Biomolecules And Key Biochemical Principles Case Study

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Your full name:

Rita Uzor Ezeobi

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Rita U Ezeobi

Date on which assessmentwas set:

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Extension - date due (if applicable):

16/05/2023

Self-set

19/05/2023

Actual word count per TAQ:

2500

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Study Skills - How to make sure your work is ‘Ready for Marking’

PART 2: Learning outcomes and assessment criteria for this unit

The following table shows the assessment criteria that your tutor will use to mark your work. To Pass a unit you must achieve all of the assessment criteria below. When all assessment criteria have been met, your tutor will use the grading descriptors shown on your assessment’s TAQ sheet to assess whether it should receive a Pass, Merit or a Distinction. When you have completed your work insert the page number/s on which, in your opinion, you have met each of the assessment criteria.

PART 3: Your comments on this assignment”

This assessment has helped me to develop in-depth idea about the cellular respiration, the Punnett diagram, Mendel’s law of dominance, law of inheritance, law of segregation. This assignment has helped me to develop knowledge about glycolysis reactions, aerobic and anaerobic reaction and so on.

What were your targets to improve from the last assessment and what have you done to address this?

From the last assessment, the improvement that I have made is to carry out thorough research before answering the question. Precision has been maintained throughout the assignment.

PART 4: Your answers to the TAQs

Please type your answers in the boxes provided below.

TAQ 1

Peptidoglycan is an essential component in bacterial cell envelops and it protects cells from bursting due to turgor. Peptidoglycan is also important to maintain cell shape. It also protects the cytoplasm from osmotic fluctuation and acts as the backbone of providing mechanical rigidity and strength to the bacterial cell. It assists in protecting bacterial cells from environmental stress and helping to preserve the cell morphology in the entire life cycle.

Word count: 400

TAQ 2

Part 1

• Structure A is the general formula of alpha-amino acid

b)

• The “X” is the bond formed between two amino acids and this bond is known as a peptide bond.
• The end product of the condensation reaction is- the formation of a dipeptide

• A protein molecule is formed from the long chain of amino acids, each amino acid link together to form the long-chain polymer.
• d) the primary structure of a protein refers to the precise sequence of amino acids in the polypeptide chain.
• e) two secondary forms of folding are α-helix and β- pleated sheet.

α-helix is formed by the in-chain of hydrogen bonds at the side chain and β-[pleated sheets are formed by the polypeptide chains which are less helical and fold back to carry hydrogen and oxygen bonds together. The structure of alpha-helices and beta-sheet is formed and stabilise by the non-covalent interaction between the hydrogen bonds

• f) The tertiary structure of Protein is the three-dimensional shape of a single major polypeptide chain and the shape of the tertiary structure of the protein is determined by the amino acid sequences and how the amino acids are cross-linked to each other within the structure.
• g) insulin as a peptide hormone helps to regulate the level of glucose in blood cells. It also acts as the binding receptor protein on the cell surface and enables to uptake the of glucose
• h) the quaternary structure of a protein involved the arrangement of multiple polypeptide chains within similar molecules. In the case of haemoglobin, the structure is made up of four sub-units to form a tetramer, which is again composed of two identical dimers. Every dimer consists of one alpha and one beta chain and two chains in one dimer held together through the hydrophobic bond. The most important function of haemoglobin is the transport of oxygen. One oxygen is attached to each of the four heme groups.

Part 2

• Alpha glucose, the fundamental function of alpha glucose is energy production in the form of ATP or adenosine triphosphate within the cell at the time of cellular respiration
• b) the disaccharide will be Sucrose. The structure of sucrose is comprised of two monosaccharides- glucose and fructose and form the chemical formula of C12H22O11
• c) a branched polysaccharide made up of many monomers of alpha glucose bonded together to form the key components of the cell surface membrane. That polysaccharide is known as Cellulose. The chemical formula of Cellulose is (C6H10O5)n. It provides the mechanical strength to the cell wall and the orientation of cellulose microfibrils can provide control on cell growth.

d)Ribose is the monosaccharide pentose sugar present in mRNA.

Part 3

• a) Cholesterol is made up of 27 carbon compounds with a unique structure of hydrocarbon tail and central sterol nucleus. The nucleus is made up of four hydrocarbon rings and a hydroxyl group. It is the main structural component of the cell membrane. It plays the role of the building block to the synthesis reaction of steroid hormones, bile acids, and vitamin D. Additionally, Cholesterol provides regulation, fluidity, and stability to cellular function.
• The phospholipid is one sub-class of the lipid which contains a phosphate group. The phosphate group is located at the hydrophilic head region, while the fatty acid chain is located at the hydrophobic region of the structure. The hydrophilic phosphate group of the phospholipid is prone to make spheres of a single lipid layer which can be termed a micelle, which contain a hydrophobic end and two tail facing away from water.

