Lab 8 DNA Coding and Protein Synthesis


Question Description:

49.99

Lab 8
DNA Coding and Protein Synthesis

Introduction: Connecting Your Learning

As covered in a previous lesson, DNA is an abbreviation for the biological molecule called deoxyribonucleic acid. DNA is found in the nucleus of cells and it stores genetic information and the code for synthesizing proteins. A DNA molecule is organized into sections called genes. Genes code for specific proteins that are vital to living organisms. Proteins are important because they are the building blocks of cells and, additionally, proteins called enzymes play a key role in catalyzing chemical reactions in living organisms. Proteins are synthesized in the cytoplasm of cells by ribosomes. Long strands of DNA along with proteins called histones are tightly wound together to form chromosomes.

In order to synthesize a protein, information must be transferred from the DNA in the nucleus of the cell to the ribosomes in the cytoplasm of the cell. The process involves two steps: transcription and translation. Transcription is the process where the code for a protein found on a specific gene located on DNA is written to another nucleic acid molecule called ribonucleic acid (RNA). This process takes place in the nucleus of the cell. The molecule of RNA then leaves the nucleus and enters the cytoplasm where the process of translation takes place. Translation is a process where the code for making a protein (found on a molecule of RNA) is translated into specific amino acids that will comprise apolypeptide (protein).

In this lab, students will simulate the process of protein synthesis by transcribing DNA molecules into RNA molecules and then subsequently, translating RNA molecules into amino acids. Students will also extract DNA from an onion.

Resources and Assignments
Multimedia Resources None
Required Assignments Lesson 10 Lab 8
Required Materials From the Lab Kit

  • DNA template cards
  • Universal genetic code reference card
  • Ribosome circle
  • Thermometer
  • Funnel
  • 100 ml graduated cylinder
  • 2 coffee filters
  • Measuring spoon (teaspoon)
  • 2 teaspoons salt

Student Provided

  • Pen or pencil
  • Standard printer paper (1 sheet)
  • Scissors
  • Onion
  • 1 small pot
  • 120 ml water
  • 2 tablespoons liquid dish detergent
  • Fork (Needed only if a blender or food processor is not available)
  • Knife
  • Cutting board or other surface for cutting the onion
  • Additional water for heating (3-4 cups tap water)
  • 2 small glass, ceramic, or metal mixing bowls
  • 2 cups ice or access to a freezer
  • 2 clear drinking glasses (12-16 oz)
  • 100 ml Isopropyl alcohol (rubbing alcohol), 71% or higher
  • Blender or food processor
  • Tape

Optional: (if the student wants to preserve the extracted DNA)

  • Small clear glass container with a lid
  • Alcohol (100-200 ml)
  • Wooden skewer or toothpick

Focusing Your Learning

Lab Objectives

By the end of this lesson, you should be able to:

  1. Identify the structure of DNA including the following: nucleotides, base pairing, hydrogen bonds, and the double helix model.
  2. Compare and contrast the structure of DNA and RNA.
  3. List and describe the steps of DNA replication.
  4. List and describe the steps of transcription and translation, and describe where each occurs in the cell.
  5. Explain the functions of messenger RNA, transfer RNA, and ribosomal RNA.
  6. Distinguish between codons and anticodons.

Background Information

DNA and RNA Structure

DNA and RNA molecules are nucleic acids. These molecules are comprised of long chains of nucleotides. Nucleotides are monomers made up of a five-carbon sugar molecule covalently bonded to a nitrogenous base and a phosphate group. Four different nucleotides make up DNA. The nucleotides differ from one another in structure due to the nitrogenous base and the sugar molecule they contain. The four nitrogenous bases found in DNA molecules include Thymine (T), Cytosine (C), Adenine (A) and Guanine (G). The nucleotides of RNA are the same as DNA with one exception: instead of thymine, RNA contains Uracil (U). The other difference between DNA and RNA is the sugar molecule each contains; DNA contains deoxyribose while RNA contains ribose. Covalent bonds between the sugar of one nucleotide and the phosphate of the neighboring nucleotide join the nucleotides together, forming a sugar-phosphate backbone. The nitrogenous bases project from the backbone. See the image below for three illustrations detailing the structure of DNA.

