Biochemical Genetics: DNA, RNA, and Protein Synthesis.

INTRODUCTION

Genetic information is coded for in GENES, which in turn are composed of a compound called DEOXYRIBONUCLEIC ACID ("DNA" for short). Genes do their work by telling a cell what type of proteins to make. Most of these proteins are enzymes and they control attributes such as eye color, hair color, and the ability to resist a particular disease (among other genetic TRAITS).

Eukaryotic organisms, like ourselves, have thread-like strands of DNA while prokaryotes (bacteria, for example) have circular loops of DNA. Eukaryotic DNA is also packaged (along with proteins) in cellular organelles called CHROMOSOMES while prokaryotic DNA is not packaged in chromosomes. Chromosomes are found in the nucleus and contain nearly all the DNA in a eukaryotic cell.

Deoxyribonucleic acid belongs to a class of organic compounds called NUCLEIC ACIDS. A related compound, RIBONUCLEIC ACID (RNA) is also classified as a nucleic acid. Ribonucleic acid does its work outside the nucleus in the cytoplasm. As we shall see, RNA is directly involved in protein synthesis.

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Simulating DNA Replication and RNA Transcription.

PURPOSE: To simulate replication and transcription using paper models.

MATERIALS NEEDED:

Scissors.

One picture of a DNA molecule with Bases.

Three dimensional models of DNA and RNA structure.

When a cell divides it must make an exact genetic duplicate of itself (except for reproductive cells- eggs and sperm, or in abnormal cell division- cancers, for example). If the copies are not exact then the DAUGHTER CELLS produced by the division would not have the complete DNA instructions to synthesize enzymes for metabolism. The parent cell must therefore copy its DNA before reproducing and thus insure that each daughter cell receives complete genetic information. DNA makes copies of itself in the nucleus of the cell; a process called REPLICATION (Fig 1).

FIGURE 1. Replication of DNA

A second class of nucleic acids, the ribonucleic acids (RNA) are also manufactured in the nucleus by a process known as TRANSCRIPTION. Transcription

uses the instructions of DNA to synthesize RNA. There are three flavors of RNA

and, although they are synthesized in the nucleus, they do their work out in

the cytoplasm (Fig 2)

FIGURE 2 Comparison of Replication, Transcription, and Translation.

A third process (TRANSLATION) is the reason for all this replication and transcription: protein synthesis. Translation uses RNA to bond together amino acids to synthesize polypeptides. In brief, TRANSFER RNA (tRNA) picks up amino acids and carries them to the site of protein synthesis at the ribosome (Fig 2). The ribosomes are composed of protein and RIBOSOMAL RNA (rRNA). There the instructions for making the protein are read from a strand of MESSENGER RNA (mRNA). The transfer RNA then releases the amino acid and is free to pick up another in the cytoplasm.

PROCEDURE:

1- Preparation. Work in pairs. Locate and remove one DNA template sheet from one of your laboratory manuals. Separate the pieces by first cutting along the green dashed lines. Each person works with one of the base sheets while the DNA molecule is shared. Separate the bases by cutting along the black dashed lines and then place them face-up on the table. Keep your bases separate from those of your partner.

2- General DNA structure. Examine the crib sheet card with the structural formulae and compare to the color drawing of the DNA molecule, and three dimensional display model(s).

Your game pieces are color-coded as follows:

Adenine: Yellow-Brown. .

Cytosine: Green. Deoxyribose: A black pentagon

Guanine: Blue. Phosphate Group (PO4): An blue circle

Thymine: Red.

3- Deoxyribonucleic Acid Replication. The base pairing between purines and pyrimidines allows DNA to serve as its own template to make accurate copies of itself. To take advantage of the base pairing, DNA must unwind and split lengthwise to expose the base pairs on each strand (Fig 1). Simulate the initial separation of your DNA by cutting the colored DNA molecule along the orange dashed line. Take one of the DNA chains

FIGURE 3. DNA Replication using the game pieces.

4- Single nucleotides are attracted to the proper exposed bases on the single-stranded DNA (Fig 3). Enzymes (the DNA POLYMERASES) are used to attach the 3' end of the deoxyribose to the previous phosphate. In this way, a new DNA strand is built using the exposed strand as a template.

5- Use your nucleotide game pieces to build a new DNA molecule. Your strand of DNA will serve as a template. Start at the 5' end of your DNA template (DNA is always read from the 5' to the 3' end) and add nucleotides. Pair up thymine (Red) with adenine (yellow) and cytosine (green) with guanine (blue; Fig 7). Compare your replicated DNA molecule with that of your lab partner. They should have identical sequences; otherwise one of you has made a mistake and have inadvertently simulated a mutation. Record your base sequences in the results section.

