Biology SME copy 1

Key unit competence: Explain nucleic acids, DNA replication, and the process of protein synthesis in eukaryotes.

Introductory activity 3

1. Three men A, B and C are at police station for being investigated as rapists. They are accused of forcing a woman to have sex with her and finally the woman gets pregnant and she is just given birth to a baby girl. All of those three men do not accept to be real father of a baby girl, the police brings men A,B and C to the forensic laboratory , the DNA samples of three men are collected and analyzed together with the one of the child and the mother in order to find out the potential father. The table below indicates the results obtained

a. Basing on the results of table who should be the real father of baby girl between A, B and C 

b. Explain why man named in a, is a real father of baby girl?

3.1 Structure of nucleic acids

Activity 3.1

Detecting components of DNA /RNA structure 

 Observe the diagrams below (a, b, c, d and d)

 Do you see the resemblance? Which parts of the DNA molecule are like the steps and the spiral of the following figures? Use the tools indicated (like thread, toothpick and balls with different colurs) and construct the structure of DNA

DNA stands for deoxyribonucleic acid and RNA for ribonucleic acid. The nucleic acids such as DNA and RNA, like proteins and polysaccharides, are macromolecules. They are also polymers made up of many similar, smaller molecules joined into a long chain. The smaller molecules from which DNA and RNA molecules are made are nucleotides. DNA and RNA are therefore polynucleotides. They are often referred to simply as nucleic acids.

3.1.1 Nucleotide

Nucleotides are made up of three smaller components. These are: 

• a nitrogen-containing base 

• a pentose sugar 

• a phosphate group.

 There are just five different nitrogen-containing bases found in DNA and RNA. In a DNA molecule, there are four: adenine, thymine, guanine and cytosine 8080 while in an RNA molecule also contains four bases, but instead of having the base thymine RNA has base called uracil rather than thymine.. These bases are often referred to by their first letters: A, T, C, G and U.

The pentose (5-carbon) sugar either ribose in RNA or deoxyribose in DNA molecules. As their names suggest, deoxyribose is ribose that has one fewer oxygen atoms in its molecule from which DNA and RNA molecules can be built up.

Figure 3.1 Nucleotides. A nucleotide is made of a nitrogen containing base, a pentose sugar and a phosphate group P

In DNA and RNA, bases are covalently bonded to the 1’ carbon of the pentose sugar. The purine and pyrimidines bases attached to pentose sugar from 81 different positions of their nitrogen bases. Purine bases use the 9th position of nitrogen to attach with 1’ carbon of pentose sugar, while pyrimidine bases use the 1st position of nitrogen to attach with 1’ carbon of pentose sugar. In both DNA and RNA, the phosphate group (PO4 2–) attaches to the 5’ carbon of pentose sugar. Thus, by attaching phosphate group to a nucleoside yields a nucleoside phosphate or nucleotide.

3.1.2 Chargaff’s rules: Rules of base pairing 

Erwin Chargaff’s rules state that DNA of any cell in organism should have a 1:1 ratio of pyrimidine and purine bases, where the amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine (C). This equivalence of purine and pyrimidine bases is known as Chargaff’s rules. This pattern is found in both strands of DNA.

Table 3.1: Difference between purine and pyrimidine

The specific base pairing of A-T bases and G-C bases is called complementary base pairs.

For example, if one strand of DNA sequence is 5’-ATATCCGGAT-3’, then the opposite strand of DNA sequence will be 3’-TATAGGCCTA-5’. Thus, by using the rules of base pairing, once we have the sequence of at least DNA strand, we can find out the opposite base sequence of that DNA. In the structure of DNA, the strong electronegative atom is the Oxygen (O) and Nitrogen (N), while H atom has positive charge. In the structure of DNA (Figure 3.5), thymine and adenine have two hydrogen bonds; while guanine and cytosine have three hydrogen bonds. Hydrogen bonds or interactions play very important role in binding the bases of the opposite strands in the DNA. Though RNA is not the genetic material in most of the cases, both single-stranded and double-stranded RNAs are the genomes of certain viruses. RNA double-stranded molecules show structural similarity to that of double-stranded DNA molecules. The similarities are:

a. Both have anti-parallel strands.

b. Both have sugar-phosphate backbones on the outside of helical molecule.

In both the cases, in the middle of the helix, a complementary base pairing is formed by hydrogen bonds.

3.1.3 WATSON and CRICK hypothesis of the nature of DNA

In 1953, James D. Watson, an American molecular biologist, and Francis H.C. Crick, a British molecular biologist, proposed a model for the physical and chemical structure of the DNA molecule. Today, their model is known as double helix model of DNA or simply the Structure of DNA.