The DNA molecule is comprised of two polynucleotide strands forming a double helix structure, which looks like a ladder. The sugar-phosphate structure forms the backbone of the ladder, while the nitrogenous bases form the rungs of the ladder. By constructing a physical model of DNA, scientists James Watson and Francis Crick were able to confirm the double helix structure of DNA. Additionally, the physical model helped Watson and Crick to establish the specific base-pairing present in DNA: adenine pairs with thymine and guanine pairs with cytosine by hydrogen bonds. The image seen above illustrates the double helix structure and base-pairing of the DNA molecule. The following base pairing occurs in a molecule of RNA: adenine pairs with uracil and guanine pairs with cytosine.

Once the structure of DNA was identified, this led to an understanding of the process of DNA replication. The key to understanding DNA replication lies in the specific base-pairing rules identified by Watson and Crick. DNA replication is initiated by proteins that attach to specific sites along the strand called the origin of replication. When the proteins attach to the DNA molecule, they separate the two strands and each half becomes a template used to assemble a complementary strand by following the base-pairing rules. For example, if one half of the DNA molecule is a strand of ACCCTG, then the following complementary strand is formed according to the base-pairing rules: TGGGAC. As a DNA molecule separates, replication proceeds in both directions, creating replication bubbles as seen in the image below.

The DNA of eukaryotic organisms has more than one replication origin, which results in many replication bubbles; this allows for DNA replication to proceed very quickly. All of the replication bubbles merge to form two daughter strands of DNA.

Nucleotides are added to the end of the daughter strand by an enzyme called DNA polymerase for the daughter strand moving toward the forked section of the parent strand. For the strand where nucleotides are being added outward from the forked section of the parent strand, the strand is synthesized in small pieces that are linked together by another enzyme called DNA ligase (see image below).

While the process of DNA replication is highly accurate, errors can happen. If an error is made that leads to a change in the nucleotide sequence of DNA, a mutation has occurred. Mutations can be limited to a single nucleotide pair, or they can involve large areas of a chromosome. Two types of mutations can occur during protein synthesis. The first type of mutation is called a point mutation. A point mutationoccurs when a nucleotide (base) is replaced with a different nucleotide (base); for example, a C is replaced with an A. Because of the redundancy in the genetic code, sometimes the substitution will not have a negative effect on an organism. For example, if the codon GGC is changed to GGA, no harmful change would result because both codons code for the protein Glycine (Gly). In some instances, however, a change of a single nucleotide (base) can code for a different protein, which can result in harmful effect. If the point mutation changes an amino acid codon into a stop codon, the result is called a nonsense mutation. If protein synthesis halts prematurely as a result of a point mutation that changes an amino acid codon to a stop codon, a nonfunctional protein results.

The second type of mutation is called a frameshift mutation. In a frameshift mutation, at least one base is added or deleted from the DNA. If a base is added to the DNA, a base insertion occurs; conversely if a base is removed from DNA, a base deletion occurs. The addition or subtraction of a nucleotide (base) will change the entire reading frame, which will cause a harmful effect. For example an original protein code reads: THE WET CAT SAT AND ATE THE FAT RAT. If a base insertion occurs and an “A” is added in the second word (WET) after the W, the new code would read: THE WAE TCA TSA TAN DAT ETH EFA TRA T. In contrast, if a base deletion occurs and the E in the second word (WET) is removed, the new code would now read: THE WTC ATS ATA NDA TET HEF ATR AT. Note that the whole reading frame of the original code has been altered when either a base insertion or a base deletion occurs.

In addition to storing genetic information, DNA contains information for directing the manufacture of proteins through a process that starts in the nucleus. The process of protein synthesis occurs in two stages: transcription and translation. The process of transcription occurs in the nucleus, while the process of translation occurs in the cytoplasm. Transcription converts the genetic information contained in DNA to a special RNA molecule, called messenger RNA (mRNA). The coded information on DNA is written in a language of nucleotide base sequences. Specific sequences of the bases make up genes along the DNA strand. The molecule of mRNA contains a sequence of complementary bases to those found on the DNA strand.

During transcription, a DNA molecule separates and one strand serves as a template for the formation of a strand of mRNA. Base-pairing rules are applied as mentioned previously (C pairs with G and U pairs with A). RNA nucleotides are linked together by an enzyme called RNA polymerase. RNA polymerase binds to the DNA molecule at a promoter. The promoter correlates to the start of a gene found on the DNA strand, and it consists of a sequence of nucleotides on the DNA that serves as a start signal for transcription to begin.

The process of transcription takes place in three steps: initiation, elongation, and termination (see image below).

