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GENETIC CODE

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?.

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GENETIC CODE

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  1. GENETIC CODE

  2. FILE 8: POINT MUTATIONS

  3. 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 isnecessary a mutation in the DNA From T T T to T T A AAA A A T Thisis a Transversion: itrefers to the substitution of a purine for a pyrimidine or vice versa, in deoxyribonucleic acid (DNA). The aminoacidPhenylalanineencodes for UUU will be substitute with Leucine (UUA). Thisis a Missensemutation. The effect on the proteinfunctionisnotpredictable.

  4. 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).

  5. 3. • A nucleotide sequence below • Knowingthat the transcription of thissequenceis from left to right (5’->3’), write the resultingmRNAsequence: • b) KnowingthatthismRNAsequencecontains the translationstart codon, identify the startingcodon and indicate the amino acid sequence of the resulting peptide. 5 10 15 20 25 30 35 5' ATTCGATGGG TGGCAG G CAAAGTGGTGATGGC 3' A T C 3' TAAGCTACCC ACCGTC C GTTTCACCACTAC CG 5' T A G 5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly

  6. C) Find which consequences will have on the amino acid sequence: - a transition of the T base pair in position 18; 5' ATTCGATGGG TGGCAG G CAAAGTGGTGATGGC 3' A T C T A G 3' TAAGCTACCC ACCGTC C GTTTCACCACTAC CG 5' In the DNA sequence:switch from TA to CG; In the mRNAsequencethereis a switchbetween U and C thusbetween the triplet UGC (Cys) to CGC (Arg) thatcauses a MIS-SENSE mutation.

  7. - a transversion of the CG base pair in position 20; 5' ATTCGATGGG TGGCAG G CAAAGTGGTGATGGC 3' A T C T A G 3' TAAGCTACCC ACCGTC C GTTTCACCACTAC CG 5' 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

  8. - an insertion of a base pair after pair 11 (AT) +1 5' ATTCGATGGG TGGCAG G CAAAGTGGTGATGGC 3' A T C T A G 3' TAAGCTACCC ACCGTC C GTTTCACCACTAC CG 5' 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.

  9. d) How can we abolish the insertion mutation in position 11? If in a position near the first insertion a deletionhappens, we can restore the correct frame of aminoacids. For example in position 15 (intragenicsuppression-suppressormutation) +1 AUG-GGA-NUG-GAG-UGC-CAA-AGU-GGU-GAU-GGC Met-Gly-aaX-Glu-Cys-Gln-Ser-Gly-Asp-Gly The secondmutationrestore the correctreading frame. Only the aminoacids locatedbetween the twomutationswill be different. DNA with insertion and subsequentsuppressormutation: 5’ ATTCGATGGGANTGGAGTGCCAAAGTGGTGATGGC 3’ 3’ TAAGCTACCCTNACCTCACGGTTTCACCACTACCG 5’

  10. 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 withthewt: thesecondaminoacidisdifferent Argcould be encodedby CGU, CGC, CGA, CGG, AGA, AGG Ilecould be encodedby AUU, AUC, AUA Itisprobablethat the codonencodingArgcould be AGA, and a mutationoccouredorginatingAUA Phecould be encodedby UUU/UUC and Thrby ACU/ACC/ACA/ACG, Thepossible DNA sequencewill be: Sequence of wt: AUG – AGA – UUPy – ACN Sequence of mutant 1: AUG – AUA – UUPy – ACN -

  11. WT : Met-Arg-Phe-Thr… Mutant 1: Met-Ile-Phe-Thr… Mutant 2: Met-Ser-Ile-Tyr We compare mutant 2 withwt: alltheamicoacidsaftertheMethionine are different. Aframeshiftmutationcausedbyaninsertion. AUG – AGA – UUPy – ACN – Ser isencoded by UCN or AGU-AGC , We can hypotesize the sequence: AUG –AGN–AUU-PyAC- With N=U/C The thirdcodon AUU = Ile The fourthcodonwill be UAC= Tyr

  12. 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

  13. 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.

  14. 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.

  15. if a deletion of a base pairoccurs in the middle of the secondintron, whichwill be the likelyeffects on the resultingpolypeptide? if a deletion of a base pairoccurs in the middle of the first exon, whichwill be the likelyeffects on the resultingpolypeptide? Wedon’thaveanyeffectif the mutationisnot in the splicing site. Wehave a frame-shifteffect or a truncatedprotein.