The main features of Watson and Crick double helix model (Figure 3.6) of DNA are:

a. Two polynucleotide chains wind around each other in a right-hand double helix (Figure 3.12). 

b. The two polynucleotide chains run side-by-side in an antiparallel fashion. This means that one strand of DNA will orient itself in a 5’ -3’ direction, whereas, the other strand will orient itself alongside the first one in a 3’-5’ direction. In this way, the two strands are oriented in opposite directions (Figure 3.13).

c. On one hand, the sugar-phosphate backbones lie outside of the double helix. On the other hand, the bases orient themselves toward the central 8484 axis of the double helix structure. The bases of one strand are bonded with the bases of the other strand of double helix by hydrogen bonds. These bonds are weak chemical bonds. Since hydrogen bonds are relatively weak bonds, the two strands can be easily separated by heating the DNA. The bonding of these bases in the double helical structure follows the Chargaff’s base pairing rules. For example—Adenine (A) will form a hydrogen bond with Thymine (T). Similarly, Guanine (G) will form a hydrogen bond with Cytosine (C). This specific base paring is called complementary base pairing (Figure 3.6).

Watson and Crick proposed the double helix model for DNA. (a) The sugarphosphate backbones are on the outside of the double helix and purines and pyrimidines form the DNA helix ladder. (b) The two DNA strands are antiparallel to each other. (c) The direction of each of each strand is identified by numbering the carbons (1 through 5) in each sugar molecule. The 5’ end is the one where carbon 5 is not bound to another nucleotide; the 3’ end is the one where carbon 3 is not bound to another nucleotide Source: 3.1.4.

3.1.4 Types and functions of RNA

The three most well-known and most commonly studied are messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA), which are present in all organisms. These and other types of RNAs primarily carry out biochemical 85 reactions, similar to enzymes. Some, however, also have complex regulatory functions in cells. Owing to their involvement in many regulatory processes, to their abundance, and to their diverse functions, RNAs play important roles in both normal cellular processes and diseases.

Major differences between DNA and RNA 

The three major structural differences of RNA from that of DNA are: 

 a. RNA contains ribose sugar instead of 2’-deoxyribose. It means that ribose has a hydroxyl group (OH) at the 2’ position, whereas, deoxyribose has hydrogen (H) at 2’ position in pentose sugar. 

1. RNA has Uracil (U), whereas DNA has thymine (T). 

2. Unlike DNA, which consists of two polynucleotide chains, in most cases, RNA is found in a single polynucleotide chain.

Table 3.2: Differences between DNA and RNA

Application activity 3.1

The diagram below represents a complex organic substance found in living cells. Observe carefully this diagram and answer the questions that follow 

1. Name that complex organic substance 

2. Name A, B and C

 3. Create a diagram showing another complex organic substance found in living cells

3.2 Mechanism of DNA replication

Activity 3.2

The diagram below shows how DNA replicates, 

a) Explain the mechanism of DNA replication basing on this diagram.

b) Using a school library, search additional information on the internet and make short summary on DNA replication.

3.2.1 Different models of DNA replication

DNA is often described as a double helix. This refers to the three-dimensional shape that DNA moleculesform. The hydrogen bonds linking the bases, and therefore holding the two strands together can be broken relatively easily. This happens during DNA replication (DNA copying) and also during protein synthesis (protein manufacture). As we shall see, the breaking of the hydrogen bonds is a very important feature of the DNA molecule that enables it to perform its role in the cell. RNA molecules, unlike DNA, remain as single strands of polynucleotide and can form very different three-dimensional structures. We will look at this later in the chapter when we consider protein synthesis.

1. Semi-conservative model

In 1953, Watson and Crick proposed their classic paper postulating a double helix for DNA. A month later, they published another paper suggesting how such base-paired structures in DNA might duplicate itself. The essence of Watson and Crick suggestion is that if DNA molecule was untwisted and the two strands get separated, each strand could act as a template for the synthesis of a new complementary strand of DNA. And this new complementary strand could then be bound to the parental strand of DNA. This model replication is known as the semiconservative model. It is because half of the parent strand of DNA is retained by newly formed daughter DNA strand. Experimental evidence of semi-conservative DNA replication

In 1958, Matthew Meselson and Franklin Stahl used two isotopic forms of nitrogen, 14N (light) and 15N (heavy), to distinguish newly synthesized strands of DNA from old strands. 

 - Initially, Meselson and Stahl grew E. coli (bacteria) for many generations in a medium containing 15N-labelled ammonium chloride (15NH4 Cl) to incorporate this heavy isotope of nitrogen into their DNA molecule. As expected, the DNA strands in the bacteria had ¹⁵N-¹⁵N (heavy) DNA (Figure 3.7). 

- In the second stage, they transferred the 15N-labelled bacteria to a medium containing nitrogen in the normal 14N form (light). Then the bacteria were allowed to reproduce for several generations. Since, the bacteria were grown in the normal 14N form, the entire newly synthesized DNA after the transfer was now labelled with ¹⁴N. 