During initiation, RNA polymerase attaches to the promoter and RNA synthesis begins. During elongation, RNA elongates as RNA synthesis continues, which allows the RNA strand to separate from the DNA template. This allows the DNA strands to come back together in the region where transcription has occurred. During the final stage of transcription (termination), RNA polymerase reaches a sequence of bases on the DNA strand called a terminator which signals the end of the gene. Once the RNA polymerase reaches the terminator, it detaches from the mRNA molecule and gene.

In prokaryotic cells, the RNA transcribed from a gene is used as-is and functions as a molecule of mRNA. In eukaryotic cells, however, the RNA molecule is first modified prior to moving into the cytoplasm where translation occurs. First, extra nucleotides are added to each end of the molecule, called the cap and tail, which help identify the molecule as a molecule of mRNA and also help protect the molecule from destructive enzymes. The second modification that occurs to the molecule involves the removal of certain noncoding regions found along the molecule which are called introns. Prior to leaving the nucleus, RNA molecules are modified–the introns are removed–leaving only the coding regions (exons) which are spliced together to form a molecule of mRNA.

After transcription, the mRNA molecule translates the nucleic acid language (bases) into the language of polypeptides. The polypeptides are made of monomers of amino acids. There are 20 amino acids that are common to all living organisms. The synthesis of proteins, as directed by genes, is based on codes related to a series of three consecutive bases called a codon. A codon consists of a three-base sequence of mRNA that codes for a specific amino acid. Since there are four bases with three possible arrangements, there are a total of 64 possible amino acid codes. However, the reason that there are only 20 amino acids instead of 64 is due to the fact that several codons code for the same amino acid, as seen in the table below. Of the 64 codons, 61 code for amino acids, and three are stop codons that signal the ribosome to end translation.

The process of translating the genetic code from acids to polypeptides involves mRNA, which has been discussed previously, as well as two other important molecules: tRNA and ribosomes. As discussed above, mRNA carries the genetic code into the cytoplasm where it can be translated into a protein. Transfer RNA, or tRNA, is a special form of RNA that takes the mRNA codons and converts the code into amino acids that string together to form a polypeptide (protein). tRNA contains a triplet of bases called an anticodon that is complementary to a codon triplet found on the mRNA molecule. tRNA identifies the codon via the complementary anticodon, interprets the information, and then picks up the respective amino acid to be added to the polypeptide chain. The final molecules involved in translation are ribosomes. Ribosomes are responsible for polypeptide synthesis and they facilitate the interaction between mRNA and tRNA. Ribosomes contain ribosomal RNA (rRNA), proteins, and two subunits.

Translation consists of the same three steps as transcription: initiation, elongation, and termination (see below).

Initiation occurs in two stages. During the first stage, an mRNA molecule binds to a small ribosomal subunit. An initiator tRNA binds to a specific mRNA codon called the start codon, where translation begins. The initiator tRNA carries an amino acid called methionine (Met). The anticodon for Met is UAC, which binds to the start codon (AUG). The second stage of initiation occurs when the large ribosomal subunit binds to the small subunit to create a functional ribosome. The initiator tRNA attaches to one of the ribosomal binding sites called the P site, which holds the growing polypeptide. The second tRNA binding site is called the A site, and it houses a tRNA molecule that carries an amino acid to be added to the polypeptide chain. See the image below for a visual representation of the two steps of initiation.

During the second step of translation, elongation, amino acids continue to be added to the polypeptide chain through three steps. In codon recognition, a molecule of tRNA carries an amino acid and pairs the anticodon of tRNA with the codon of mRNA in the A site of the ribosome. During peptide bond formation, the polypeptide separates from the tRNA in the P site and attaches by a peptide bond to the amino acid carried by the tRNA in the A site; thereby increasing the chain. During translocation, the tRNA in the P site leaves the ribosome and the tRNA in the A site moves over into the P site. The mRNA and tRNA move as a unit, which brings the next mRNA codon into the A site to allow for another amino acid to be added to the polypeptide chain. Elongation continues until a stop codon moves into the A site. If one of the three stop codons (UAA, UAG, or UGA) reaches the A site, termination occurs and the polypeptide is released from the tRNA and exits the ribosome, which causes the ribosome to split into subunits.

