Assignment Code: BCHET-149/TMA/2023

Course Code: BCHCT-149

Assignment Name: Molecules of Life

Year: 2023

Verification Status: Verified by Professor

Part A

Q1) (a) Describe the method for fractionation of subcellular organelles.

Ans) Fractionation of subcellular organelles is a technique that involves breaking open cells and separating organelles based on their size, shape, density, and solubility. The steps involved include homogenization of the cells, centrifugation to separate organelles based on size and density, and multiple rounds of differential centrifugation to purify the organelles. Density gradient centrifugation and other methods, such as immunoprecipitation and affinity chromatography, can also be used to isolate specific organelles or proteins of interest. Overall, fractionation of subcellular organelles is a powerful technique that allows researchers to study the structure, function, and composition of different organelles in a cell.

Q1) (b) What is the role of Golgi bodies in protein processing? Illustrate with the help of a diagram.

Ans) The Golgi apparatus, also known as the Golgi body, plays a crucial role in protein processing, modification, and sorting. It is a stack of flattened, membrane-bound sacs or cisternae that are organized into distinct regions or compartments, each with a specific function.

Proteins that are synthesized in the endoplasmic reticulum are transported to the Golgi apparatus for further processing and modification. The Golgi modifies proteins in various ways, such as adding carbohydrates, lipids, and phosphate groups, and folding the protein into its final conformation.

In this diagram, we can see how proteins are synthesized in the endoplasmic reticulum and then transported to the Golgi apparatus for further processing and modification. The Golgi has a cis face, a medial region, and a trans face. As the proteins move through the Golgi, they are modified by the Golgi enzymes in each compartment, and the modified proteins are sorted into vesicles that bud off from the trans face of the Golgi and move to their final destinations.

Q2) (a) Name the storage polysaccharides present in animals and plants. Indicate the structural differences between these.

Ans) The storage polysaccharide in animals is glycogen, while the storage polysaccharide in plants is starch. The structural difference between glycogen and starch is that glycogen is highly branched, while starch is mostly unbranched.

Glycogen is a highly branched polymer of glucose that is stored in the liver and muscles of animals. It is similar in structure to amylopectin, which is a component of starch in plants. However, glycogen has more frequent and shorter branches, which allow for rapid breakdown and release of glucose when needed.

Starch, on the other hand, is a mixture of two types of polysaccharides: amylose and amylopectin. Amylose is a linear polymer of glucose, while amylopectin is a highly branched polymer. Starch is stored in specialized plant organelles called amyloplasts, where it serves as an energy source for the plant.

Q2) (b) (i) What are lipoproteins? Name the major groups of lipoproteins.

Ans) Lipoproteins are complex molecules that transport lipids in the blood. They are made up of a core of lipids and a shell of proteins. The proteins in the shell allow lipoproteins to dissolve in the watery environment of the blood and move freely throughout the body.

The major groups of lipoproteins include chylomicrons, very low-density lipoproteins, low-density lipoproteins, and high-density lipoproteins. Chylomicrons are produced by the small intestine after a meal and transport dietary fats to other parts of the body. VLDLs are produced by the liver and transport triglycerides to other tissues.

LDLs are often called "bad" cholesterol because they can build up in the arteries and contribute to atherosclerosis. HDLs are often called "good" cholesterol because they help remove excess cholesterol from the bloodstream and carry it back to the liver for processing and excretion.

Q2) (b) (ii) What is the significance of phospholipids in the membrane structure and function.

Ans) Phospholipids are a key component of the membrane structure and function. They are amphipathic molecules with a hydrophilic (water-loving) head and a hydrophobic (water-hating) tail. In a membrane, the hydrophilic heads of the phospholipids face outward towards the aqueous environment, while the hydrophobic tails face inward towards each other, creating a lipid bilayer.

The lipid bilayer serves as a barrier that controls the movement of molecules in and out of the cell. The hydrophobic interior of the membrane prevents the free diffusion of hydrophilic molecules, such as ions and polar molecules, while allowing the free diffusion of hydrophobic molecules, such as oxygen and carbon dioxide.

Q3) (a) Why are the amino acids called -amino acids? The amino acids tyrosine and tryptophan have nonpolar side chains; still these are placed in the uncharged polar group, why? Write the advantages of polar side chains in the amino acids.