- Samples of E. coli were taken at various time periods as the bacteria continued to reproduce in the medium. The DNAs from these bacteria were extracted and analyzed to determine its density. They determined the density of extracted DNAs by using equilibrium density gradient centrifugation technique. This technique uses Cesium Chloride (CsCl), a heavy metal salt that forms solutions of very high density. Thus, they analyzed the extracted DNA by simply mixing it with a solution of cesium chloride and then centrifuged at high speed.

In ¹⁵NH₄ Cl compound, the normal isotope of nitrogen, ¹⁴N, is replaced with ¹⁵N, which is a heavy isotope. Since the density is equal to weight divided by volume, the ¹⁵N which has one extra neutron in its nucleus is 1/14 times denser than ¹⁴N. 

As a density gradient of cesium chloride is established by the centrifugal force, the DNA molecules float “up” and sink “down” within the gradient to reach their equilibrium density positions. The difference in density between the heavy (¹⁵N) DNA and the light (¹⁴N) DNA causes DNA molecules to rest at different positions by forming bands in the gradient (Figure 3.7)

Final observations

- First generation (After one replication cycle)

When the observation was made after one replication cycle in the ¹⁴N medium, the entire DNA had a density that was exactly intermediate between that of ¹⁵N-¹⁵N DNA and that of ¹⁴N-¹⁴N DNA. The intermediate composition was ¹⁵N-¹⁴N DNA.

- Second generation (After two replication cycles)

 Again, when the observation was made after two replication cycles, half of the DNA was that of intermediate density (¹⁵N-¹⁴N DNA) and half was that of the density of ¹⁴N-¹⁴N DNA.

The observations made in this experiment exactly tested and proved the predication of the semi-conservative model. Therefore, through this experiment it has been known that DNA replication follows semi-conservative model. At the same time, it disproved the claim that DNA replication follows either conservative or dispersed replication models.

2. Conservative DNA replication model

 In this model, the two parental DNA strands come together right after replication; and as a whole, these two parental DNA strands serve as template for the synthesis of completely new daughter DNA strands. As a result, one daughter DNA molecule contains parental DNA strands, while the other daughter DNA molecule contains newly synthesized DNA strands 

 3. Dispersive DNA replication model

 In this model, the parental double helix is broken or cleaved into double-stranded DNA that acts as templates for the synthesis of new double helix molecules. The segments then reassemble into complete DNA double helices, each with parental and daughter DNA segments interspersed. After the replication, although the two daughter DNA molecules are identical in their base pair sequence, the parental double stranded DNA has become dispersed throughout both in the daughter DNA molecules

3.2.2 Importance of DNA replication

Genes duplicate themselves very accurately by DNA replication. The three main importance of DNA replication are:

Reproduction: One of the most fundamental properties of all living things is the ability to reproduce. It is through reproduction that parents faithfully pass on their genetic information specifying their structure and function to their young ones. At organism level, organisms reproduce either by asexual or sexual reproduction methods. At cellular level, cells duplicate by cellular division. And at the genetic level, the genetic material duplicates by DNA replication.

Repair: DNA is the centre of instructions that govern the cell’s protein production, growth, and many other activities in the cells. With this enormity of precise responsibility, any minor mistakes in the replication process can bring 93 potentially harmful changes in the cell’s behaviour or for that matter, the whole organism. Therefore, DNA employs various error repair mechanisms to ensure accurate DNA replication.

Growth: DNA Replication is required for the growth of organisms. DNA replication occurs in two different forms of cellular division. They are mitosis and meiosis. In mitosis, a single parent cell divides and gives rise to two identical daughter cells. Each of the daughter cells has the exact amount of genetic material. For example, growth of limbs, organs, hair etc. On the other hand, in meiosis, cells divide and give rise to two haploid sex cells. Thus, DNA replication plays a vital role in both mitosis and meiosis.

3.2 3 Enzymes and proteins involved in DNA replication

a. DNA polymerases

In 1955, Arthur Kornberg and his colleagues were the first ones to identify an enzyme that could synthesize DNA. Back then this enzyme was originally called Kornberg enzyme. But now it is called DNA polymerase I. (enzymes that catalyzes the synthesis of DNA). There are five DNA polymerases: DNA polymerase I, DNA polymerase II, DNA polymerase III, DNA polymerase IV and DNA polymerase V

On one hand, DNA polymerase I and III are functionally required for replication. DNA polymerase III along with other DNA polymerases (I and II) has the capability to elongate an existing DNA strand. But cannot initiate DNA synthesis, DNA polymerases can polymerize nucleotides only in 5′ → 3′ direction.

b. DNA helicase 

DNA helicase is an enzyme that unwinds or unzips the double stranded DNA by breaking the hydrogen bonds between the complementary bases.

