0,6 1
p
2
= 0,3
2
= 0,09 2pq =
0,42
q
2
= 0,7
2
=
0,49
1 x 0,09 = 0,09 0,6 x 0,42 =
0,252
1 x 0,49 =
0,49
[
0,42
+ 2(
0.49
)]/2(0.09+0,42+0,49)=0,7
[
0,252
+ 2(
0,49
)]/2(0,09 + 0,252 + 0,49) = 0,74
q1 – q0 = 0,74 – 0,7 = 0,04
The recessive allele will be fixed and the dominant allele will be delete from the population
- What is the final result of this selection?
Slide68genotypes
AA
Aa
aa
fitness
Genotypic frequency
Gametic
contribution
Frequency
of
q0
Frequency of q1
D
q
13 – The population is composed of: 640 AA, 320 Aa and 40 aa,
The fitness of homozygous dominant is 0,8 and the fitness of homozygous recessive is 0,1.
Try to figure out the variation of genetic frequency of after a generation of selection
0,8 1 0,1
640/1000=0,64 320/1000=0,32 40/1000=0,04
0,64 x 0,8=0,51 0,32 x 1 = 0,32 0,04 x 0,1=0,004
[320+2(40)]/2(1000) = 0,2
[0,32+2(0,004)]/2(0,51+0,32+0,004) = 0,19
0,19 – 0,20 = -0,01