Ans) Amino acids are called α-amino acids because the amino group (-NH2) is attached to the α-carbon atom, which is adjacent to the carboxyl group (-COOH). This distinguishes them from other amino acids, such as γ-amino acids, which have the amino group attached to a carbon atom that is farther away from the carboxyl group.

Tyrosine and tryptophan are placed in the uncharged polar group because although they have nonpolar side chains, they also have functional groups (-OH and -NH) that can form hydrogen bonds and interact with water molecules, making them more soluble and polar than other nonpolar amino acids such as phenylalanine and leucine.

The polar side chains in amino acids have several advantages. They can form hydrogen bonds with water molecules, which enhances solubility and helps stabilize protein structure. They can also form hydrogen bonds with other polar side chains or with charged side chains, allowing for the formation of specific interactions and functional groups within a protein structure. Polar side chains can also contribute to protein-protein interactions and help regulate enzymatic activity by providing binding sites for substrates or cofactors.

Q3) (b) Describe the stereochemistry of the peptide bond and explain how this is significant in restricting the number of conformations of a polypeptide chain.

Ans) Amino acids are called α-amino acids because the amino group (-NH2) is attached to the α-carbon atom, which is adjacent to the carboxyl group (-COOH). Tyrosine and tryptophan are placed in the uncharged polar group because although they have nonpolar side chains, they also have functional groups (-OH and -NH) that can form hydrogen bonds and interact with water molecules, making them more soluble and polar than other nonpolar amino acids such as phenylalanine and leucine. The advantages of polar side chains in amino acids include enhanced solubility, stabilization of protein structure, specific interactions, protein-protein interactions, and regulation of enzymatic activity by providing binding sites for substrates or cofactors.

Q4) (a) How does the pH affect the activity of enzymes in the biological systems? Illustrate your answer.

Ans) The activity of enzymes is highly dependent on pH, which can affect the enzyme's shape and charge distribution, ultimately influencing its function. The optimal pH range for an enzyme is the pH at which it exhibits maximum activity. Outside this range, the enzyme activity may decrease, and at extreme pH values, the enzyme can become denatured.

For example, the enzyme pepsin, which is found in the stomach, has an optimal pH range of 1.5 to 2.0. At this pH, pepsin is highly active and can break down proteins into smaller peptides. However, in more neutral or basic environments, pepsin activity decreases, and the enzyme can become denatured, losing its shape and function.

The pH is a critical factor that affects the activity of enzymes. The optimal pH range for each enzyme is specific and reflects its function and the environment in which it operates. Maintaining appropriate pH levels is essential for optimal enzymatic function and, in turn, the biological processes they drive.

Q4) (b) What is the significance of vitamins and minerals in the living system? Write one biochemical reaction in which the coenzyme of cyanocobalmin is associated and explain the mechanism of the reaction.

Ans) Vitamins and minerals are essential nutrients that play important roles in various biochemical reactions in living systems. Vitamins are organic compounds that act as coenzymes, which are necessary for the functioning of enzymes that catalyse reactions in cells. Minerals, on the other hand, are inorganic elements that serve as cofactors or structural components of enzymes.

Examples:

Conversion of homocysteine to methionine.

Conversion of Propionyl CoA to succinyl-CoA via Methyl Malonyl CoA.

One biochemical reaction in which the coenzyme of cyanocobalamin (vitamin B12) is associated is the conversion of homocysteine to methionine, which is a critical step in the metabolism of amino acids and nucleotides. The coenzyme of cyanocobalamin is adenosylcobalamin, which binds to the enzyme methionine synthase and facilitates the transfer of a methyl group from 5-methyltetrahydrofolate to homocysteine to form methionine.

The mechanism of this reaction involves the transfer of the methyl group from 5-methyltetrahydrofolate to the cobalt atom of adenosylcobalamin, which creates a highly reactive intermediate. This intermediate then reacts with homocysteine, transferring the methyl group to form methionine and regenerate adenosylcobalamin. The reaction is important for maintaining proper levels of methionine and preventing the accumulation of homocysteine, which is associated with various health problems.

Q5) (a) When is a reaction called spontaneous? Differentiate between  and describe their significance in predicting the direction of a biochemical reaction.

