GenesCodeForProteins

Transcription

1 GenesCodeForProteins The genetic code is responsible for the construction of proteins, which may be structural components of cells or metabolism.ontrollinq enzymes, The various levels of genetic instructions are illustrated below, together with their 'protein equivalents', Nucleotides are the simplest basic unit of genetic information, that are read in groups of three (called triplets), One triplet provides information to bring in a single amino acid during protein construction, Series of triplets in a long string allow the synthesis of polypeptide chains and are called genes, Some triplets have a special controlling function in the making of a polypeptide chain. The equivalent of the triplet on the mrna molecule is the codon.three codons can signify the end point of polypeptide chain construction in the mrna : UAG. UAA and UGA (also called STOP codons). The triplet ATG is found at the beginning of every gene (codon AUG on mrna) and marks the starting position for reading the gene. Several polypeptide chains may be needed to form a functional protein, The genes required to do this are collectively called a transcription unit. Thispolypeptidechain Thispolypeptidechain formsonepartofthe Functlcnal formstheotherpartof functionalprotein, ~ protein thefunctionalprotein, 5' (0 (0 (0 (0 8 8 (0 ~ ~ ~! t t! i,, 1 i START Triplet Triplet Triplel Triplel Triplet Triplel Triple, Atriplet codesforone aminoacid STOP START (0 (0 (0 (0 (0 (0 + Aminoacids t : t ~ ~ ~ i i,, i Triplet Triplet Triplel Triplet Triplet ~ Proteinsynthesis: transcriptionand translation Triplet STOP 3'... DNA Note: Thit start code is fer the codingstrandofthednathe templatednastrandfromwhich themrnais madewouldhave thesequence:tac. Transcription unit Three nucleotides make up a triplet nmodelsofnucleicacids, nucleotidesaredenoted bytheirbasetetter. 1, The following exercise is designed to establish an understanding of the terms used in describing protein structure and the genetic information that determines them. Your task is to consult the diagram above and match the structure in the level of protein organization with its equivalent genetic information: (a) Nucleotide codes for: (b) TriPlet/eoaon codes for: a N'\ 0. N"Y"-o''oA=> a.. cd J. (c) Gene codes for: ~l~f2c"fh A.sz..~ (y~c;.:) (d) Transcription unit codes for: a. ~c*i5>na- t pn*~ C~ oemqaa c:s cd-~ ~) 2, Name the basic building blocks for each of the following levels of genetic information: (a) Nucleotide...(o-tLu,.y,~R J. ~ ~~'C4 ~f' ~~l OY"" u..ro..~ l) l.o::tcil '''1 is made up of: p~bo.1s.l s,""'(f\. o.x"l l ~ 4: n,~ kns.c4 c..ons.oc.c.url t&cm... --tbo* a...ce:. (b) Triplet is made up of: 3 V't- ~-n('h ba.~eo rtead ~etbl< m Ck (c) Gene is made up of: a. r.eo,\m>ao' a (d) Transcription unit is made up of : -two ttf- 1T\~ld-~, t±cu"+i~ wj"",, a. idnk't c.e:x4. O!... tu-~on (s-lnp) e.ccnt. themrnais madewouldhave 1 nmodelsofnucleicacids,