The action of DNA helicase can be compared with a zipper. When we open a zip, the zipper runs on a zip and makes a Y-shape structure with the two strands of interlocking teeth. In the same way, DNA helicase unzips the double stranded DNA and form a Y-shaped fork known as a replication fork.

c. Single-strand DNA-binding proteins (SSB) 

In DNA replication when helicase unwinds the double stranded DNA, the two separating strands of DNA have the tendency to reform or reanneal into double stranded DNA. A protein called single-strand DNA-binding (SSB) proteins bind to each single-strand DNA and stabilize them, so that the separating two strands of DNA do not reform double stranded DNA by complementary base pairing (Figure 3.9). 

d. DNA ligase

 At the end of DNA replication right after the DNA Pol I is removed and replaced all the RNA primer nucleotides with DNA nucleotides, normally as single-strand nick (gap) is left between the two DNA fragments (Figure 3.9). This nick is the point where the sugar-phosphate backbone between adjacent nucleotides is unconnected. So, what DNA ligase does is that it joins the two fragments resulting into a longer and continuous DNA strand.

 Chemically, DNA ligase catalyzes the formation of a phosphodiester bond between the 3′-OH and the 5′-phosphate groups on either side of a nick. As a result, it seals the nick (gap).

Figure 3.10: Flow diagram showing DNA ligase sealing the gap in a new DNA strand

3.2.4. Genes and chromosome 

 The DNA molecule is packaged into thread-like structures called chromosomes in cell. Chromosomes are not visible in the cell’s nucleus, not even under a microscope, when the cell is not dividing. Each chromosome is made up of DNA tightly coiled many times around proteins called histones that support its structure. 

A gene is the basic physical and functional unit of heredity. Genes are made up of DNA. Some genes act as instructions to make molecules called proteins. However, many genes do not code for proteins.

3.2.5 Significance of telomere in permitting continued replication

A telomere is a region of repetitive nucleotide sequences at each of a chromosome. It protects the end of the chromosome from being deleted or from fusion with neighboring chromosomes. In vertebrates, the repetitive sequence of nucleotides in telomeres is TTAGG. In humans, this sequence is repeated about 2500 times.

Telomeres and cancer 

Telomeres maintain genomic integrity in normal cells, and their progressive shortening during successive cell divisions induces chromosomal instability. In the large majority of cancer cells, telomere length is maintained by telomerase. Shortening of telomeres is associated with each round of cell division owing to the inability of conventional DNA polymerases to replicate the ends of linear chromosomes, the so-called ‘end replication problem’. Cancer cells are characterized by their rapid and uncontrollable division of cells. These cells have active telomerase to help them divide uncontrollably and become immortal. The enzyme telomerase is used to extend the life span of cancer cells. In the absence of telomerase, the cancer cells would become inactive and would stop dividing resulting into death of the cancer cells because of Shortening of telomeres after each division, which causing the cell to die. Cancer therapies can take advantage of this concept by designing drugs that can inhibit telomerase activity, thereby killing the cancer cells. Telomere biology is an important aspect of human cancer. Many scientists are hoping and working hard to understand the best way to use anti-telomerase therapy and advance the treatment of cancer.

Application activity 3.2

Observe the diagram carefully and answer the following questions

1. Label this diagram

2. On the diagram the colors are different? Explain why?

3. On the diagram, the colors are arranged in this order e.g sky blue is paired with yellow, green –red, explain

a. why sky blue is not paired with red? 

b. Why red is not paired with yellow?

(Source: adapted from https://www.compoundchem.com/2015/03/24/dna/ )

3.3 Nature of gene, genetic code and protein synthesis 

Activity 3.3 

Research activity 

Make a research on gene, genetic code and process of protein synthesis using different biology books and internet. Make short notes on them.

3.3.1. Protein synthesis 

A complete DNA set of an organism is called genome. Protein synthesis begins with genes. A gene is a functional segment of DNA that provides the genetic information necessary to build a protein and it is also DNA that encodes for a particular trait (can be physical appearance or hidden feature kown as genetic). For example, black hair, brown hair Genes are located on the chromosomes. Each particular gene provides the code necessary to construct a particular protein.

Gene expression which is a process to transform the information coded in a gene to a final gene product, ultimately dictates the structure and function of a cell by determining which proteins are made. 

 DNA which specifies protein synthesis is confined in the nucleus while the process of protein synthesis takes place in the cytoplasm.

DNA holds all of the genetic information necessary to build a cell’s proteins. The nucleotide sequence of a gene is ultimately translated into an amino acid sequence of the gene’s corresponding protein.

Application activity 3.3

Draw and label:

1. A diagram showing transcription

2. A diagram showing all steps of transcription

A. Transcription (From DNA to RNA) 

It is known that DNA is housed within the nucleus, and protein synthesis takes place in the cytoplasm, thus there must be some sort of intermediate messenger that leaves the nucleus and manages protein synthesis. 99 This intermediate messenger is messenger RNA (mRNA) which is a single-stranded nucleic acid that carries a copy of the genetic code for a single gene out of the nucleus and into the cytoplasm where it is used to produce proteins. A gene is made up of a sequence of nucleotides that forms part of a DNA molecule that codes for a specific polypeptide.

