Transcript Lac A

GENETIC CODE
FILE 8:
POINT MUTATIONS
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?
To switch from UUU to UUA is necessary a mutation in the DNA
From
TTT
to
TTA
AAA
AAT
This 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 for UUU will be
substitute with Leucine (UUA).
This is a Missense mutation. The effect on the protein
function is not predictable.
2.
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…
The deletion causes a frameshift of the codon code. All the aminoacids after will
be different.
Compare the two previous mutations:
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).
3.
A nucleotide sequence below
5
5'
3'
10
15
20
25
30
35
ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC
TAAGCTACCCTACCGTCACGGTTTCACCACTACCG
3'
5'
a) Knowing that the transcription of this sequence is from left to right (5’->3’),
write the resulting mRNA sequence:
5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC
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.
Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly
5'
3'
ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC
TAAGCTACCCTACCGTCACG GTTTCACCACTACCG
3'
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.
5'
3'
ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC
TAAGCTACCCTACCGTCACG GTTTCACCACTACCG
3'
5'
- a transversion of the CG base pair in position 20;
Case 1: from CG to GC
mRNA: switch between C to G
From UGC (Cys) to UGG (Trp)
MIS-SENSE MUTATION: protein with one
different aminoacid
Case 2: from CG to AT
mRNA: switch between C to A
From UGC (Cys) to UGA (STOP)
NON SENSE MUTATION thus a truncated
protein
5'
3'
ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC
TAAGCTACCCTACCGTCACGGTTTCACCACTACCG
3'
5'
- an insertion of a base pair after pair 11 (AT)
+1
AUG GGA NUG 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.
d) How can we abolish the insertion mutation in position 11?
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 suppressionsuppressor mutation)
+1
AUG-GGA-NUG-GAG-UGC-CAA-AGU-GGU-GAU-GGC
Met-Gly-aaX-Glu-Cys-Gln-Ser-Gly-Asp-Gly
The second mutation restore the correct reading frame.
Only the aminoacids located between the two mutations will be different.
DNA with insertion and subsequent suppressor mutation:
5’ ATTCGATGGGANTGGAGTGCCAAAGTGGTGATGGC 3’
3’ TAAGCTACCCTNACCTCACGGTTTCACCACTACCG 5’
This 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-Tyr
We compare mutant 1 with the wt: the second aminoacid is different
Arg could be encoded by CGU, CGC, CGA, CGG, AGA, AGG
Ile could be encoded by AUU, AUC, AUA
It is probable that the codon encoding Arg could be AGA, and a mutation
occoured orginating 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
Sequence of mutant 1: AUG – AUA – UUPy – ACN -
WT
: Met-Arg-Phe-Thr…
Mutant 1: Met-Ile-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-PyACWith N=U/C
The third codon AUU = Ile
The fourth codon will be UAC = Tyr
5An 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
We can hypothesize a single base substitution
First base G could switch to C or A
GGU->CGU; GGC->CGC; GGA->CGA; GGG->CGG;
GGA->AGA; GGG->AGG
:
CGU
CGC
CGA
CGG
AGA
AGG
In 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.
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?
Exon
INTRON
mRNA
Structure with loops corresponding to introns.
if a deletion of a base pair occurs in the middle of the second intron, 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.
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 have a frame-shift effect or a truncated protein.
FILE 9
A 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
Karyotype
21
Pairing of
homologous
chromosomes
during meiosis
14-21 21
14
GAMETES
14 14-21
14-21
Parent with translocation
Normal Parent
Normal Gamete
Gametes (First parent)
Gametes (First parent)
14-21 21
14 14-21
14
21 Trisomy : Down’s Syndrome
2 chromosomes 14,
3 chromosomes 21
21 Monosomy : not compatible with life
2 chromosomes 14,
1 chromosomes 21
14 Trisomy : not compatible with life
3 chromosomes 14,
2 chromosomes 21
Monosomy 14 n:ot compatible with life
1 chromosome 14,
2 chromosomes 21
21
Zigote healthy carrier: normal phenotype
14 21
14-21
Normal Zigote: normal phenotype
2 chromosomes 14,
2 chromosomes 21
2 chromosomes 14,
2 chromosomes 21
Which will be the consequences on the progeny, if this man has a child together
with a normal woman?