2 P":.....&.\ RAE) GeneExpression The process of protein synthesis is fundamental to the process of transferring the information encoded in a gene to its understanding of how a cell can control its activities. Genetic functional gene product is called gene expression. t is divided instructions, in the form of DNA, are used as a blueprint for up into two distinct stages:transcription and translation.these designing and manufacturing proteins. Some of these proteins are summarized below and detailed in the following pages. For are the enzymes that control the complex biochemical reactions the sake of simplicity, the involvement of introns in gene in the cell, while others take on a variety of other roles. The expression has been omitted from the following pages. Chromosomal DNA DNA ntron ntron ntron ntron ntron Double stranded DNA contains the master copy ot ] molecule of genomic DNA all the genetic information to produce proteins for the cell. Exon Exon Exon Exon Exon Exon Most eukaryotic genes contain segments of coding sequences H~ Reverse transcription occurs when retroviruses (e.g. (exons) interrupted by non HV) invade host cells. Their viral RNA is converted Transcription coding sequences (introns). to DNA and spliced into the host's genome by an enzyme called reverse transcriptase. Primary RNA Transcript Both exons and introns are transcribed to produce a long primary RNA transcript.._----- Primary RNA -~ Exons are spliced together -, r ntrons ::::~., Messenger RNA ntrons in the DNA (also copied to the primary The introns are then removed by RNA) are long sequences of codons that have splicing to form a mature mrna. Translation (as yet) no apparent function. They may be the Messenger RNA is an edited remnants of now unused ancient genes. t has copy of the DNA molecule (now been suggested that they might facili tate axctuding the introns) that codes recomb ination between protein-codinq regions for the making of a single protein. (exons ) of different genes; a process known as exon shuffling. Th is may accelerate evolution. 'f Y Y Y y Structural Regulatory mmunological Transport Catalytic proteins proteins proteins proteins proteins 1. The hypothesis known as the central dogma of biology states that: "genetic information can only flow in the direction of DNA to proteins and not in the opposite directiort'. Accounting for the ideas in the diagram above, form a discussion group with 2-3 of your classmates and discuss the merits of this statement. Summarize your group 's response below : ~ CJ.Preoo i C\l"\ ',~ -tb.t. f20cc;oo ~ wh'tc..h..fut. c..cll (or c~.~ rf'o,n\a.tto..~ rtcm,o ~ ~ ~ fln:d..mc~. ~A ~ -+bt.."lnftx~s>00j2 c0c4" or "ost<udi.ori foe ~'Pc1d~ a.. ~. 'Thnk \nb=n"'\a.noo o\",crj...a "s. -h-o.j(\$m"~ ~~c.r'f'\i~ ~ -tco.m.s.~s*'\)v..u\~ majk'~ 0. ~K'\. 1'h;.. ~ crl ~ ~ \s ',mflb~~ "...,nt~ ~~q... Q... l~ da NoT ccv1l ""ceri~u..vu ~ Explain the significance of introns and exons found in DNA and primary RNA: (a) ntron:.s<lt. k\...-cj:d"""""",--_-i+cx;t""""'-=-:.l.-~""""--l LUo""""''''''''~'''''''''''''''''--''-...L..L~ W'\~\ \oe.!ca.-tr()n\.s.lo~, '1h\s \s CO l f 9 d... -=-. _ (b) Exon: rcjrra ~ \n -tb.l mt(.na. o.~ ~ -fur a..."' - " _ _. _~:_ : : : : ~ - - " '-- " ---~. _.. _.. _... ~ r r ---_._~

3 Coding strand of n NA h~~ ~ ~~L " '~~L_ ~~~.. Transcription is the process by which the code contained in the DNA molecule is transcribed (rewritten) into a mrna molecule. Transcription is under the control of the cell's metabolic processes which must activate a gene before this process can begin. The enz yme that directly controls the process is RNA polymerase, which makes a strand of mrna using the single antisense (template ) strand of DNA as a template. The enzyme Transcripti on transcribes onl y a gene length of DNA at a time and therefore recognizes star t and stop signals (codes ) at the beginning and end of the gene. Only RNA polymerase is involved in mrna synthesis ; it causes the unwinding of the DNA as well. t is common to find several RNA polymerase enzyme mo lecules on the same gene at anyone time, allowing a high rate of mrna synthesis to occur. 5' Single-armed enrornosomeas found in non-dividingcell. Freenucleolides usedto construct the mrnastrand. RNApolymerase enzyme A copy of the genetic information for making a protein is made in the form of messenger RNA (mrna). Many mrna copies may be made from a single gene on the DNA molecule. Once the mrna is complete and has been released from the chromosome, it travelsto the edgeof the nucleus where it gains access to the cytoplasm through a tiny hole called a nuclear pore. n prokaryoticcells (bacteria) thereis no nucleus, and thechromosomes are in directcontact with the cytoplasm.this means that the next stage (translation ) can begin immediately, with the mrna still being synthesized by enzymes on the DNA molecule. Template strand of DNA containstheinformationfor the constructionof a protein. Coding strand of ~ DNAhasa nucleotidesequence -, complementary to the templatestrand. Pore(hole) in the nuclear mp.mhr;:l:ne throuah which "", the mrnapassesto enter.,~ ". the cytoplasm. -. :'?~~j '. "'-'" X ' The two strands of DNAcoil up into a helix. Formationof a singlestrandof mrnathatis complementaryto the templatestrand(thereforethesame "message" as the codingstrand). Nuclear membranethat enclosesthenucieus. Oncein thecytoplasm, the mrnawill engage ribosomesto beginthe nex1stagein protein synthesis: translation 5' Nucleus Cytoplasm 1. Explain the role of messenger RNA (m R NA) in protein synthesis : m ~NA COKrieo 0.. ~ of -tht.. ~c. \~~.fn:m. -l-bddna ',0. +Vw.o.~p~m. 'W\. fun n"cj~ ~ r 'W;;;.~ 2. The genetic code contains punctuation codons to mark the starting and finishing points of the code for synthesis of polypeptide chains and proteins. Consult the mrna-amino acid table earlier in this manua l and state the codes for : (a) Start codon: --LA...u,l""'-l""c:.r04- _ (b) Stop (termination ) codons : (A.M, laag",ugra For the following triplets on the DNA, determine the~ sequence for the mrna that would be synthes ized : (a ) Triplets on the DNA: T A C ~A G C C G C G A T T T Codons on the mrna: AlA c:r iluc.. GGcC GrCL.l AAJA... (b) Triplets on the DNA : T A C A A G C C T A T A A A A Codons on the mrn A : ALJ.G laue GMA laal\. tala.l\.