The genetic code is the set of rules by which information encoded in genetic material (DNA or RNA sequences) is translated into proteins (amino acid sequences) by living cells using ribosome machinery. Gene expression begins with the process called transcription, which is the synthesis of a strand of mRNA that is complementary to the gene of interest. This process is called transcription because the mRNA is like a transcript, or copy, of the gene’s DNA code. 

 During transcription, a specific region on the DNA molecule un zips/ unwinds by breaking hydrogen bonds between complementary bases. This is catalyzed by an enzyme DNA helicase. The DNA strands act as a template.

Transcription begins in a fashion somewhat like DNA replication, in that a region of DNA unwinds and the two strands separate, however, only that small portion of the DNA will be split apart. The triplets within the gene on this section of the DNA molecule are used as the template to transcribe the complementary strand of mRNA (Figure 3.11). A codon is a three-base sequence of mRNA, so-called because they directly encode amino acids. Like DNA replication, there are three stages to transcription: initiation, elongation, and termination.

In the first of the two stages of making protein from DNA, a gene on the DNA molecule is transcribed into a complementary mRNA molecule

Stage 1: 

 Initiation. A region at the beginning of the gene called a promoter—a particular sequence of nucleotides—triggers the                                  start of transcription.

Stage 2: 

 Elongation. Transcription starts when RNA polymerase unwinds the DNA segment. One strand, referred to as the coding                                  strand, becomes the template with the genes to be coded. The polymerase then aligns the correct nucleic                                   acid (A, C, G, or U) with its complementary base on the coding strand of DNA. RNA polymerase is an enzyme                                 that adds new nucleotides to a growing strand of RNA. This process builds a strand of mRNA.

Stage 3:

 Termination. When the polymerase has reached the end of the gene, one of three specific triplets (UAA, UAG, or UGA)                                        codes a “stop” signal, which triggers the enzymes to terminate transcription and release the mRNA                                                  transcript.

Requirements for transcription 

- DNA molecule to act as a template.

- The appropriate enzymes i.e. DNA helicase and RNA polymerase.

- Free RNA nucleotides.

ATP as source of energy

B. Translation (From RNA to protein)

Activity 3.4 

a. The ribosome binds to the mRNA molecule to start translation of its code into a protein. What happens to the small and large ribosomal subunits at the end of translation? 

b. Using charts showing the process of translation and explain how translation takes place. 

c. In your free time, watch a video on translation and make short summary on the process of translation.

This step is sounding like translating a book from one language into another, where the codons on a strand of mRNA must be translated into the amino acid alphabet of proteins. Translation is the process of synthesizing a chain of amino acids called a polypeptide. 

 Translation requires two major aids: first, a “translator,” the molecule that will conduct the translation, and second, a substrate on which the mRNA strand is translated into a new protein, like the translator’s “desk.” Both of these requirements are fulfilled by other types of RNA. The substrate on which translation takes place is the ribosome.

Remember that many of a cell’s ribosomes are found associated with the rough ER, and carry out the synthesis of proteins destined for the Golgi apparatus. Ribosomal RNA (rRNA) is a type of RNA that, together with proteins, composes the structure of the ribosome. Ribosomes exist in the cytoplasm as two distinct components, a small and a large subunit. When an mRNA molecule is ready from 5’ end to 3’ end to be translated, the two subunits come together and attach to the mRNA. The ribosome provides a substrate for translation, bringing together and aligning the mRNA molecule with the molecular “translators” that must decipher its code. Generally, ribosome is composed of two dissociable subunits called the large and small subunits. In prokaryotes (bacteria), ribosome has a sedimentation coefficient of 70S; it is made up by 30S small subunit and 50S large subunit (Figure 6.10). In eukaryotes, ribosome has a sedimentation coefficient of 80S; it is made up of 40S small unit and 60S large unit.

The other major requirement for protein synthesis is the translator molecules that physically “read” the mRNA codons. Transfer RNA (tRNA) is a short single stranded and backed twisted helix. It has paired bases in some nucleotides and its folding form three main loops. It is also a type of RNA that ferries the appropriate corresponding amino acids to the ribosome, and attaches each new amino acid to the last, building the polypeptide chain one-by-one. Thus, tRNA transfers specific amino acids from the cytoplasm to a growing polypeptide. The tRNA molecules must be able to recognize the codons on mRNA and match them with the correct amino acid.

The tRNA is modified for this function. On one end of its structure is a binding site for a specific amino acid. On the other end is a base sequence that matches the codon specifying its particular amino acid. This sequence of three bases on the tRNA molecule is called an anticodon.

