1 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 ID: 915397
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Slide1
GENETIC CODE
Slide2FILE 8: POINT MUTATIONS
Slide31.
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
P
henylalanine
encodes
for UUU
will
be
substitute
with Leucine (UUA).
This
is
a
Missense
mutation
. The
effect
on the
protein
function
is
not
predictable
.
Slide42.
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 probably will not be active. (
FrameShift
Mutation
).
Slide53.
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
Slide65'
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 CGC
(Arg) that causes a MIS-SENSE mutation.
Slide75'
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)
MIS-SENSE 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
Slide85'
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.
The frame-shift creates a stop codon.
Slide9+1
Met-Gly-
aaX-Glu
-Cys-Gln-Ser-Gly-Asp-Gly
d) How can we abolish the insertion mutation in position 11? AUG-GGA-
NUG-GAG-UGC-CAA-AGU-GGU-GAU-GGCThe
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’ TAAGCTACCCTN
ACCTCACGGTTTCACCACTACCG 5’
Slide10This 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 aminoacid is
different Arg could be encoded by CGU, CGC, CGA, CGG, AGA, AGGIle
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 – AC
NSequence of mutant 1: AUG – AUA – UUPy – ACN - It
is probable that the codon encoding Arg could be AGA, and a
mutation occoured orginating AUA
Slide11WT :
Met-Arg-Phe-Thr
…
Mutant 1:
Met-Ile-Phe-Thr…Mutant 2: Met-Ser-Ile-TyrWe 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 –AG
N–AUU-PyAC-
With N=U/CThe third codon AUU = IleThe fourth codon will be UAC = Tyr
Slide125-
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 GGGArg Codons : CGU CGC CGA CGG AGA AGGWe can hypothesize a single base substitutionFirst base G could switch to C or AGGU->CGU; GGC->CGC; GGA->CGA; GGG->CGG;GGA->AGA; G
GG->AGG
Slide13In 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.
Slide14Exon
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.
Slide15if
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.
Slide16FILE 9
Slide17A 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
Slide1814
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, 3 chromosomes 21Parent with translocation
Normal Parent Normal Gamete Gametes (First parent)
Gametes (First parent) 21 Trisomy : Down’s Syndrome21 Monosomy
: not compatible with life14 Trisomy : not compatible with lifeMonosomy 14 n:ot compatible with life
Zigote healthy carrier: normal phenotypeNormal Zigote: normal phenotype
Slide191/6+1/6+1/6:
ZIGOTEs
not compatible with
life1/6 1/6 1/6Which will be the consequences on the progeny, if this man has a child together with a normal woman?
Slide202-
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 incorrectlyThe gametes either are missing or have an extra chromosome 21It is caused by the NONDISJUNCTION of chromosome 21 during meiosis.
Slide213.
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
.
Verranno prodotti i GAMETI PARENTALI ma mancheranno i gameti che hanno subito un evento di ricombinazione all’interno della regione invertita.
A B C D
E F G H
a b c d
g f e h
Parental
gametes
Slide22Does
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
duplication
is
not
extended
.
RECOMBINATION
(CROSSING-OVER)
A B C D
E f g
h
e F G H
d c b a
RECOMBINANT GAMETES
Slide234.
Deletion of a small region on Y chromosome in humans can prevent the individual development as a male. How can you explain this result?
The deletion is on a locus of Y chromosome where the SRY gene (
Sex determining Region Y) is located. It encodes for the TDF, Testis Determining Factor.Missing of this gene prevent the development as a male, thus the individual develops as a FEMALE
Slide24AA (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 a gamete n + gamete 2n. C
hromosomes A, B e C gamete 2n= AABBCCgamete n = ABCindividual 3n=AAABBBCCC
Slide257-
Asiatic
cotton and American cotton have both 26 chromosomes. The cultivated cotton, that is derived from the previous species by
alloploydia, has 52 chromosomes. Explain, with a scheme, how it originates. We hypotize that both species 2n = 26ASIATIC COTTON (A) = 13 chromosomesAMERICAN COTTON (B) = 13 chromosomesIf in the hybrid a doubling of
chromosomes occours, we have an alloploid that is fertile
because
each
chromosome
has
its
homologous
.
2 A + 2 B = 26 + 26 = 52 Chromosomes
Slide26FILE 10
Slide271.
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.
PROMOTEROPERATORSTRUCTURAL GENESREPRESSORDNA site 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.
Slide282.
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 costitutive 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, LacI
s)
Slide293.
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.Thus, we have the constitutive expression of the genes. The mutant in operator constitutive lacOc
(cis DOMINANT)
Slide304.
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
)
Slide314.
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)
Slide32CIS
dominant
mutation: it expresses the dominant phenotype
but it affects only the expression of genes on the same DNA molecule where the mutation occurs. LacOc, affects only neighbouring genes (plasmid)We construct an heterozygote with:The mutation lacZ- located
in cis to lacOc (no production of b-galattosidase)The gene lacZ+
in trans (
plasmid
)
Phenotype
in
absence
of
induction
:
mutation
is
cis-dominant -> no b-gal activity
mutation is trans-dominant -> b-gal activity
5. What does it mean that the Oc
mutation is dominant in cis? How can I demonstrate it?
Slide33I
complement
them: I produce bacteria carrying both
mutations one on a plasmid, the other on the chromosome.7. Two bacteria have a Trp- phenotype, cioe?The Trp- bacteria are unable to synthesize TryptophanHow do I verify whether the two mutations are in the
same gene?To analyze the phenotype I
plate
them
on a medium
without
Tryptophan
CASE 1:
If
bacterias
grow, the two mutations
complement each other, because they affect two different genes
.CASE 2: If bacterias don’t grow, the
two mutations do not complement each other, because they affect the
same gene.