6- Simulating a Mutation. In this exercise the colored DNA strand labeled for person #1 (the one with the plain blue phosphates) will be used as a template for replication. We will simulate our mutations at the sixth base from the 5' end (thymine). Move in and pair up five nucleotide bases as before (read from the 5' end). At the sixth position insert a thymine instead of the correct base, and then continue your replication as before. Note the "dimple" in your double-stranded DNA. This type of genetic damage is common when cells are exposed to high levels ultraviolet radiation. Ultraviolet light is used in some bacterial sterilizers and is responsible for causing sun burns and increasing the incidence of skin cancer in humans (especially for those of you who enjoy basking like lizards in the summer sun). The double thymine condition is called a THYMINE DIMER. Other types of radiation or chemicals can also cause changes in the DNA sequence.

7- Remove the long colored DNA strand and then use your mutated strand as a template for further DNA replication. Compare your finished DNA molecule to the original. What will happen when the mutated strand is replicated? This simulation demonstrates how mutations can become fixed in the DNA. Record your base sequences in the results section.

8- Transcription of Ribonucleic Acid. DNA serves not only as a template for its own replication, but also provides instructions for RNA synthesis. (Transcription; Fig 2). Ribonucleic Acid differs from DNA in several ways: While the sugar molecule for DNA is deoxyribose, that for RNA is RIBOSE. Ribose differs from deoxyribose in the substitution of a hydroxyl group (OH) for a hydrogen (H; the change is shown in violet on the crib sheet). RNA is often found in a single-stranded form while DNA is mostly double- stranded. RNA also lacks thymine (the base URACIL is substituted). Uracil is structurally-related to thymine (it is a single-ringed pyrimidine), but it has a hydrogen instead of a methyl group (CH3) on one of the ring carbons. (see the crib sheet).

9- Transcription follows more-or-less the same sequence of you simulated for DNA replication: DNA molecule unwinds to expose a strand. The DNA template of one strand (the TRANSCRIBED STRAND) is read in a 5' to 3' direction. The other DNA strand (the NONCODING STRAND or NONTRANSCRIBED STRAND) is not read. A class of enzymes called "RNA polymerases" bond RIBONUCLEOTIDES (ribose attached to phosphates and a base) to one another to form single-stranded RNA. The base-pairing is the same as that for DNA except that adenine on the transcribed strand of the DNA attracts uracil ribonucleotides rather than thymine.

10- Since RNA transcription is so similar to DNA replication, we won't bother with a simulation. Answer the questions and exercises in the results section.

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Simulating Translation.

PURPOSE: To simulate translation and introduce the student to the genetic code.

MATERIALS NEEDED:

Scissors.

Color sheet of mRNA, tRNA, and base code table

Amino acid/ribosome sheet

Now that you have seen how DNA and RNA are synthesized, we will continue with a simulation of translation (protein synthesis).

PROCEDURE:

1- Preparation. Locate and remove the color sheet of messenger RNA (mRNA), transfer RNA (tRNA), base code table, and full-color souvenir bookmark. Also remove the amino acid/ribosome sheet. Cut along the dashed lines and place the game pieces face-up. Cut the box out of the ribosome and cut two slots in the paper as indicated.

2- Decoding the Message and Simulating Translation. Before translation can be simulated you should first understand the relationship between the mRNA codons and tRNA anticodons. You should also know the procedure used to decode the message carried by the mRNA (Fig 4 and Base Code Table). The abbreviations for the amino acids are listed in Table 1.

Amino AcidAbbr.Amino AcidAbbr.Amino AcidAbbr.Amino AcidAbbr.

AlanineALAGlutamic AcidGLULeucineLEUSerineSER

ArginineARGGlutamineGLNLysineLYSTheonineTHR

AsparagineASNGlycineGLYMethionineMETTryptophanTRP

Aspartic AcidASPHistadineHISPhenylalaninePHETyrosineTYR

CysteineCYSIsoleucineILEProlinePROValineVAL

TABLE 1. Amino Acid Abbreviations.

FIGURE 4. Relationship between codons, anticodons, and the genetic code.

3- Slide the long mRNA strand through the right-hand slot of the ribosome. Position the start codon on the mRNA (AUG) at the left-hand side of the open box on the ribosome (Fig 5A). This lines the start sequence (AUG) up with the first groove on the ribosome. The first mRNA codon (AUG) is the "START" command and corresponds to the amino acid methionine (Fig 4, Base Code Table). The tRNA that interacts with a AUG codon is one with an anticodon sequence of UAC (Fig 5A). In this diagram the UAC tRNA is shown with an attached methionine. It is in position at the first groove of the ribosome and is lined up with the start codon on the mRNA. At this point you have simulated activation of the ribosome and protein synthesis can continue (activation is actually more complex than this; see your text for details).

FIGURE 5. Reading the genetic code.