By nature in both eukaryotes and prokaryotes, the 5’ to 3’ nucleotide sequence of the coding DNA strand exactly corresponds or specifies the same N-terminal to C-terminal amino acid sequence of the encoded polypeptide (Figure 3.12).

During translation, the mRNA transcript is “read” by a functional complex consisting of the ribosome and tRNA molecules. tRNAs bring the appropriate amino acids in sequence to the growing polypeptide chain by matching their anti-codons with codons on the mRNA strand.

With four different nucleotides (A, C, G, U), a three-letter code (codon) can give 64 different possible codons (i.e. 43 = 64) or (4 × 4 × 4 = 64). These 64 possible codons are more than enough to code for the 20 amino acids found in living cells. The genetic code allows an organism to translate the genetic information found in its chromosomes (by m-RNA) into mature functional proteins.

3.3.2 Characteristics of genetic code

The following are some characteristics of genetic code: 

1. Genetic code as a triplet codon: A codon consists of a group of three nucleotides. And each codon codes for a specific amino acid in a polypeptide chain with some exceptions. 

2. Genetic code is used without comma: The three nucleotides in a codon are read in a continuous fashion without any comma. Examples: AUG, UAG, UGA and UAA. 

3. Genetic code is non-overlapping: The codons in the m-RNA sequence are read successively without overlapping.

 4. Genetic code is almost universal: For many long years, it was thought that the genetic code is universal, which led us into believing that all living organisms have the same genetic code. However, recent studies have revealed that there are some organisms where there is difference in genetic code (Table 3.3). That is the reason why it is appropriate to use the phrase “almost universal” rather than the word “universal.” The examples of organisms or organelles where genetic codes have different meanings:

Table 3.3: Genetic code

5. Genetic code is “degenerate”: A codon is thought to code for a particular amino acid. That is one codon for one amino acid. But more than one codon can code for a particular amino acid, with two exceptions of AUG and UGG. This multiple coding by a single codon is called the degeneracy or redundancy of the code. Example: UUU and UUC codons code for the same specific phenylalanine amino acid. In the same way, CAU and CAC codons code for the same specific histidine amino acid (Figure 6.2).

Application activity 3.2

Use the following genetic code and translate the following segment of RNA into a sequence of five amino acids: GUC-GCG-CAU-AGC-AAG

How to read the genetic code?

The genetic code is an mRNA code read in three letter blocks (codons) in the 5′ → 3′ direction. Example: Say you want to find out the codon of ACG. 

- Firstly, locate and read letter “A” on the first base column on the left-hand side of the table.

- Secondly, locate and read the letter “C” on the second base row and locate the same letter down the column where it pairs with the previous first letter “A”. So you have AC bases now.

 - Thirdly, locate and read the letter “G” on the third base column where this letter makes the third letter with the previous AC bases. Now you have the codon ACG, which codes for threonine (Thre) amino acid.

N.B: Out of the total 64 codons, 61 sense codons specify one of the 20 amino acids. The other three nonsense codons are Stop Codons and, therefore, do not specify any amino acid. The sense codon AUG, which specifies Methionine, is a Start Codon

6. Genetic code has start and stop codons: Out of 64 codons, only 61 codons are called sense codons (Figure 6.2). The other three codons are called nonsense codons or stop codons or chain-terminating codons. These three codons are UAG, UAA, and UGA; they do not specify any amino acid, and there are no t-RNAs to carry the appropriate anticodons. The AUG codon, which code for methionine, is most of the time the start codon or initiation codon for protein synthesis in both eukaryotes and prokaryotes. 

7. Wobble hypothesis: Francis Crick has pointed out that the complete set of 61 sense codons can be read by fewer number than 61 t-RNAs. The simple reason being, the pairing properties in the bases in the anticodons are wobble in nature. Here, the word “wobble” simply means “fluctuating” or “unsteady.”

For example: The two different leucine codons (CUC, CUU) can be read by the same leucine t-RNA molecule, contrary to regular base-pairing rules (Figure 6.3).

DNA is extremely stable and replicates accurately

According to central dogma concept, m-RNA is copied from DNA and m-RNA is then translated to form proteins. Therefore, it is critical to maintain the integrity of DNA to accurately produce the desired and correct amino acids (proteins).

DNA is the repository of genetic information gathered over millions of years and it is stored in a stable form inside the cell. The stability of DNA is a property critical to the maintenance of the integrity of the gene.