Word count: 450

TAQ 3

• One of the common chemical reactions that occur in the human body is the alteration(conversation) of glucose to pyruvic acid during glycolysis. The glycolysis reaction takes place in the cytoplasm of the cell. The substrate of this reaction is glucose and the enzyme involved in this reaction is hexokinase. The Glucose is phosphorylated by hexokinase and forms glucose 6 phosphates. The first phase of the reaction involved the breakdown of glucose and the production of Energy as ATP and then the phosphate group is transferred from ATP to glucose to form the end product and ADP. The reaction is :

Glucose + ATP= Glucose-6-Phosphate +ADP

• B) The process does not completely utilise enzymes since they do not function as reactants. Instead, they unite with the substrate to produce an enzyme-substrate complex that speeds up the reaction. The enzyme is released undamaged when the reaction is finished and is then ready for another reaction. By reducing a chemical reaction's activation energy, enzymes operate as biological catalysts. This suggests that the presence of an enzyme reduces the amount of energy required to accomplish the reaction. Furthermore, the ratio of substrate molecules to enzymes is not necessary for the process to take place. It is considered “Catalytic Enzymatic Activity” and initially, it is increasing the chemical reaction rate without them or has been altered by this reaction permanently. After that, they are increasing the reaction rate without chemical equilibrium alteration between products and reactants.

(Figure: Activation energy with the enzyme)

The above figure has illustrated the chemical reaction which is also represented with the help of a combustion reaction and the reaction has been completed with Oxygen (O2) and Glucose (C6H12O6). According to the statement of Othman et al. (2021), it has been stated that enzymes have been considered as proteins which are acting based on substrate molecules and reducing the energy activation level. It is also necessary for the chemical reaction for being occurred with the help of transition rate stabilisation. This kind of stabilisation speeds up the rate of the chemical reaction and making them to happen at the rate of physiologically significant (Wei et al. 2019). Energy has been released at the time of enzymatic reaction which speeds up enzymatic reaction through lowering the activation energy required for the initiation of the reaction.

The induced fit theory suggests that the enzyme changes its shape in the presence of the substrate. This model proposes that the binding of substrate to an enzyme causes a change in the shape of that enzyme and can enhance or inhibit the reaction. The induced fit theory resembles the lock-and-key theory. It allows the active site of the enzyme to become complementary to fit the substrate. When the enzyme binds to the substrate, it forms the enzyme-substrate complex. This interaction can cause a mild shift of the enzyme structure which in turn can confirm the binding arrangement. This binding can increase the ability of the enzyme to catalyze the reaction and activation energy.

Induced Fit has been considered as a model which is proposing that all bindings are substrate and other molecules to enzymes which have caused a change in enzyme shape hence; it is enhancing or inhibiting the activity. As per the words of Guengerich et al. (2019), it has been stated that Adenylate Kinase is an accurate example of an Induced fit enzyme. Likewise substrate binding in the active site, they are changing like a little and grasps substrate in a huge amount which is also bound tightly to prepare in catalysing the reaction. The induced fit model is also described as a "Lock and Key Model" and interaction has been linked between enzyme and substrate which is really specific to lock (Peeples and Rosen, 2021). It has also been depicted as rather rigid and static forms in interaction. It is also necessary for the chemical reaction for being occurred with the help of transition rate stabilisation. This kind of stabilisation speeds up the rate of the chemical reaction and making them to happen at the rate of physiologically significant.

(Figure: Induced Fit theory- according to this theory, both enzyme and substrate go through the conformational change reaction. The enzyme contorts the substance into the transition state and increases the rate of reaction).

The above figure has highlighted the entire procedure of “Induced Fit enzymatic reaction” and it is indicating that all enzymes have been shown that all enzymes have a flexible structure and it is an active site which is reshaping the interaction with the substrate (Sheehan et al. 2021). In addition, it is entirely bound. Induced Fit has been considered as a model which is proposing that all bindings are substrate and other molecules to enzymes which have caused a change in enzyme shape. For this reason, it is enhancing or inhibiting the activity.