Procedures

  1. Extract DNA from an onion.

DNA is found in the cells of all living organisms. This procedure will provide an opportunity to extract and observe DNA from an onion. Plant cells like those found in onions contain a cell wall that needs to be broken down in order to release the DNA. In order to accomplish this, liquid dish detergent combined with mechanical disruption will be utilized. The dish detergent contains substances that break down the fats that make up the cell wall thereby releasing the cell contents (including DNA). Additional cell wall disruption is accomplished by mashing the onion, or blending it in a blender or food processor. A combination of detergent, mechanical disruption, and heat facilitates the breakdown of fats and proteins that comprise the cell wall. Salt is also needed in the experiment to help the DNA precipitate out of solution. Once DNA has been released from the cells into a solution of detergent, water, and salt, it must be extracted from the solution. In order to extract the DNA from solution, alcohol is added. Alcohol is used because DNA is not soluble in alcohol, and it precipitates into the alcohol and becomes visible as cloudy white strands.

    1. Place the alcohol in the freezer to chill. Alternatively, if a freezer is not available, create a water bath for the alcohol by placing 2-3 cups of ice into a mixing bowl and then placing the jar of alcohol into the ice.
    2. Using a knife and cutting board, coarsely peel and then chop an onion (discarding the peel).
    3. Place a piece of tape on each of two large drinking glasses. Using a marker or pen, label the first glass “heat” and the second glass “room temperature.”
    4. Prepare a warm water bath by placing 2 cups of tap water into a small pot. Carefully lower the glass labeled “heat” into the water. Be sure that there is not too much water in the pot; the water should only be about halfway up the side of the glass. Adjust the level of tap water in the pot, if needed, so that the water is about halfway up the glass (remove or add water, as needed). Remove the drinking glass. Place the pot of water on the stove and begin heating on low heat. Using the thermometer, check the temperature of the water. Optimal temperature is 55-60°C (130-140°F). If the temperature is too high, it will destroy the DNA.

Procedures for extraction using a blender/ food processor ***Note: The DNA yield will be greater if a blender or food processor is used so this is the preferred method.

    1. Place the onion pieces into a blender or food processor.
    2. Using the 100 ml graduated cylinder, measure out 120 mL of water and pour it into the blender or food processor.
    3. Using the teaspoon, add 2 teaspoons of salt to the onion and water. Blend or process the mixture for one minute.
    4. Pour the onion, salt, and water into a mixing bowl.
    5. Add 1 tablespoon of liquid detergent to the onion mixture and stir. Let stand for 10 minutes.
    6. Place a coffee filter or several folded paper towels into the funnel and then place the funnel over the cup labeled “heat.”
    7. Strain half of the onion mixture by pouring it slowly into the filter-lined funnel and allow the mixture to drain for a minute. Using a teaspoon, gently mix the solution to allow as much liquid as possible to drain out.
    8. Discard the onion and filter paper in the trash and retain the liquid
    9. Using the thermometer, check the temperature of the water bath and ensure it is between 55-60°C (130-140°F). If the water is not at the correct temperature, adjust the heat (e.g., increase the heat level to raise the temperature, or add some ice to the water and turn off the heat to decrease the temperature).
    10. Place the glass labeled “heat” into the water bath for 5 minutes. Check the temperature to ensure it does not rise above 60°C (140°F).
    11. After 5 minutes, mix the onion solution, recheck the temperature of the water, and then leave the glass in the water bath for another 5 minutes.
    12. After the glass has been in the water bath for 10 minutes, gently remove the glass from the water bath and place it on a heat resistant surface. Allow the glass to come to room temperature for 5 minutes.
    13. Remove the container of alcohol from the freezer or ice bath. Using the graduated cylinder, measure out 50 ml of chilled alcohol and then return the alcohol to the freezer or ice bath.
    14. Add the alcohol to the onion, detergent, and salt solution by tipping the glass and then slowly pouring the alcohol down the inside of the glass. The alcohol should form a layer on top of the onion mixture; if the alcohol mixes with the onion solution, the DNA will not be extracted from the solution. The DNA is soluble in water, but not in alcohol, and therefore, the DNA should be extracted from the onion solution into the alcohol. Let the mixture stand for 5-10 minutes. The DNA will appear as stringy, cloudy, white masses or threads in the alcohol-onion solution interface.
    15. While waiting, prepare the rest of the onion for extraction.
    16. Repeat steps M and N for the second glass labeled “room temperature.”
    17. If desired, the DNA can be preserved and removed for closer examination by placing it into a sealed container filled with alcohol. Pour some alcohol into a sealed container and then use a wooden skewer or toothpick to slowly pull the DNA out of the alcohol, along the inside of the glass and then place the DNA into the sealed container.