Ans) A reaction is called spontaneous when it occurs without any external energy input and proceeds naturally in the direction of the product formation. In a spontaneous reaction, the reactants have higher free energy, and the products have lower free energy. The spontaneous reaction occurs until a state of equilibrium is reached, where the rate of the forward and reverse reactions are equal.

ΔG and ΔG° are both thermodynamic properties that are used to predict the spontaneity of a biochemical reaction. ΔG represents the change in free energy of the reaction, and ΔG° represents the standard free energy change. ΔG° is the free energy change that occurs when all reactants and products are at standard conditions, which are defined as 1 M concentration, 1 atm pressure, and 25°C temperature.

The significance of ΔG and ΔG° is that they provide information about the spontaneity and direction of the reaction. If ΔG is negative, the reaction is spontaneous, meaning the reaction proceeds naturally in the forward direction. If ΔG is positive, the reaction is non-spontaneous, meaning the reaction requires energy input to proceed in the forward direction. If ΔG is zero, the reaction is at equilibrium.

ΔG° is used to calculate the actual free energy change, ΔG, under non-standard conditions using the following equation:

where R is the gas constant, T is the temperature, and Q is the reaction quotient, which is the ratio of the concentrations of the products to the concentrations of the reactants at any given time. If Q is greater than 1, the reaction proceeds in the reverse direction, and if Q is less than 1, the reaction proceeds in the forward direction.

Q5) (b) What is meant by the turnover of ATP? How is oxidative phosphorylation related to it? Write the steps involved in the phosphorylation process.

Ans) ATP (adenosine triphosphate) turnover refers to the continuous cycle of ATP production and consumption in cells. It is a process by which the energy stored in the phosphate bonds of ATP is utilized by the cell for various metabolic processes. When ATP is used, it is hydrolysed to ADP (adenosine diphosphate) and inorganic phosphate (Pi), and the released energy is used for cellular work. The ADP is then recycled back to ATP through a process called oxidative phosphorylation.

Oxidative phosphorylation is a process in which ATP is synthesized by the coupling of electron transport chain and ATP synthase in the mitochondria. It involves the transfer of electrons from NADH and FADH2 to oxygen through a series of redox reactions, which releases energy that is used to pump protons across the inner mitochondrial membrane, creating a proton gradient. The energy stored in this gradient is then used by ATP synthase to produce ATP from ADP and Pi.

The steps involved in the phosphorylation process are:

1. Electrons are transferred from NADH and FADH2 to the electron transport chain in the inner mitochondrial membrane.

2. The electrons pass through a series of protein complexes (I-IV), which pump protons from the mitochondrial matrix to the intermembrane space.

3. The movement of protons creates a proton gradient across the inner mitochondrial membrane.

4. The energy stored in the proton gradient is used by ATP synthase to convert ADP and Pi to ATP.

5. Oxygen is the final electron acceptor in the electron transport chain, and it combines with protons to form water.

The oxidative phosphorylation process is highly efficient, as it can generate a large amount of ATP from a single molecule of glucose. It is an important process in cellular respiration and provides the energy needed for various cellular processes, including muscle contraction, active transport, and biosynthesis. Disruption in oxidative phosphorylation can lead to various metabolic disorders and diseases, such as mitochondrial diseases and neurodegenerative disorders.

Part-B

Q6) (a) Differentiate between anabolism and catabolism. Write the mechanism of the reaction of conversion of glyceraldehyde-3-phosphate to pyruvate.

Ans) Anabolism and catabolism are two opposing metabolic processes that occur in living organisms. Anabolism refers to the building up of complex molecules from simpler ones, while catabolism refers to the breakdown of complex molecules into simpler ones.

Anabolic reactions are endergonic, which means they require energy input to proceed, while catabolic reactions are exergonic, which means they release energy. Anabolism is the process of biosynthesis and is involved in the formation of new tissues, while catabolism is the process of breaking down stored nutrients for energy production.

The conversion of glyceraldehyde-3-phosphate to pyruvate is a catabolic reaction that occurs in the glycolysis pathway. The reaction takes place in two steps, which are:

1. Oxidation of Glyceraldehyde-3-Phosphate: The first step involves the oxidation of glyceraldehyde-3-phosphate to 1,3-bisphosphoglycerate by glyceraldehyde-3-phosphate dehydrogenase. This reaction is coupled with the reduction of NAD+ to NADH and a release of free energy.