4- Determine the next amino acid required in the sequence by decoding the next three bases. In our example the GCA codon corresponds to the amino acid alanine (pay attention only to the text figures at this time, not the colored game pieces). A tRNA with a complimentary CGU anticodon is picked from the tRNA pool, and is used to transfer an alanine to the second groove (Fig 5A,B; Fig 4). When the methionine and alanine are brought close to one another, enzymes connect the two amino acids and release a molecule of water. A hydrogen (H) and hydroxyl group (OH) are set off from your amino acid game pieces by dotted lines to remind you of this. Note that, like DNA, the mRNA is read from the 5' to the 3' end.

5- Following the formation of a bond between the two amino acids, the "older" of the two tRNAs uncouples from its amino acid and leaves the ribosome (Fig 5). The newly-released tRNA can pick up another amino acid of the proper type and be re-used. The ribosome slides toward the 3' end of the mRNA strand and exposes the next codon on the mRNA and a third tRNA with an the proper anticodon and attached amino acid is moved in (Fig 5A). This process continues (Fig 5B) until the entire mRNA strand is read and a stop code is encountered (UAA in our example). The polypeptide and mRNA are then released and may be re-used (this depends on the organism)

6- Run through the translation simulation as described above. Use Figures 2, 4, and 5 as guides. The color codes follow the same pattern as your DNA replication simulation:

Adenine: Violet. Guanine: Blue. Cytosine: Green. Uracil: Red.

RNA anticodons must always match up with the codon on the mRNA. Adenine (violet) pairs with Uracil (red) while the blue guanine and green cytosine form a pair. The base code table is also color coded. Remember that the base code table is keyed to the codon on the mRNA, not the anticodon on the rRNA. Record the resulting sequence of amino acids in the results section and answer the appropriate questions.

REPORT SECTION

by __________________________________

To complete your assignment for this exercise, fill in all the information

requested in the RESULTS section, tear out at the perforations and hand in.

This constitutes your Laboratory Report for this Experiment.

RESULTS ____________________________________________________________________________

Simulating DNA Replication and RNA Transcription.

Deoxyribonucleic Acid Replication. Record the sequence of bases for both strands of your replicated DNA molecule and that of your partner.

Are the base sequences of the two DNA molecules equivalent to each other and

the original? _______________________________________________________________

Simulating a Mutation. Record the sequences for both mutated DNA molecules.

Are these sequences different from the original? ____________________________

What will all future DNA molecules replicated from the above pair look like (assuming that there are no further mutations)? _________________________________________________

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Transcription of Ribonucleic Acid. Show that you understand the relationship between DNA and RNA by filling in the blanks below the DNA strand with a properly transcribed molecule of RNA (remember uracil):

3' Single-stranded DNA to be transcribed

5'

T---A---C-----G---G---G-----T---T---T------C---A---T-------G---G---C----A---T---C

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

5' end 3' end

of RNA

of RNA

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Simulating Translation.

Decoding the Message and Simulating Translation.

Assume that the RNA you transcribed in the previous question is messenger RNA.

Decode the codons for the above base sequence and indicate the resulting

polypeptide:

___________ ___________ ___________ ___________ ___________ ____________

We will call this peptide "ENZYME 1"

Consequences of Mutations at the Level of DNA. In this section you will simulate the effects of a DNA mutation on translation. The underlined bases of the DNA molecule in the previous section that transcribe codons two and three of the above mRNA molecule will be affected.

Starting with the following mutated single strand of DNA, indicate the mRNA that would be transcribed and the protein that would be translated (an adenine has been substituted for the third base (guanine) that transcribes for mRNA codon 2):

3' Single-stranded DNA to be transcribed 5'

T---A----C----G---G---A----T----T---T----C---A----T----G---G---C-----AT---C

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

5' end 3' end

of mRNA of mRNA

Protein Translated from the Above mRNA

___________ ___________ ___________ ___________ ___________ ____________

We will call this peptide "ENZYME 2".

Does the protein made from the mutated strand (ENZYME 2) differ from the unmutated protein (ENZYME 1)? Are there any base substitutions that can be made at position three that will affect enzyme activity? Explain. ________________________________________

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Assume that the first base coding for the third codon (thymine) has been exchanged for an adenine:

3' Single-stranded DNA to be transcribed 5'

T----A---C----G---G---A----A----T---T-----C---A----T----G---G---C----A---T---C

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

5' end 3' end

of mRNA of mRNA

Protein Translated from the Above mRNA

___________ ___________ ___________ ___________ ___________ ____________

Do you think this third enzyme will be functional (why or why not)? _________________

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Working Backwards. A protein is sequenced and found to have the following order of amino acids:

Methionine-Proline-Glycine-Cysteine-Alanine-Valine

Indicate a possible mRNA codon sequence that may have produced this protein:

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

5' end 3' end

of mRNA of mRNA

Indicate the base sequence of the DNA strand that transcribed your mRNA

strand:

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

3' end 5' end

of DNA of DNA

And, just to finish things up, what does the other strand of your DNA look

like?

___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___

5' end 3' end

of DNA of DNA