Word count: 450

TAQ 4

Part 1

Law of Dominance- Mendel’s law of dominance states that when parents with pure and different traits cross together, only one form of the trait appears in the allele. The hybrid offsprings exhibit the dominant traits in phenotype. Working with garden pea plants, Medel found that the cross between two parental peas can provide offspring with different traits in F1 generation. While parents are with pure and contrasting traits have been crossed and only a form of a trait is appearing in the upcoming generation. This offspring is exhibiting a dominant trait in phenotype. While parents are with pure and contrasting traits have been crossed and only a form on a trait is appearing in the upcoming generation. This offspring is exhibiting a dominant trait in phenotype. More specifically, according to this law, in a monohybrid cross between two pairs of contrasting characters, one dominant parental character will express in the F1 generation and both the parental character will express in the F2 generation at a 3:1 ratio.

In the Punnett diagram,

D is the dominant trait (green colour) and d is the recessive trait (yellow colour), so as per the law:

DDx dd = gametes D d

DXd------ Dd

F1 generation: ½ D and ½ d

F2 generation

Genotypic ratio: 1DD:2Dd:1dd

Phenotypic ratio: 3:1

Law of Segregation:

Mendel's law of segregation states that pairs of alleles segregate during the meiosis cell division. Therefore, one allele will always be present in every gamete. In the case of a monohybrid cross, both alleles are expressed in F2 generation. Based on this law, only a gene copy is present in organism and it is distributed in each gamete which is making the allocation of all gene copies in a random manner. Then two allele pair differently- one is expressed in the dominant state and the other as recessive traits.

As per the Punnett Diagram:

Y represent the purple colour of the flower and y represents the White colour of the flower. Each homozygous parent in the P generation produces one kind of gamete: Y, y

In the F1 generation, the heterozygous offspring produced two kinds of gametes Y, and y

In the F2 generation, self-pollination of F1 offspring produces F2 offspring with the ratio of 3:1 Purple to White flower.

Law of Independent assortment

The law of independent assortment states that two-hybrid characters will cross with the genotypic character of YyRr. Before crossover, there will be gametes. The alleles will be separated during cell division. The copy of each chromosome is expressed in different gametes. This refers that regardless of the parental phenotype, the alleles inherit differentcombinations of traits. The law of independent assortment states that during the dihybrid cross, the assortment of each pair of traits remains independent from each other. As per the above Punnett Square diagram,

Y- Purple flower

y- white flower

R- round seed

r-wrinkled seed

so, according to the F2 generation,

YYRR- Purple flower with round pea seed

yyrr-White flower with wrinkled seed

YYRr- Purple flower and round seed, YyRr- purple flower with round seed, YYrr- Purple flower with wrinkled seed, Yyrr-Purple flower with wrinkled seed, yyRr- white flower with round seed

Part 2

• to be born with sickle cell anaemia, the child has to inherit the cope of the Sickle cell gene from both parents. This can happen when both parents are carriers of the sickle cell gene.

In order to inherit Sickle Cell Gene from parents and it has happened while both parents are carrying Sickle Cell Gene and have Sickle Cell Trait. In case a parent is passing two copies of the altered form of a child then it is possible for a phenotypical parent to have a child who is suffering from Sickle Cell Anaemia.

According to the Punnett diagram:

Table 1: Punnett Square

(Source: Self-developed)

• According to the Punnett diagram,

AS is the carrier

AA is the no sickle cell

SS is the sickle cell

ASx AS----- AA, AS AS, SS

So, in the F1 generation, the ratio of offspring born with sickle cell anaemia will be 1:2:1 (1 AA:2AS:1SS)

Sickle cell is an autosomal recessive disorder. Both parents have to carry the gene responsible for the disease. The inheritance of Sickle Cell diseasedepends on what genes the parent is carrying. The carriers of the disease will be labelled as AS then there is a 25% chance that the child will carry normal Haemoglobin or AA and 50% chance the child will be born with the disease AS and a 25% chance the child would have the disease SS.

c)healthy gene can be represented as AA, and one parent carrying the gene of the disease can be represented as AS

according to the Punnett diagram

Table 2: Punnett Square

(Source: Self-developed)

So, according to the diagram, the chances of the offspring carrying the sickle cell gene will be 25% or ¼ (AS). In case an individual or a parent has “Sickle Cell Trait” or HbAS and do not sickle hemoglobin in all HbAA then no children are suffering from Sickle Cell Anaemia. There is only a single chance which has been given to the child for receiving an HbS gene copy hence, it has Sickle Cell Trait.

