Protein Synthesis

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BACKGROUND

DNA carries the information for the synthesis of all the proteins of an organism. Proteins are large and complex molecules, composed of hundreds of amino acids. There are twenty different amino acids. The sequence of the nucleotides in the DNA molecule determines the sequence of amino acids in a protein molecule.

The information encoded in the nucleotide sequence of DNA is not used directly in protein synthesis. Instead, the information carried in the DNA molecule is encoded, or transcribed, into a special type of RNA, messenger RNA (mRNA). This process is called transcription. In transcription, the nucleotide thymine is replaced by uracil, so the pairing of nucleotides becomes A-U, while C-G remains the same. Once transcription has finished, the mRNA passes out of the nucleus and travels to a ribosome. Here the mRNA is read, and the correct amino acids are brought into specified places by another type of RNA, transfer RNA (tRNA). The amino acids are then linked together by peptide bonds to form the protein originally specified by DNA.

OBJECTIVES

In this activity you will:

  1. Learn how the transcription of DNA occurs during protein synthesis.
  2. Become familiar with the code by which the information in mRNA is translated.
  3. Use paper models to see how translation of mRNA occurs during protein synthesis.

MATERIALS

1/2-in transparent tape

Glue Sticks

PROCEDURES AND OBSERVATIONS

During transcription, the DNA double helix unzips. As the hydrogen bonds between the two DNA strands break, nucleotides present in the cell line up next to the nucleotides of one DNA strand to form mRNA. RNA contains uracil in place of the thymine in DNA. Thus in the pairing of RNA nucleotides with DNA, uracil pairs with adenine and cytosine with guanine. The nucleotides in the newly formed mRNA are complementary to the nucleotides of the DNA segment on which it formed. For example, where the DNA contained guanine, the RNA, contains cytosine. After the mRNA is formed, it passes out of the nucleus and into the cytoplasm.

a. One strand of DNA has the base sequence:

C G A T T G G C A G T C A T. Determine the sequence of bases in the complementary strand of mRNA that would form next to this DNA strand.

  1. Write the sequence of bases in the complementary mRNA strand below.

The information carried on the mRNA is in a code the genetic code. A group of three nucleotides on a molecule of mRNA is called a codon; each codon specifies one of the twenty amino acids, except for three codons that are stop, or termination, signals. There are sixty-four codons in the genetic code. Most amino acids are coded for by more than one codon. The genetic code and the abbreviations for the twenty amino acids are given in Figure 1.

GENETIC CODE

 

U

C

A

G

Abbreviations for the Amino Acids

UU

phe

phe

phe

phe

 

 

UC

ser

ser

ser

ser

gly

ala

val

ile

leu

ser

thr

pro

asp

glu

lys

arg

asn

gin

cys

met

trp

phe

tyr

his

stop

glycine

alanine

valine

isoleucine

leucine

serine

threonine

proline

aspartate

glutamate

lysine

arginine

asparagine

glutamine

cysteine

methionine

tryptophan

phenylalanine

tyrosine

histidine

termination

UA

tyr

tyr

stop

stop

UG

up

up

stop

trp

CU

leu

leu

leu

leu

CC

pro

pro

pro

pro

CA

his

his

gin

gin

CG

arg

arg

arg

arg

AU

stop

stop

stop

met

AC

thr

thr

thr

thr

AA

asn

asn

lys

lys

AG

ser

ser

arg

arg

GU

val

val

val

val

GC

ala

ala

ala

ala

GA

asp

asp

glu

glu

GG

gly

gly

gly

gly

  1. Use the genetic code to read the codons below. Find the first two letters of the codon in the left column of the chart. When you have found them, follow the row across to the column beneath the last letter of the codon. This is the abbreviation of the amino acid specified. Find the name of amino acid and write it in the space provided. (If there is more than one amino acid, separate the names by dashes.)
  1. U U A:
  2. G A G:
  3. UAU CUA:
  4. AUC UUG: .
  5. AAG AGU UCG:
  6. AAA UUU GGG:
  7. CCA GCU AGA GGG UGG CUG UCA:
  8. Molecules of tRNA are formed in the nucleus and migrate into the cytoplasm. There are twenty different types of tRNA, one for each type of amino acid. Each type of tRNA can grab one kind of amino acid molecule. Each tRNA has a three-base segment called an anticodon, which is complementary to a codon on the mRNA. In protein synthesis, a tRNA molecule carrying an amino acid molecule becomes attached briefly to the mRNA at a codon complementary to its anticodon. Then a tRNA complementary to the adjacent codon attached on the mRNA. A peptide bond forms between the two amino acid molecules carried by the tRNAs. Amino acids are added one at a time to the growing chain as the mRNA strand is "read." This continues until a termination (stop) codon is encountered. After each peptide bond is formed, the tRNA is released to go and pick up a new amino acid in the cytoplasm.