4 Translati on The diagram below shows the translation phase of protein more of the mrna than the ribosome to the left. The antisynthesis. The scene shows how a single mrna molecule codon at the base of each trna must make a perfect can be 'serviced' by many ribosomes at the same time. The complementary match with the codon on the mrna before ribosome on the right is in a more advanced stage of the amino acid is released. Once released, the amino acid is constructing a polypeptide chain because it has 'translated' added to the growing polypeptide chain by enzymes. dbd Thr RNA unlob:t: i unloa~ ""~~ in an early stage of synthesis This chain is in an advanced stage of synthesis. unload:t;ed Thr RNA -'---" ' " i 5' JJdJjbtDjjoiilllUl.nllillUlJlllJdlltlJ mrna_.~ =- ---:>... Ribosomes rnovmq n this direction trna molecules move into the ribosome. bringing in amino acids 10add to the polypeptide cham under construction. Amino acid attachment site Ribosome Large subunit Small subuni t L=--.J Anticodon t i The anticodon is the site of the 3 base sequence that 'recognizes' and matches up with the codon on the mrna molecule. Ribosomes are maoe up of a complex of ribosomal RNA (rrna) and proteins. They exist as two seoarate sub-units until they are attracted to a binding site on the mrna molecule. when they join together. Ribosomes have binding sites that attract trna molecules loaded with amino acids. The transfer RNA (trna) molecules are about 80 nucleotides in length and are made under the direction ot genes in the chromosomes. There is a different RNA molecu le for each of the differenl possible anticodons (there may be up to six different trnas carrying the same amino acid). 1, For the following cedens on the mrna, determine the an1l;codons for each trna that would deliver the amino acids: Codons on the mrna: U A C U A G C C G C G A U U U Ant i-codons on the trnas: 2. There are many different types of trna molecules, each with a different anti-codon (HNT: see the mrnatable ). (a) State how many different trna types there are, each with a unique anticodon :.J2J b) Give a reason for your answer in ra) above: :fb.a..<e _ c::vye. <ell'f'b~; bls..~ -mr: Wll..NA} bu* 3 C1J«..-s-k>f~. Col vxt.ot'\$ ~r m tna rr:qu: G'=21~NAs etalh.we&.0.. ea_ <;.oc;ton.s bringing in amino acids 10add to the attachment site - n