Word count: 500

TAQ 5

Part 1

Part 2

• Anaerobic respiration is significant because it gives cells a way to generate energy without oxygen. Glycolysis, which happens in the cytoplasm and involves the conversion of glucose into pyruvate, is the first stage of anaerobic respiration. A little quantity of ATP is produced during this process, along with NADH, a coenzyme that transports highly energetic electrons.

Fermentation is the process by which pyruvate is transformed into lactic acid in the absence of oxygen. Although large amounts of lactic acid might be hazardous, anaerobic respiration produces lactic acid for a very specific reason. It permits the creation of NAD+ from NADH, which is necessary for the continuation of glycolysis. Glycolysis would stop without this regeneration, substantially reducing the amount of ATP that could be produced.Therefore, lactic acid generation during anaerobic respiration is essential for maintaining ATP production and ensuring the continuing operation of cells even though its buildup might be harmful.

b)

Part 3

Glycolysis:

In Glycolysis, glucose, a six-carbon sugar undergoes a series of chemical transformations. At the end of the reaction, the glucose is converted to 2 molecules of pyruvate. In this reaction, ATP is made along with NAD+ is converted to NADH, This reaction is occurred in the cytoplasm, under the aerobic condition, the pyruvate originated from glucose and enters the mitochondria to take part in oxidative phosphorylation. In glycolysis, ATP is used to add two phosphates to the glucose, and 6-carbon sugar phosphate breaks down to form 3-carbon sugar phosphate or triose phosphate. In the matrix, the carbon is removed from pyruvate and it combines with the oxygen to form carbon dioxide whichdiffuses down in mitochondria.

Energy Investment Phase of Glycolysis:

The Glycolysis Energy Investment Phase consist two different ATP molecules which are activating glucose in general. It is breaking down all molecules and glucose producing molecules of “Glyceraldehyde 3-phosphate or G3P. Energy in the form of ATP must be provided to this phase because it does not produce ATP on its own.The energy investment phase is followed by the energy payoff phase. The two G3P molecules are further digested in this stage, resulting in the generation of ATP, NADH, and pyruvate. Through a series of enzymatic processes, each G3P molecule generates two molecules of ATP through substrate-level phosphorylation. Two molecules of NADH are also produced, carrying high-energy electrons for utilization in the subsequent steps of the electron transport chain.

Hence, it is acting a whole in this phase of energy investment in Glycolysis and ATP activates consumption procedure. It has also been observed in “Energy Payoff Phase”, generate ATP, pyruvate and NADH. Along with that, ATP is serving as an effective energy resource or NADH contribution and cellular respiration to aerobic respiration such as Oxidative Phosphorylation and Kreb’s Cycle.

LINK REACTION:

The mitochondria are the site of the link reaction, commonly known as pyruvate conversion to acetyl CoA. Each pyruvate molecule produced by glycolysis is oxidized and changed into acetyl coenzyme A (acetyl CoA) during this process.

Pyruvate undergoes oxidation and loses a carbon dioxide (CO2) molecule in the link reaction, creating the two-carbon complex acetyl CoA. NAD+ is also converted during the process to NADH.

Since it is linking with the process of Glycolysis and Krebs cycle which is also addressed as TCA cycle or “Citric Acid Cycle” hence, it can be stated that both are interlinked each other. In addition, this kind of conversion of “Pyruvate into acetyl coA” has been considered as an essential phase in the Aerobic Respiration. While acetyl coA is taking entry in Krebs cycle which is continuing of being oxidised and resulting to produce additional NADH, ATP, FADH2 and CO2. So, overall in link reaction, pyruvate converts to the Acetyle Co A and produces NADH. This process acts as a bridge between glycolysis and the Krebs cycle

KREBS CYCLE

The decarboxylation of Pyruvate for the Acetate. CO2 has initiated Acetate production and it has been attached with Coenzyme A for forming Acetyl CoA which is the first outcomeKrebs Cycle. The role of oxygen in the Krebs cycle is, in the absence of oxygen, the electron transport chain may stuff with electrons. For this reason, NAD+ and FAD may not be produced which causes the glycolysis and production of lactic acid instead of pyruvate. In Krebs Cycle, the glucose is oxidized to carbon dioxide, and oxygen is reduced to water. The energy released in the process is stored in the form of ATPs. 36 to 38 ATPs are formed from every glucose molecule.Aerobic glycolysis is a series of reactions wherein oxygenis required to re-oxidize the NADH to NAD+.