    Determine the anticodon for each codon below. Write it in the space provided.

  9. G G U:
  10. C G C:
  11. A U G:
  12. U C G:
  13. A A A:
  14. C U G:
  1. Cut out the tRNA models with amino acids attached on page 7. Spread them out on the table in front of you. Then cut out the mRNA strands and tape them together.
  2. Starting from the left of the mRNA strand, find a tRNA molecule with an anticodon complemen-tary to the first codon. With a small piece of tape, attach the tRNA to the mRNA strand, anticodon to codon.
  3. Then go on to the next codon and find a tRNA with the complementary anticodon. Tape the tRNA in place to the mRNA. Also, use a small piece of tape between the two amino acids to represent a peptide bond.
  4. Once the peptide bond has been formed, the tRNA molecule attached first is released. Carefully cut the tape attaching the first tRNA to the mRNA, and cut the line that separates the tRNA and the amino acid. You may set the tRNA model aside and discard it later.
  5. There are three termination codons in the genetic code. When the termination codon is read, the strand of amino acids is released. In the cell, the strand folds and twists to assume the final, complex structure of the protein. The mRNA is destroyed.

  6. Repeat steps c and f along the mRNA strand. When you have used up all the tRNA-amino acid models provided, you will notice that there is one codon left on the mRNA—a termination codon. Cut the tape between the mRNA and the tRNA, and the line between the last tRNA and amino acid, thus releasing the chain of amino acids.
  1. Write the sequence of the amino acids formed by translation of the mRNA strand.

CONCLUSIONS AND APPLICATIONS

  1. Write the mRNA that would be transcribed from the following strand of DNA. Then read the mRNA, and write the names of the amino acids coded for.
  2. DNA G T A T A C C A G T C A T T T G T C

    mRNA

    amino acids

  3. Suppose the bases of the DNA in question 1 were not transcribed correctly and the mRNA read: CAAAUGGUCAGUAAACAG How many mistakes were made in transcription? Write the amino acids that would be formed by translation of this mRNA. '
  4. Suppose the bases of the DNA were not transcribed correctly and the mRNA read: CACAUGGUUAGUAAGCAG How many mistakes were made in transcription? Write the amino acids that would be formed by translation of the mRNA.
  5. Sometimes a mistake occurs in the translation of an mRNA strand. Suppose that the reading of the mRNA strand in question 1 began, by mistake, at the second nucleotide instead of the first. The first codon would be AUA. Write the sequence of amino acids that would be formed.

GENETIC CODE CHART

 

U

C

A

G

 

U

UUU

Phe

UCU

Ser

UAU

Tyr

UGU

Cys

U

UUC

Phe

UCC

Ser

UAC

Tyr

UGC

Cys

C

UUA

Leu

UCA

Ser

UAA

Stop

UGA

Stop

A

UUG

Leu

UCG

Ser

UAG

Stop

UGG

Trp

G

C

CUU

Leu

CCU

Pro

CAU

His

CGU

Arg

U

CUC

Leu

CCC

Pro

CAC

His

CGC

Arg

C

CUA

Leu

CCA

Pro

CAA

Gln

CGA

Arg

A

CUG

Leu

CCG

Pro

CAG

Gln

CGG

Arg

G

A

AUU

Ile

ACU

Thr

AAU

Asn

AGU

Ser

U

AUC

Ile

ACC

Thr

AAC

Asn

AGC

Ser

C

AUA

Ile

ACA

Thr

AAA

Lys

AGA

Arg

A

AUG

Met

ACG

Thr

AAG

Lys

ACG

Arg

G

G

GUU

Val

GCU

Ala

GAU

Asp

GGU

Gly

U

GUC

Val

GCC

Ala

GAC

Asp

GGC

Gly

C

GUA

Val

GCA

Ala

GAA

Glu

GGA

Gly

A

GUG

Val

GCG

Ala

GAG

Glu

GGG

Gly

G

 

GGU

AUC

GUU

GAA

GAG

UGU

UGC

GCU

 

 

 

 

 

 

 

 

UCC

GUG

UGC

AGU

CUG

UAC

CAA

 

 

 

 

 

 

 

 

 

CUA

GAA

AAC

UAC

UGC

AAU

UAA