2. Formation of Pyruvate: In the second step, 1,3-bisphosphoglycerate is converted to pyruvate by phosphoglycerate kinase and pyruvate kinase. This reaction involves the transfer of a phosphate group from 1,3-bisphosphoglycerate to ADP, producing ATP and 3-phosphoglycerate. The 3-phosphoglycerate is then converted to pyruvate, releasing one molecule of ATP.

Q6) (b) Describe the process of biosynthesis of fatty acids catalysed by fatty acid synthase.

Ans) Fatty acid synthase is a multienzyme complex that catalyses the biosynthesis of fatty acids in cells. The process of fatty acid biosynthesis involves the conversion of acetyl-CoA and malonyl-CoA to palmitate, a 16-carbon saturated fatty acid. The FAS complex contains several enzyme activities, including acyl carrier protein, ketoacyl synthase, ketoacyl reductase, and enoyl reductase, which work together to synthesize the fatty acid.

The biosynthesis of fatty acids by FAS occurs in a series of steps that are repeated in cycles, known as the FAS cycle or the fatty acid elongation cycle. The FAS cycle consists of the following steps:

1. Condensation: The first step involves the condensation of acetyl-CoA and malonyl-CoA, which are attached to the ACP domain of FAS. The KS domain of FAS catalyses the formation of a β-ketoacyl intermediate.

2. Reduction: The β-ketoacyl intermediate is then reduced by the KR domain of FAS to form a β-hydroxyacyl intermediate.

3. Dehydration: The β-hydroxyacyl intermediate is dehydrated by the dehydratase domain of FAS, producing a trans-2,3-double bond.

4. Reduction: The trans-2,3-double bond is then reduced by the ER domain of FAS, producing a saturated fatty acid intermediate.

These steps are then repeated multiple times until a 16-carbon saturated fatty acid, palmitate, is produced. The final product, palmitate, is then released from the ACP domain of FAS.

Q7) (a) How is feedback regulation different from allosteric regulation? Name the most important regulatory enzyme and its allosteric effector in the glycolytic pathway. Also write the reaction catalysed by this enzyme.

Ans) Feedback regulation and allosteric regulation are two different mechanisms of enzyme regulation in living organisms.

Feedback regulation involves the control of enzyme activity by the end-product of a metabolic pathway. When the concentration of the end-product exceeds a certain threshold, it can bind to the enzyme and inhibit its activity, preventing further production of the end-product. This mechanism is often used to maintain homeostasis and prevent excessive accumulation of metabolic intermediates.

Allosteric regulation, on the other hand, involves the binding of an effector molecule to a specific site on the enzyme, which can either enhance or inhibit its activity. This mechanism is used to regulate enzyme activity in response to changes in the cellular environment, such as the availability of substrates and energy.

One of the most important regulatory enzymes in the glycolytic pathway is phosphofructokinase-1 (PFK-1). PFK-1 is an allosteric enzyme that is regulated by several factors, including ATP, ADP, and fructose-2,6-bisphosphate (F2,6BP). F2,6BP is an important activator of PFK-1 and is produced by a separate enzyme, phosphofructokinase-2 (PFK-2), which is also regulated by several factors.

The reaction catalysed by PFK-1 is the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate, which is a key regulatory step in the glycolytic pathway. This reaction is irreversible and is also the rate-limiting step in the pathway. The binding of F2,6BP to PFK-1 enhances its activity, leading to increased production of ATP through the glycolytic pathway. The binding of ATP and other inhibitors to PFK-1, on the other hand, inhibits its activity, slowing down the glycolytic pathway and preventing excessive ATP production.

Q7) (b) Define photophosphorylation. Differentiate between substrate level and oxidative phosphorylation.

Ans) Photophosphorylation is the process by which light energy is used to generate ATP (adenosine triphosphate) in photosynthetic organisms, such as plants, algae, and some bacteria. During this process, light energy is absorbed by chlorophyll pigments in the photosystems of thylakoid membranes and is used to drive the transfer of electrons from water to NADP+, which generates a proton gradient across the thylakoid membrane. This proton gradient is then used by the ATP synthase enzyme to produce ATP from ADP and inorganic phosphate.

Substrate-level phosphorylation and oxidative phosphorylation are two different mechanisms by which ATP is generated in cells.