Procedures for extraction without a blender/food processor

    1. Using the knife, chop the onion into medium sized pieces.
    2. Place the onion into a mixing bowl.
    3. Using a graduated cylinder, measure out 120 ml of water and pour it into the mixing bowl with the onion.
    4. Add 2 teaspoons of salt to the onion and water.
    5. Using a tablespoon, add 1 tablespoon of liquid detergent to the onion, water, and salt. Stir the mixture.
    6. Using the back of a spoon or a fork, press the onion against the side of the mixing bowl; crushing the onion to help break apart the cell walls. Let the mixture stand for 10 minutes.
    7. Place a coffee filter or several folded paper towels into the funnel and then place the funnel over the cup labeled “heat.”
    8. Strain half of the onion mixture by pouring it slowly into the filter-lined funnel and allow the mixture to drain for a minute. Using a teaspoon, gently mix the solution to allow as much liquid as possible to drain out.
    9. Retain the balance of the onion mixture but discard the filter paper in the trash.
    10. Using the thermometer, check the temperature of the water bath and ensure it is between 55-60°C (130-140°F). If the water is not at the correct temperature, adjust the heat (e.g., increase the heat level to raise the temperature, or add some ice to the water and turn off the heat to decrease the temperature).
    11. Place the glass labeled “heat” into the water bath for 5 minutes. Check the temperature to ensure it does not rise above 60°C (140°F).
    12. After 5 minutes, mix the onion solution, recheck the temperature of the water, and then leave the glass in the water bath for another 5 minutes.
    13. After the glass has been in the water bath for 10 minutes, gently remove the glass from the water bath and place it on a heat resistant surface. Allow the glass to come to room temperature for 5 minutes.
    14. Remove the container of alcohol from the freezer or ice bath. Using the graduated cylinder, measure out 50 ml of chilled alcohol and then return the alcohol to the freezer or ice bath.
    15. Add the alcohol to the onion, detergent, and salt solution by tipping the glass and then slowly pouring the alcohol down the inside of the glass. The alcohol should form a layer on top of the onion mixture. If the alcohol mixes with the onion solution, the DNA will not be extracted from the solution. The DNA is soluble in water, but not in alcohol and therefore the DNA should be extracted from the onion solution into the alcohol. Let the mixture stand for 5-10 minutes. The DNA will appear as stringy, cloudy, white masses or threads in the alcohol-onion solution interface.
    16. While waiting, prepare the rest of the onion for extraction.
    17. Repeat steps N and O for the second glass labeled room temperature.
    18. If desired, the DNA can be preserved and removed for closer examination by placing it into a sealed container filled with alcohol. Pour some alcohol into a sealed container and then use a wooden skewer or toothpick to slowly pull the DNA out of the alcohol, along the inside of the glass and then place the DNA into the sealed container.
  1. Prepare the DNA template cards and a data sheet for the results.
    1. Use a tabletop to complete this exercise. Using a pair of scissors, cut out each of the 14 DNA template cards and the ribosome. The first DNA card should look like the following:

#1

DNA =

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

mRNA=

    1. Prepare a sheet of paper for the purpose of recording the results of the laboratory. On a sheet of printer paper, create a numbered list from one to fourteen along the length of the paper to correspond to each DNA template card.
  1. Simulate the process of transcription.
    1. Select DNA code card #1. The sequence of nucleotide bases is shown in the example above (2.A.).
    2. With a pencil or pen, mark off the letters into groups of three, representing individual DNA base triplets. In this example, separate the DNA as follows:

TAC / TTA / CCG / AGA / TTC / TTG / TTT / ATC

    1. Transcribe the DNA language into mRNA language. Using the space below the mRNA, convert the DNA code into the corresponding mRNA code. Using a pencil or pen, convert each DNA base into the corresponding mRNA base. Remember that in RNA, the base uracil is substituted for the DNA base thymine.
    2. If the step was performed correctly, mRNA should read:

AUG / AAU / GGC / UCU / AAG / AAC / AAA / UAG

    1. Fold over the template card in half, along the line located in between the DNA code and the mRNA code. Refer only to the mRNA code for the remaining steps.
    2. Repeat steps under procedure 3 for DNA template cards numbered 2 through 14.
  1. Simulate the process of translation.
    1. Simulate the processes of translation by converting a DNA code into an amino acid sequence.
      1. Select DNA template card #1. Place the ribosome on the mRNA start codon. If this step was performed correctly, the following letters should appear in the hole: AUG. AUG is the start codon. AUG codes for the amino acid methionine (Met).
      2. Move the ribosome to the right along the strand of mRNA.
      3. Identify the amino acid that corresponds to the codon. To find the correct amino acid that the tRNA is carrying, consult the Universal Genetic Code card in the lab kit, or the corresponding figure in the textbook. Using GAA as an example, GAA codes for the amino acid Glutamic acid (Glu).
      4. Using a pen or pencil, record the name and three letter abbreviation of the amino acid on the data sheet, next to the number 1. Work along the rest of the mRNA strand in a similar fashion and complete the polypeptide by converting each mRNA codon into the complementary amino acid that the codon codes for. If this step is completed correctly, a list of amino acids should be provided similar to the following example: start(Met)-asparagine(Asn)-glycine(Gly)-serine(Ser)-lysine(Lys)-asparagine(Asn)-lysine(Lys)-stop
      5. Repeat steps under Procedure 4.a. for the DNA template cards 2-14 and record the chain of amino acids for each card on the data sheet.

