1. MUTANTS in MICROORGANISMS

2. The 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

3. Minimum media containing
A 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
Arabinose
n. 1 +
n. 2 -
n. 3
n. 4 wt
Lac-
Gal- Lac-
Ara-

4. Minimum media+ glucose and…
A 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-

5. A 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
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?
30°C grow
42°C not grow

6. Medium with streptomycin
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)?
Medium with streptomycin
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?
Synthesize methionine
S1 S2 methionine
a b
Genes: met met2

7. Assign the phenotypes at the mutants ……………..
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-
No because the pathways of galactose degradation is composed by different genes. If only one of the genes is inactive the galactose is not metabolized.

8. FILE 8: POINT MUTATIONS

9. GENETIC CODE

10. 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 T T T to T T A
AAA A A T
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 by UUU will be substitute with Leucine (UUA).
This is a Missense mutation. The effect on the protein function is not predictable.

11. 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…
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).

12. 5 10 15 20 25 30 35 5' ATTCGATGGG A TGGCAG T G C CAAAGTGGTGATGGC 3'3.
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'
ATTCGATGGG A
TGGCAG T G C
CAAAGTGGTGATGGC 3'
TAAGCTACCC
ACCGTC
GTTTCACCACTAC CG
5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC
Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly

13. 5' ATTCGATGGG A TGGCAG T G C CAAAGTGGTGATGGC 3' TAAGCTACCC ACCGTC
GTTTCACCACTAC CG
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.

14. 5' ATTCGATGGG A TGGCAG T G C CAAAGTGGTGATGGC 3' TAAGCTACCC ACCGTC
GTTTCACCACTAC CG
- 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 UGG (Trp)
MIS-SENSE MUTATION: protein with one different aminoacid
mRNA: switch between C to A
From UGC (Cys) to UGA (STOP)
NON SENSE MUTATION thus a truncated protein

15. 5' ATTCGATGGG A TGGCAG T G C CAAAGTGGTGATGGC 3' TAAGCTACCC ACCGTC
GTTTCACCACTAC CG
- 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.

16. 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 suppression-suppressor 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’

17. This is a polypeptidic sequence of a protein
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 amino acid 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 Py=C o T
Sequence of mutant 1: AUG – AUA – UUPy – ACN -

18. 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-PyAC-
With N=U/C
The third codon AUU = Ile
The fourth codon will be UAC = Tyr

19. 5-
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 substitution
First base G could switch to C or A
GGU->CGU; GGC->CGC; GGA->CGA; GGG->CGG;
GGA->AGA; GGG->AGG

20. In the Escherichia coli metA gene a base substitution occurred
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.

21. Structure with loops corresponding to introns.
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.

22. 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?
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.

23. FILE 9

24. Pairing of homologous chromosomes during meiosis
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
GAMETES 14 21
14-21 14 21
14-21
Pairing of homologous chromosomes during meiosis 14 21
14-21
Karyotype

25. 2 chromosomes 14, DOWN 3 chromosomes 21 SYNDROME
Parent with translocation
Normal Parent
Normal Gamete
Gametes (First parent) 14 21
14-21
2 chromosomes 14, DOWN chromosomes 21 SYNDROME
21 Trisomy : Down’s Syndrome
2 chromosomes 14, chromosomes 21
NOT VITAL
21 Monosomy : not compatible with life 14 21
14-21
3 chromosomes 14, chromosomes 21
NOT VITAL
14 Trisomy : not compatible with life
1 chromosome 14, chromosomes 21
NOT VITAL
Monosomy 14 n:ot compatible with life
2 chromosomes 14, chromosomes 21
CARRIER
TRASLOCATION 14 21
14-21
Zigote healthy carrier: normal phenotype
2 chromosomes 14, chromosomes 21
NORMAL
Normal Zigote: normal phenotype

26. 1/6+1/6+1/6: ZIGOTEs are not compatible with life
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 are not compatible with life
1/ / /6

27. 2-
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 21
It is caused by the NONDISJUNCTION of chromosome 21 during meiosis (homologous chromosomes or sister chromatids).

28. A man carries a heterozygous paracentric inversion.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
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

29. Does the presence of the mutation change the fertility of this man?
A B C D
a b c d g f e G F E h H
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
Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC FRAGMENT CHROMOSOME (lost).
No, if inversion is not extended and is limited to a small part of chromosome.

30. 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

31. From the cross between gamete n + gamete 2n. Chromosomes A, B and C
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 = ABC
individual 3n=AAABBBCCC
In a triploid cell which gametes are produced?
AA (1/2)
A (1/2)
BB (1/2)
B (1/2)
CC (1/2)
C (1/2)
CC(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)
2/8 are gametes that could generate an individual 6/8 are not compatible with life

32. We hypotize that both species 2n = 26 7-
Asiatic cotton and American wild cotton have both 26 chromosomes. The American cultivated cotton, that is derived from the previous species by alloploidy, has 52 chromosomes. Explain, with a scheme, how it originates.
Alloploidy=A hybrid individual having two or more sets of chromosomes derived from two different species.
We hypotize that both species 2n = 26
ASIATIC COTTON (A) = 13 chromosomes
AMERICAN COTTON (B) = 13 chromosomes
If in the hybrid we have 26 chromosomes from two different species the plant is vital but it is sterile.
If in the hybrid a doubling of chromosomes occurs, we have an allopolyploid that is fertile because each chromosome has its homologous.
2 A + 2 B = = 52 Chromosomes

33. FILE 10

34. 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
OPERATOR
DNA region where RNA polymerase sits to start transcription.
DNA locus where the repressor could bind to stop transcription.
REPRESSOR
STRUCTURAL GENES
gene that encodes for a protein that negatively regulates transcription.
Genes that are usefull for a cellular function; for example metabolism of lactose.

35. 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.
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)

36. 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. 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 lacOc (cis DOMINANT)
Cis indicates that the mutation acts on the nearest genes.

37. 4.
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 - c
Genotype 1
Genotype 2
Genotype 3
Genotype 4
Genotypes 3 and 4 -> constitutive transcription of Lac operon (no repression)

38. 4.
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 - c
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

39. To demonstrate it we construct an heterozygote (diploid) with:
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 (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 lacO
Phenotype in absence of induction:
mutation is cis-dominant -> no b-gal activity
mutation is trans-dominant -> b-gal activity
lacOc
lacZ-
lacO+
lacZ+
IPTG is a synthetic and structural analogue of allolactose
Genotypes
Genotypes
BGalactosidase

40. 6.
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+ /LacIS
LacI+ 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 unable the protein to interact with lactose (constitutive repression).
LacIS repressor is always bound on the Operator repressing the transcription of Operon Lac : the operon Lac genes are not transcribed and expressed
In 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. (dominant mutation)
The wt bacteria (control) express the structural genes and grow on Lactose.

41. 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?
I complement them: I produce bacteria partial diploid carrying both mutations: one on a plasmid with all the operon Trp, the other on the chromosome.
To analyse the phenotype I plate them on a medium without Tryptophan
CASE 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.