5 and the number of different codons that can code tor each amino acid (the first amino acid has been done for vou), TheGeneticCode The genetic information that cooes for the assembly of amino each amino acid. there may be more than one codon. Most of acias is stored as three-ietter codes. calied codons. Each codon this aegenerac y involves the third nucleotide of a codon. The ioresents one of 20 amino acids used n the construction of genetic code is universal ; all living organisms on Earth, from olyoeptide chains, The mana amino acid table (bottom 01 viruses and bacteria, to plants and humans, share the same page) can be used to identify the amino acid encoded by each of genetic code book (with a few minor exceotions representing the mrna cocons. NOTethat the code is degenerate in that for mutations tnat have occurred over the iong history of evolution), Ala Arg Alanine Arginine GCU, GCC, GCA, GCG 4 Leu Asn Asparagine AA.<A, AAC 12 Lys Leucine Lysine Met Methionine Asp Asparticacid Phe Phenylalanine Cys Cysteine tagfu, \AGrC 12. Gin Glutamine Glu Glutamicacid GrAA..Gt'AGr Z. Gly His r' so Glycine Histidme soleucine CAu,C,Ac.. AUU., N..AC,AUA Pro Proline Ser Thr Try Tyr Val Serine Threonine Tryptophan Tyrosine Valine UAA, UUGr, CW--\.. r's u: r'dl c:.ul::r AAA,A~ UtA.U.., \..llac, 12. l,agla, uc.c. v.la, (J.f r...a (;ret A~c.. b ~, Ace, k..a. Ac&lf ta.aca laac 1 2 Use tne mrna amino acid table (below ) to us: n the taole above all tne codons that code for each of the amino acids and the number of different codons that can code tor each amino acid (the first amino acid has been done for you). 2. (a) State how many amino acids could be coded for if a codon consisted of just TWO bases: --l/... b...,. _ (b) Explain wh y this number of bases is inadequate to code for the 20 amino acids required to make proteins : ~-~<. ~ c:la N::>t pn:>vido enr.::.wt' G<:)tnbiOQh'~ - ~ q,xc: 2-0 c:a. f ~ a~d o.~-ela 3, There are multiple codons for a single amino acid, Comment on the significance of this with respect to point mutations: ~kv'&- p:(m ~~n,q ""o"od ~-t ~ -#u. ~ ac:'ol-e&pc-c.t~ 'rf..fha. 0"wM2trv6t>1') (~'l\ 'n +tu b)± ba.t(., 6.,r ~ CecA opt - ~tz.~..~p,r"j if *'--~'. c..od6'nchart \r>,kf " \ ) ~ ~r'"" mrna-amino Acid Table Read secon d Read lhrrd lener nere Second Letter e)here A. Td' Head firs' letter nere ~ u c A G ~f\. How to read the table: The table on the right S used to 'oecods' the genetic code as a UUU Pne UCU se- UAU Ty' UGU e VE UUC sequence of amino acids in a polypentide Pn" UCC Ser UAC i yr UGC c."s UUA Leu UCA Se t UAA STOP UGA STOo chain. from a given mrna sequence To work UUG Leu UCG Ser UAG STOP UGG Try out whic h amino acid is code d tor by a codon (trip let of oases ) look for the first tette r of the CUU Leu CCU Pro CAU!-1 1 ~ CGU Ar" cue L et, CCC F:,c CAe H S CGC codon in tne row label on the iei! hand side. CUA Lel. CCA Pre CAf... Gl... CGA A'9 A r~ ]ben lo ok for the column that ntersects the CUG Le u CCG " rc CAG Gte CGG A,g ----e row from above tnat matches tne.ronc base. Finally, locate the third base in AUU rsc ACU Tnr AAU.A.Sr. AGU Se r the codon by looking along the row from the AUC SO ACC in - AAC p,sn AGC Ser AUA so ACA n: AAA LYS AGA Arg.,...,~ rrgnt nand end tha t matcnes you r codon, AUG Me: ACG AAG LYS AGG Arg Example : Determine CAG GUU va GCU Al e GAU ASL GGU Gry GUC Va GCe AlE: GAC Aso GGC Gl,! C on the left row, fl.on the too colu mn, G GUA Va GCA Ali GAt. Glu GGA G'y on the right row GUG ",jo GCG Ali GAG G... GGG Gr.' CAG is Gin,glutamine )

6 ... 11'0. ProteinSynthesisSummary ~.. :.0.. : ~ 6,' ~ b.'.tj '0-. ' ~aq ~ V ee 8eO.. ltlr J ~... Cytoplasm The diagram above shows an overview of the process of protein synthesis. t is a combination of the diagrams from the previous two pages. Each of the major steps in the process are numbered, \~hil ' structures :r9 labeled with letters. 1. Write a brief description of each numbered process in the diagram above: (a) Process 1: \AK\wi ~ DNA (b) Process 2 : -tra.,owr'ipt6>n-mitna s..~w -r\a.lcj,dic44 Q"Q(dp.,t 10 ~"3 mlenf\ ::: (c) Process :::::a:::;i:::!':::::::: 3: DNA feb>" rd.o :::::: ~ :::~::=~r~ (f) Process 6: ~ ~t... ~h 4eJlW*c~crxAAn f Q'>clAnkn'L -pmlf a.rn;(1() "1«~ (g) Process 7 : l:,(?na lea.v'co i'h.q ";~ny" o;:.~.l'".-j ~ '~-...Lt..- 'd (h) Process 8: ~...NA u rc.c:~"" Wl=rn ~ ~ 0.(.4 2. dentify each of the structures marked with a letter and write their names below in the spaces provided : (a) Structure A: ])... NL.1Lf\~ _ (f) Structure F: N.ACg De,., fdr"-' (b) Structure B: "FCc.e. W'\M("ld ertickoe (g) Structure G:_"'=-""""""e..!!!<L...N::..oA-=- _ (c ) Structure c RNA Pl)l~ (h) Structure H: ~Q o.cac::lo (d) Structure D:... m~r"""""n.3.j'p.::l- _ (i) Structure : ~ 1r::c~r'Y'A- (e) Structure E "'uddl1#' ~~ Po\~r~h c.k\a.,,:o U) Structure J : /p~ 3 Explain the purpose of protein synthesis (gene expression): 7~k+,An c:.wn:. (J'!) fs'tlj.j bla. -ftir: -:E'N\..~J ~p>rt Prb~J So ~ fu.mca-ic)n1wi:twn CA. W.l ~in.e-hl(a..i ~} ~'!LJ """"-KnCdat(f. d ~ Go.Dec h..ey ~ ~ -tho. n"nc4-\c*"-uf ~ c..em - PAMt'9V\ vs rcea,blotd ~tl. l')n ~,~.J