The stability of DNA can be explained and evidently supported by the fact that DNA has been extracted from Egyptian mummies and extinct animals such as the woolly mammoth and it can also be extracted from dried blood sample or from a single hair at a crime scene which is old enough. DNA molecule is a stable structure and replicates accurately in order to avoid any mutation or change in nucleotides sequences in DNA. The stability of DNA can be attributed to important factors — Hydrogen bonds and base stacking

Hydrogen bonds

Hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom (O) of one molecule and an electronegative atom (N) of a different molecule (Figure 6.4). In the structure of DNA, the strong electronegative atom is the oxygen (O) and Nitrogen (N), while H atom has positive charge. In the structure of DNA (Figure 6.4), thymine and adenine have two hydrogen bonds; while guanine and cytosine have three hydrogen bonds. Hydrogen bonds play very important role in binding the bases of the opposite strands in the DNA. Hydrogen bonds are very weak by themselves. But in a DNA sequence, there will be thousands of these H-bonds which make DNA very stable.

Base stacking

In DNA, the stacked base pairs also attract to one another through Van der Waals forces. The energy associated with a single Van der Waals interaction has small significant to the overall DNA structure. But the large amounts of these interactions help to stabilize the overall structure of the helix

3.4 Effects of alteration of nucleotide sequence

 Activity 3.5 

Make a research using books from school library and internet on effects of alteration of nucleotide sequence, write a short summary on your research and discuss the effects of alternation of nucleotide sequence.

Change in nucleotide (mutation) sequence leads to change in polypeptides 

Amino acids (proteins) are the ultimate product of the nucleotide sequence present in genes (DNA). Thus, any change in the nucleotide sequence of a gene can result into producing wrong or different polypeptide chain. In other words, gene mutation is a change in sequence of nucleotides that results in change in the synthesis of polypeptide chains.

One of the best examples is Sickle-cell anaemia. In this disease, the nucleotide “T” in the DNA sequence is replaced by “A” nucleotide. 

 The minor substitution in the nucleotide sequence is transcribed as a mutant codon on the m-RNA. And during translation, due to mutant codon on the m-RNA, valine is synthesized instead of glutamic acid. Valine distorts red blood cells and cause sickle-cell anaemia. You will be studying next about it in next section. 

Another example is Albinism. Albinism occurs due to mutation in the gene for tyrosinase, an enzyme which coverts tyrosine to DOPA (dihydroxyphenylalanine) Melanin, skin pigment, is derived from DOPA. Melanin absorbs light in the ultraviolet (UV) range and protects the skin against harmful UV radiation from the sun. People with albinism produce no melanin. Therefore, they have white skin, white hair, eyes with red iris, and they are very sensitive to light. Mutation in the gene of tyrosinase

Sickel cell anaemia

 In 1910, J. Herrick first described sickle-cell anaemia. He found out that in conditions of low oxygen tension, the normal disc-shaped red blood cells of people with sickle-cell anaemia get distorted into sickle-shaped red blood cells. Sickle-cell anaemia is a genetic disease that affects haemoglobin molecules. Haemoglobin is a protein found in red blood cells and is responsible for the transportation of oxygen through the body. Haemoglobin, the molecule affected in sickle-cell anaemia, consists of four polypeptide chains:

Two a-globin polypeptides and two b-globin polypeptides-each of which is associated with a haeme group (a non-protein chemical group involved in oxygen binding and added to each polypeptide after the polypeptide is synthesized).

Cause 

The mutation causing sickle cell anaemia is a single nucleotide substitution (A to T) in the DNA of haemoglobin coding gene. The change in a single nucleotide is transcribed as a codon for valine amino acid (GUG) on the m-RNA instead of glutamic acid (GAG) Eventually, due to change in the codon, valine amino acid is translated instead of glutamic acid at the 6th position from N-terminus of the haemoglobin polypeptide chain. This defective form of haemoglobin in persons with sickle cell anaemia is referred to as HbS.

Symptoms 

The sickled red blood cells are fragile and break easily, resulting in the anaemia. Normal red blood cells normally squeeze and pass through blood capillaries smoothly. However, sickled cells are not flexible and therefore have the tendency to get clogged in capillaries. As a result, blood circulation is impaired and tissues become deprived of oxygen. Oxygen deprivation occurs at the extremities, the heart, lungs, brain, kidneys, gastrointestinal tract, muscles, and joints.

Sickle cell anaemia is an autosomal recessive disorder that affects 1 in 500 African americans and is one of the most common blood disorders and in the United States. By autosomal disorder, it means that in order for full disease symptoms to manifest in an individual they must carry two copies (homozygous genotype = SS, HbS & HbS) of the HbS gene. However, the individuals who are heterozygous (genotype = AS, i.e., HbA and HbS) have what is referred to as sickle cell trait, a phenotypically dominant trait.

Although heterozygous (AS) individuals are clinically normal, their red blood cells can sickle under very low oxygen pressure. Their red blood cells may sickle when they are at high altitudes in airplanes with reduced cabin pressure

Application activity 3.4 

Show your understanding of the causes and symptoms of sickle cell anaemia.