OXIDATIVE PHOSPHORYLATION

The final phase of aerobic respiration takes place in the inner mitochondrial membrane and is known as oxidative phosphorylation. Through a sequence of electron carrier molecules in the electron transport chain (ETC), electrons from NADH and FADH2 (produced in earlier stages) are transferred.

Proton Gradient has been produced with the help of pumping proton like H+ across mitochondrial membranes like electrons is travelling across ETC. On that note, it is releasing energy in this process and this kind of mechanism is called as chemiosmosis which is driving ATP synthase for producing ATP.

While FADH2 generates about 1.5 ATP molecules during oxidative phosphorylation, NADH from glycolysis and the Krebs cycle gives roughly 2.5 ATP molecules per molecule. Depending on the source of NADH and FADH2, the precise ATP yield can change. The conversion of pyruvate to Acetyl CoA is important is an important step in aerobic respiration. Acetyl CoA enters to Krebs cycle where it further undergoes through oxidation to produce NADH, FADH2, ATP, and CO2.

Word Count:700

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References

Journals

Biner, O., Fedor, J.G., Yin, Z. and Hirst, J., 2020. Bottom-up construction of a minimal system for cellular respiration and energy regeneration. ACS synthetic biology, 9(6), pp.1450-1459.

Eckardt, N.A., Birchler, J.A. and Meyers, B.C., 2022. Focus on plant genetics: Celebrating Gregor Mendel’s 200th birth anniversary. The Plant Cell, 34(7), pp.2453-2454.

Guengerich, F.P., Wilkey, C.J. and Phan, T.T., 2019. Human cytochrome P450 enzymes bind drugs and other substrates mainly through conformational-selection modes. Journal of Biological Chemistry, 294(28), pp.10928-10941.

Küffner, A.M., Prodan, M., Zuccarini, R., Capasso Palmiero, U., Faltova, L. and Arosio, P., 2020. Acceleration of an enzymatic reaction in liquid phase separated compartments based on intrinsically disordered protein domains. ChemSystemsChem, 2(4), p.e2000001.

Lobritz, M.A., Belenky, P., Porter, C.B., Gutierrez, A., Yang, J.H., Schwarz, E.G., Dwyer, D.J., Khalil, A.S. and Collins, J.J., 2015. Antibiotic efficacy is linked to bacterial cellular respiration. Proceedings of the National Academy of Sciences, 112(27), pp.8173-8180.

Othman, A.R., Hasan, H.A., Muhamad, M.H., Ismail, N.I. and Abdullah, S.R.S., 2021. Microbial degradation of microplastics by enzymatic processes: a review. Environmental Chemistry Letters, 19, pp.3057-3073.

Peeples, W. and Rosen, M.K., 2021. Mechanistic dissection of increased enzymatic rate in a phase-separated compartment. Nature chemical biology, 17(6), pp.693-702.

Richard, J.P., 2019. Protein flexibility and stiffness enable efficient enzymatic catalysis. Journal of the American Chemical Society, 141(8), pp.3320-3331.

Sheehan, F., Sementa, D., Jain, A., Kumar, M., Tayarani-Najjaran, M., Kroiss, D. and Ulijn, R.V., 2021. Peptide-based supramolecular systems chemistry. Chemical Reviews, 121(22), pp.13869-13914.

Tong, X., Shi, S., Tong, C., Iftikhar, A., Long, R. and Zhu, Y., 2020. Quantum/carbon dots-based fluorescent assays for enzyme activity. TrAC Trends in Analytical Chemistry, 131, p.116008.

Wei, R., Song, C., Gräsing, D., Schneider, T., Bielytskyi, P., Böttcher, D., Matysik, J., Bornscheuer, U.T. and Zimmermann, W., 2019. Conformational fitting of a flexible oligomeric substrate does not explain the enzymatic PET degradation. Nature communications, 10(1), p.5581.

Wikstrom, M., Sharma, V., Kaila, V.R., Hosler, J.P. and Hummer, G., 2015. New perspectives on proton pumping in cellular respiration. Chemical reviews, 115(5), pp.2196-2221.

Yu, M.H., Space, B., Franz, D., Zhou, W., He, C., Li, L., Krishna, R., Chang, Z., Li, W., Hu, T.L. and Bu, X.H., 2019. Enhanced gas uptake in a microporous metal–organic framework via a sorbate induced-fit mechanism. Journal of the American Chemical Society, 141(44), pp.17703-17712.