Assessing Your Learning

Compose answers to the questions below in Microsoft Word and save the file as a backup copy in the event that a technical problem is encountered while attempting to submit the assignment. Make sure to run a spell check. Copy the answer for the first question from Microsoft Word by simultaneously holding down the Ctrl and A keys to select the text, and then simultaneously holding down the Ctrl and C keys to copy it. Then, click the link on the Lab Preview Page to open up the online submit form for the laboratory. Paste the answer into the online dialog box by inserting the cursor in the box for question one and simultaneously holding down the Ctrl and V keys. The answer should now appear in the box. Repeat this process for each question. Review all work to make sure that all questions have been completely answered and then click on the Submit button at the bottom of the page.

  1. How does DNA differ from RNA? (2 points)
  2. Where in the cell does each of the following processes take place? (2 points)
    1. Transcription
    2. Translation
  3. The following is a base sequence on a strand of DNA. When the strand replicates, what is the base sequence on the complementary strand of DNA? (2 points)

AATCGCATACCCGGTCAG

  1. Transcribe the following DNA molecule into mRNA: TAACCTGGACTACAAATC. (2 points)
  2. What is a promoter? (2 points)
  3. What is the role of the enzyme DNA polymerase? (2 points)
  4. What is the end product of transcription? (2 points)
  5. Why would it be impossible to extract DNA from cooked foods? (2 points)
  6. Which DNA extraction procedure (warm water versus room temperature water) produced a greater quantity of DNA? Explain why the results turned out the way that they did. (6 points)
  7. Provide the amino acid sequence for DNA template cards 1 through 5 (from Procedure 4 in the lab)? (10 points)
    1. DNA template card # 1:
    2. DNA template card # 2:
    3. DNA template card # 3:
    4. DNA template card # 4:
    5. DNA template card # 5:
  8. What is an exon? (2 points)
  9. What three molecules are needed for the process of translation? (3 points)
  10. What are the building blocks of proteins called? (2 points)
  11. List the three stop codons. (3 points)
  12. Provide the amino acid sequence for DNA template cards 6 through 10 (from Procedure 4). (10 points)
    1. DNA template card # 6:
    2. DNA template card # 7:
    3. DNA template card # 8:
    4. DNA template card # 9:
    5. DNA template card # 10:
  13. Define the following terms: (2 points)
    1. Point mutation
    2. Frameshift mutation
  14. Name the type of mutation that occurred below. (2 points)

Original strand of DNA: CTAGGCCTAACTGCC

Replicated strand of DNA: CTAGGCCAAACTGCC

  1. Provide the amino acid sequence for DNA template cards 5 and 11 through 14 (from Procedure 4). (10 points)
    1. DNA template card #5:
    2. DNA template card # 11:
    3. DNA template card # 12:
    4. DNA template card # 13:
    5. DNA template card # 14:
  2. DNA template cards 11 through 14 represent mutations that occurred in template card five. Review the amino acid sequences for cards 5 and 11 through 14 and answer the following: (10 points)
    1. What type of mutation occurred in cards 11 and 12?
    2. Did the mutation in card 11 result in a change in the protein synthesized?
    3. Did the mutation in card 12 result in a change in the protein synthesized?
    4. What type of mutation occurred in cards 13 and 14?
    5. Would the mutations that occurred in cards 13 and 14 result in a change in the protein synthesized?
  3. A base deletion of base eight (frameshift mutation) occurs in the following nucleotide sequence: CTCAATGAAGGCCTA. (4 points)
    1. Write the codons for the nucleotide sequence.
    2. Write the codons for the nucleotide sequence following the base deletion.
  4. (Application) How might the information gained from this lab pertaining to protein synthesis be useful to a student employed in a healthcare related profession? (20 points)

 

Answer

49.99