Substrate-level phosphorylation is a type of ATP synthesis that occurs when a high-energy phosphate group is transferred from a phosphorylated substrate to ADP, generating ATP. This mechanism occurs during the breakdown of glucose in the cytoplasm during glycolysis, where the enzymes phosphoglycerate kinase and pyruvate kinase catalyse the transfer of phosphate groups from 1,3-bisphosphoglycerate and phosphoenolpyruvate to ADP, respectively.

Oxidative phosphorylation, on the other hand, is the process by which ATP is synthesized in the mitochondria of eukaryotic cells through the transfer of electrons from NADH and FADH2 to the electron transport chain, which generates a proton gradient across the mitochondrial inner membrane. This proton gradient is then used by ATP synthase to produce ATP from ADP and Pi.

Q8) (a) Describe the structure and role of ribosome in protein synthesis.

Ans) Ribosomes are complex macromolecular machines that play a crucial role in protein synthesis. They are found in all living cells, from bacteria to eukaryotes, and are composed of both ribosomal RNA and proteins.

The ribosome consists of two subunits, the small subunit and the large subunit, which come together during protein synthesis to form a functional ribosome. The small subunit contains the decoding center, which recognizes and binds to the messenger RNA (mRNA) that carries the genetic code for protein synthesis, while the large subunit contains the catalytic center, which catalyses the formation of peptide bonds between amino acids during protein synthesis.

The ribosome works in a cyclic process during protein synthesis, which involves several steps:

1. Initiation: The small ribosomal subunit binds to the mRNA at the start codon, which is typically AUG. The initiator tRNA, carrying methionine, then binds to the start codon with the help of initiation factors.

2. Elongation: The ribosome moves along the mRNA in the 5' to 3' direction, and as it does so, new tRNAs carrying amino acids bind to the ribosome, which adds the amino acid to the growing polypeptide chain by catalysing the formation of peptide bonds.

3. Translocation: The ribosome moves along the mRNA in a 5' to 3' direction to the next codon, which brings the next tRNA into the ribosome's catalytic center.

4. Termination: The ribosome recognizes the stop codon on the mRNA, which signals the end of protein synthesis. The completed polypeptide chain is then released from the ribosome.

The ribosome is an essential component of the protein synthesis machinery, and its structural and functional features are highly conserved across all living organisms. The structural features of ribosomes are optimized for their function in protein synthesis, and its catalytic activity is a key determinant in the efficiency and accuracy of protein synthesis.

Q8) (b) What are the advantages of using immobilized enzymes? How is the production of enzymes from microorganisms carried out?

Ans) Immobilized enzymes are enzymes that are physically or chemically bound to a support matrix, such as a resin, gel, or membrane, which can be used to improve their stability, reusability, and performance in biocatalytic applications.

There are several advantages of using immobilized enzymes, including:

1. Increased Stability: Immobilized enzymes are less susceptible to environmental factors, such as changes in temperature or pH, and are more resistant to proteolytic degradation.

2. Reusability: Immobilized enzymes can be used repeatedly for several cycles without losing their activity, which can reduce the cost of enzyme production and increase process efficiency.

3. Enhanced Performance: Immobilized enzymes can exhibit higher catalytic activity, selectivity, and specificity, due to their high concentration and favourable microenvironment.

4. Easy Separation: Immobilized enzymes can be easily separated from the reaction mixture, which can simplify downstream processing and increase product purity.

The production of enzymes from microorganisms is carried out by culturing the microorganism in a suitable growth medium that contains nutrients, such as sugars, amino acids, and minerals, under controlled conditions of temperature, pH, and aeration. The microorganisms are then harvested, and the enzymes are extracted and purified by various methods, such as precipitation, chromatography, and ultrafiltration.

Q9) (a) Describe the role of DNA polymerase in DNA replication.

Ans) Immobilized enzymes are enzymes that are physically or chemically bound to a support matrix, such as a resin, gel, or membrane, which can be used to improve their stability, reusability, and performance in biocatalytic applications.

There are several advantages of using immobilized enzymes, including:

1. Increased Stability: Immobilized enzymes are less susceptible to environmental factors, such as changes in temperature or pH, and are more resistant to proteolytic degradation.

2. Reusability: Immobilized enzymes can be used repeatedly for several cycles without losing their activity, which can reduce the cost of enzyme production and increase process efficiency.