1/6+1/6+1/6: ZIGOTEs not
compatible with life
1/6
1/6
1/6
2In 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 21
- It is caused by the NONDISJUNCTION of chromosome 21 during meiosis.
3.
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
F
E
ABCD
Draw only one cromatide for each
chromosome.
f
e
g
G
H
h
abcd
Which gametes are produced (in particular which gametes are missing)? Explain
why.
Parental gametes
A B C D
E
F
G
H
a b c d
g
f
e
h
Verranno prodotti i GAMETI PARENTALI ma mancheranno i gameti che hanno
subito un evento di ricombinazione all’interno della regione invertita.
F
RECOMBINATION
(CROSSING-OVER)
E
ABCD
f
e
g
G
H
h
abcd
A B C D
RECOMBINANT GAMETES
h
E
f
g
e
F
G
d c b a
H
Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC
FRAGMENT CHROMOSOME (lost).
Does the presence of the mutation change the fertility of this man?
No, if duplication is not extended.
4.
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
6how 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.
Chromosomes A, B e C
gamete 2n= AABBCC
gamete n = ABC
individual 3n=AAABBBCCC
In a triploid cell which gametes are produced?
BB (1/2)
CC (1/2)
C (1/2)
AABBCC (1/8)
AABBC (1/8)
B (1/2)
CC (1/2)
C (1/2)
AABCC (1/8)
AABC (1/8)
BB (1/2)
CC (1/2)
C (1/2)
ABBCC (1/8)
ABBC (1/8)
B (1/2)
CC(1/2)
C (1/2)
ABCC (1/8)
ABC (1/8)
AA (1/2)
A (1/2)
2/8 are gametes that could generate an individual 6/8 are not compatible with life
7Asiatic 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 = 26
ASIATIC COTTON (A) = 13 chromosomes
AMERICAN COTTON (B) = 13 chromosomes
If 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
FILE 10
1.
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
DNA site where RNA
polymerase sits to start
transcription.
REPRESSOR
gene that encodes for a protein that
negatively regulates transcription.
OPERATOR
DNA locus where the repressor could bind
to stop transcription.
STRUCTURAL GENES
Genes that are usefull for a cellular
function; for example metabolism of
lactose.
2. What would be the result of a base substitution that inactivates the following genes:
LacZ-
Since the gene encodes for b-galattosidase enzyme, a mutation in this gene
probably inactivates the function of the gene.
LacIThis 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, missense), 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, LacIs)
3.
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)
4.
Which of the following genotypes will be able to produce β-galactosidase and/or
permease in the presence of lactose?
bgal perm.
Genotype 1 Lac I+
Lac P+
Lac O+
Lac Z+
Lac Y+
Lac A+
+ …..
+
……
Genotype 2 Lac I+
Lac P+
Lac O+
Lac Z+
Lac Y-
Lac A+
+ …..
……
Genotype 3 Lac I-
Lac P+
Lac O+
Lac Z+
Lac Y+
Lac A+
+ ..….
+
……
Genotype 4 Lac I+
Lac P+
Lac Oc
Lac Z+
Lac Y+
Lac A+
+ ……
+
……
Genotypes 3 and 4 -> constitutive transcription of Lac operon (no repression)
4.
If the lactose is not present, in which mutants the expression of genes change?
bgal perm.
Genotype 1 Lac I+
Lac P+
Lac O+
Lac Z+
Lac Y+
Lac A+
- -…..
……
Genotype 2 Lac I+
Lac P+
Lac O+
Lac Z+
Lac Y-
Lac A+
- …..
……
Genotype 3 Lac I-
Lac P+
Lac O+
Lac Z+
Lac Y+
Lac A+
+ ..….
+
……
Genotype 4 Lac I+
Lac P+
Lac Oc
Lac Z+
Lac Y+
Lac A+
+ ……
+
……
Genotypes 1 and 2 will not express the genes (REPRESSION)
5.
What does it mean that the Oc mutation is dominant in cis? How can I
demonstrate it?
CIS 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
7.
Two bacteria have a Trp- phenotype, cioe?
The Trp- bacteria are unable to synthesize Tryptophan
How do I verify whether the two mutations are in the same gene?
I complement them: I produce bacteria carrying both mutations one on a
plasmid, the other on the chromosome.
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.