  16. FILE 9

  17. 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 14 21 14-21 14-21 14-21 GAMETES 14 21 14-21 14 21 Pairing of homologouschromosomesduring meiosis Karyotype 21

  18. Parent with translocation Normal Parent Normal Gamete Gametes (First parent) Gametes (First parent) 2 chromosomes 14, 3 chromosomes21 21 Trisomy : Down’s Syndrome 14 14 21 14-21 14-21 14-21 2 chromosomes 14, 1 chromosomes 21 21 Monosomy : not compatible with life 3 chromosomes 14, 2 chromosomes 21 14 Trisomy : not compatible with life 14 21 1 chromosome14, 2 chromosomes 21 Monosomy 14 n:ot compatible with life 2 chromosomes14, 2 chromosomes21 Zigote healthy carrier: normal phenotype 2 chromosomes14, 2 chromosomes21 21 Normal Zigote: normal phenotype

  19. 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: ZIGOTEsnot compatible withlife 1/6 1/6 1/6

  20. 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 gametesincorrectly • The gameteseither are missing or have an extra chromosome 21 • Itiscaused by the NONDISJUNCTION of chromosome21 duringmeiosis.

  21. 3. A man carries a heterozygous paracentric inversion. E F G H g f e h E F G H g f e h A B C D a b c d a b c d A B C D Draw a scheme of homologous chromosomal pairing during meiosis Which gametes are produced (in particular which gametes are missing)? Explain why. Drawonlyonecromatide for eachchromosome. F f E e g G A B C D H h a b c d Parental gametes Verranno prodotti i GAMETI PARENTALI ma mancheranno i gameti che hanno subito un evento di ricombinazione all’interno della regione invertita.

  22. Does the presence of the mutation change the fertility of this man? RECOMBINATION (CROSSING-OVER) F RECOMBINANT GAMETES f E e Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC FRAGMENT CHROMOSOME (lost). g d c b a E f g A B C D G A B C D H h e F G H h a b c d No, ifduplicationisnotextended.

  23. 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

  24. 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. Chromosomes A, B e C gamete 2n= AABBCC gamete n = ABC individual 3n=AAABBBCCC In a triploidcellwhich gametes are produced? AA (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) 2/8 are gametesthatcouldgeneratean individual 6/8 are not compatible withlife

  25. 7- 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. Wehypotizethatbothspecies 2n = 26 ASIATIC COTTON (A) = 13 chromosomes AMERICAN COTTON (B) = 13 chromosomes If in the hybrid a doubling of chromosomesoccours, wehave an alloploidthatis fertile becauseeachchromosomehasitshomologous. 2 A + 2 B = 26 + 26 = 52 Chromosomes

  26. FILE 10

  27. 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 locus where the repressorcouldbind to stop transcription. DNA site where RNA polymerasesits to start transcription. REPRESSOR STRUCTURAL GENES gene thatencodes for a proteinthatnegativelyregulatestranscription. Genesthat are usefull for a cellularfunction; for examplemetabolism of lactose.

  28. 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 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)

  29. 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 repressorcould be unable to recognize the locus. Thus, wehave the constitutiveexpression of the genes. The mutant in operator constitutivelacOc(cis DOMINANT)

  30. b gal perm. …… ….. Genotype 1 + + + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A 4. Which of the following genotypes will be able to produce β-galactosidase and/or permease in the presence of lactose? - - • Genotype2 …… ….. + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A - - • Genotype3 + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A …… ..…. • Genotype4 + + + + c c + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A …… …… + + + - + + + + Genotypes 3 and 4 -> constitutivetranscription of Lacoperon (no repression)

  31. b gal perm. …… ….. Genotype 1 + + + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A 4. If the lactose is not present, in which mutants the expression of genes change? - - • Genotype2 …… ….. + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A - - • Genotype3 + + + + + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A …… ..…. • Genotype4 + + + + c c + + + + + + Lac I Lac I Lac P Lac P Lac O Lac O Lac Z Lac Z Lac Y Lac Y Lac A Lac A …… …… - - - - + + + + Genotypes 1 and 2willnot express the genes (REPRESSION)

  32. 5. What does it mean that the Oc mutation is dominant in cis? How can I demonstrate it? CIS dominantmutation: itexpresses the dominantphenotypebutitaffectsonly the expression of geneson the same DNA moleculewhere the mutationoccurs. LacOc, affectsonlyneighbouringgenes (plasmid) Weconstruct an heterozygote with: The mutationlacZ-located in cis to lacOc (no production of b-galattosidase) The gene lacZ+ in trans (plasmid) Phenotype in absence of induction: mutationis cis-dominant -> no b-galactivity mutationis trans-dominant -> b-galactivity

  33. 7. Two bacteria have a Trp- phenotype, cioe? TheTrp- bacteria are unabletosynthesizeTryptophan How do I verifywhetherthetwomutations are in thesame gene? I complementthem: I produce bacteriacarryingbothmutationsone on a plasmid, the other on the chromosome. To analyze the phenotype I platethem on a medium withoutTryptophan CASE 1: Ifbacteriasgrow, the twomutationscomplementeachother, becausetheyaffecttwodifferentgenes. CASE 2: Ifbacteriasdon’tgrow, the twomutations do notcomplementeachother, becausetheyaffect the same gene.

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