Skills lab 3

Extracting DNA from banana

Apparatus and reagents: Banana,10% Saline solution, Ethanol, Mortar, Filter paper, Test tube, Beaker, Soap (Dish wash) and Glass rod

Procedure:

1. Put one banana into the mortar. Crush banana completely, making sure it is completely pulverized

2. Pour 10% saline solution into mortar and stir gently to avoid creating foam

3. Insert a filter into a clean beaker so it does not touch the bottom of the cup

4. Pour the mixture from step 3 into the filter. After a few minutes, some liquid, called the filtrate, should have collected in the bottom of the cup. Remove the filter and set it aside

5. Add a few drops of soap into the filtrate 4 and stir gently

6. Get a test tube of cold alcohol. Use a pipette or eyedropper to collect your filtrate. Add it to the alcohol. Let it sit undisturbed for about 5 minutes. Do not shake, the white material coming out of solution as a precipitate is DNA 

7. Dip the glass rod into the tube, slowly rotating it to spool out the banana’s DNA

Questions 

Prove that your results would be different if you were to use a fruit or a vegetable other than bananas? Explain

End unit assessment 3

Section A: multiple choice questions

1. Use the figure below to answer the questions that follow

i. The above diagram refers to a molecule of 

a. mRNA

b. tRNA

c. DNA

d. rRNA

ii. Item labeled A refers to

a. The codon

b. Amino acid attachment binding site

c. The anticodon

d. Hydrogen bond

iii.Item labeled B refers to

a. The codon

b. Amino acid attachment site

c. Hydrogen bond

d. The anticodon

2. DNA Replication is the process of 

a. Copying DNA from RNA b Copying DNA from proteins 

c Copying DNA from DNA d Copying DNA from ribosome 

3. Nitrogenous bases of the two strands of DNA are linked with

a Hydrogen bonds b Covalent bonds 

c Ionic bonds d Phosphodiester bonds 

4. DNA does not have 

a. Adenine b Cytosine 

c. Guanine d Uracil

Section B: short answer type questions

5. The drawing shows a polyribosome

a. Name X, Y and Z.

b. In which direction are the ribosomes moving? Explain how you were able to decide on their direction of movement.

6. Suggest why:

a. A mutation in which one nucleotide of a triplet code is altered often makes no difference to the protein molecule coded by the DNA.

b. The addition or deletion of three nucleotides in the DNA sequence of a gene often has less effect on the encoded protein than the addition or deletion of a single nucleotide.

7. The table shows all the messenger RNA (mRNA) codons for the amino acid leucine. Copy the table and write it in, for each codon, the transfer RNA (tRNA) anticodon that would bind with it and the DNA triplet from which it was transcribed.

AUGGCCUCGAUAACGGCCACCUAA

a. What is the maximum number of amino acids in the polypeptide for which this piece of mRNA code? 

b. How many different types of tRNA molecule would be used to produce 

a polypeptide from this piece of mRNA? 

c. Give the DNA sequence which would be complementary to the first 6 bases in this piece of mRNA?

d. Name the process by which mRNA is formed in the nucleus.

e. Give two ways in which the structure of a molecule of tRNA differs from the structure of a molecule of mRNA.

Section C: long answer type questions

1. a) Describe the role of RNA polymerase in transcription.

b) Which other enzyme is involved in transcription and what is its role?

c) Why is splicing of pre-mRNA necessary?

d) A sequence of bases along the template strand of DNA is ATGCAAGTCCAG.

i. What is the sequence of bases on a messenger RNA molecule that has been transcribed from this part of the DNA molecule?

ii. How many amino acids does the sequence code for?

iii.A gene is made up of 756 pairs. The mRNA that is transcribed from this gene is only 254 nucleotides long. Explain why there is this difference.

2. a) Describe why the genetic code is described as

i. Universal

ii. Degenerate

iii. Non-overlapping

b. State three ways in which the molecular structure of RNA differs from DNA.

c. Distinguish between a codon and an anticodon

d. Explain why:

i. DNA needs to be chemically stable

ii. mRNA needs to be easily broken down (chemically unstable)

3. The following codon dictionary shows all 64 triplet codons which may occur in mRNA and the amino acids that are coded, as well as the chain termination codon which are labelled stop.

Use this dictionary to answer questions about the diagram below which summarizes the processes of protein synthesis

a. Which is the first codon used in protein synthesis from this mRNA?

b. What is the sequence of the first 4 amino acids from the amino terminal of the growing polypeptide?

c. What is the anticodon sequence in tRNA 1?

d. Give the codon which is recognized by tRNA 2.

e. Explain what changes will occur in the translation apparatus to allow codon 6 to be translated.

f. What are the possible codon sequences for codon 6?

g. The figure above gives information about the seven amino acids of an 80 amino acid polypeptide. 

What would be the effect on this polypeptide if there was a base substitution in the DNA sequence of the gene so that the UAC codon in the diagram became a UAG codon?