3. Enhanced Performance: Immobilized enzymes can exhibit higher catalytic activity, selectivity, and specificity, due to their high concentration and favourable microenvironment.

4. Easy Separation: Immobilized enzymes can be easily separated from the reaction mixture, which can simplify downstream processing and increase product purity.

The production of enzymes from microorganisms is carried out by culturing the microorganism in a suitable growth medium that contains nutrients, such as sugars, amino acids, and minerals, under controlled conditions of temperature, pH, and aeration. The microorganisms are then harvested, and the enzymes are extracted and purified by various methods, such as precipitation, chromatography, and ultrafiltration.

Q9) (b) Describe the process of unwinding of the double helix during DNA replication.

Ans) DNA replication is the process of copying genetic information from one DNA molecule to another. It is a crucial process for cell division and growth and involves many complex molecular mechanisms. One of the key steps in DNA replication is the unwinding of the double helix, which separates the two strands of DNA and allows for the replication machinery to access the template strand.

The unwinding of the double helix is catalysed by an enzyme called helicase, which binds to the DNA at the replication fork and uses ATP hydrolysis to break the hydrogen bonds between the two complementary strands. This creates a replication bubble, where the two strands of DNA are separated, and the template strand is exposed for replication.

To stabilize the single-stranded DNA, a protein called single-stranded binding protein binds to the exposed DNA and prevents it from reannealing. This allows the replication machinery to access the template strand and synthesize a new complementary strand.

The unwinding of the double helix during DNA replication is a complex process that involves multiple enzymes and proteins working together to separate the two strands of DNA and create a replication bubble. This process is essential for DNA replication and is tightly regulated to ensure the accurate copying of genetic information.

Q10) (a) Name two carbohydrates that are metabolised by glycolysis. Write the reactions and enzymes involved in their entry into the glycolytic pathway.

Ans) Two carbohydrates that are metabolized by glycolysis are glucose and fructose.

For Glucose

1. The first step is the phosphorylation of glucose to glucose-6-phosphate by the enzyme hexokinase, using ATP as a phosphate donor:

2. Glucose-6-phosphate is then converted to fructose-6-phosphate by the enzyme glucose-6-phosphate isomerase:

3. The next step is the phosphorylation of fructose-6-phosphate to fructose-1,6-bisphosphate by the enzyme phosphofructokinase, using ATP as a phosphate donor:

4. Fructose-1,6-bisphosphate is then split into two three-carbon molecules, glyceraldehyde-3-phosphate, and dihydroxyacetone phosphate, by the enzyme aldolase.

For Fructose

1. Fructose is first phosphorylated by the enzyme fructokinase to form fructose-1-phosphate:

2. Fructose-1-phosphate is then cleaved by the enzyme aldolase B to form dihydroxyacetone phosphate and glyceraldehyde:

Both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate then enter the later steps of glycolysis, which involves the oxidation and further breakdown of these molecules to produce ATP and pyruvate.

Glycolysis is an important pathway for the metabolism of carbohydrates and provides the energy needed for many cellular processes.

Q10) (b) Describe the pathways involved in the removal of amino group from amino acids.

Ans) The removal of an amino group from an amino acid is called deamination. There are several pathways by which deamination can occur, depending on the specific amino acid and the physiological conditions.

1. Transamination: In this pathway, the amino group is transferred to an alpha-keto acid, which is then converted to another amino acid by transamination. For example, alanine can be converted to pyruvate by transamination with alpha-ketoglutarate, producing glutamate.

2. Oxidative Deamination: This pathway occurs mainly in the liver and involves the removal of the amino group from an amino acid, producing ammonia and an alpha-keto acid. The reaction is catalysed by the enzyme glutamate dehydrogenase, which converts glutamate to alpha-ketoglutarate and ammonia.

3. Deamination by Amino Acid Oxidases: Some amino acids can be deaminated by amino acid oxidases, which are enzymes that use molecular oxygen to remove the amino group, producing ammonia and an alpha-keto acid. This pathway is important for the metabolism of some aromatic amino acids, such as tyrosine.

4. Deamination by Deaminases: Some amino acids can be deaminated by specific deaminase enzymes, which remove the amino group, producing ammonia and an alpha-keto acid. For example, histidine can be deaminated to produce urocanic acid and ammonia by the enzyme histidase.