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Part A

1.1-Hexanamine and triethylamine have the same molecular weight of 101.190 amu. Despite this similarity, the boiling points of 1-Hexanamine and triethylamine are 131.5° C and 89.5° C respectively. Explain the reason for this variation in the boiling points of these two similar molecular weight compounds.

2.The boiling points of propanol and propanal are 97°C and 48°C respectively. Explain the reason for this variation in the boiling points of these two compounds with similar molecular weights. Discuss the role of intermolecular attractions between molecules to support your answer.

3.Stearic acid and linoleic acid are both long chain carboxylic acids with 18 carbon atoms.However, there is a vast difference in the melting points of these fatty acids. Stearic acid (18:0) melts at 70°C and linoleic acid (18:2) melts at -5°C. Explain why there is such a large difference in the melting points of these fatty acids.

4.Some amides are reported to have antibacterial activity. Discuss the feature that is common to amides with antibacterial activity, mechanism of action and applications.

Carbohydrates, proteins and lipids play a vital role in the efficient functioning of the human body. Knowledge of the structures of these biomolecules assists in understanding their functions in the human body. Provide an overview of the structural classification of carbohydrates and levels of organization of protein structure and explain the role of lipids in maintaining the structure of the cell membrane.

1.Cyanide inhibits a key enzyme in cellular respiration. Discuss the mechanism and effect ofenzyme inhibition.

2.Sequence 1 and sequence 2 are short sequences of DNA with a message. To decipher the message, you will need to first transcribe and then translate the sequences. Using the single letter code of each amino acid obtained upon translating the sequence, crack the message contained. Please note that there are only 20 amino acids. Therefore, you may need to insert any/all of the letters B, J, O, U, X, Z to complete the message. 

DNA Sequence 1: 3'- TAAAATCAGCTCTAGACGGTACTCTACTAGTCATGGTCCATG- 5'

DNA Sequence 2: 3'- AGGACGTAGCTTTTAACGCTCTAAAGGAAATTA- 5'

3.A part of the DNA sequence of normal hemoglobin and sickle cell hemoglobin are shown below. Using your knowledge of transcription and translation, determine the mRNA and amino acid sequence for both normal and sickle cell haemoglobin DNA. 

Normal hemoglobin DNA C A C G T G G A C T G A G G A C T C C T C T T C

Sickle cell hemoglobin DNA C A C G T G G A C T G A G G A C A C C T C T T C

Comment on the difference between the amino acid sequences obtained and state if this difference would affect the role of haemoglobin as an oxygen carrying protein.

4.State whether the following compounds are present in DNA or RNA or both.

a) Adenine

b) Guanine

c) 2-deoxy-D-ribose

d) Cytosine

e) Thymine

f) D-ribose

g) Uracil

5.A part of the aminoacid sequence in Cytochrome-C protein from 6 different species is given in the table. Rank the organisms from 1 to 5 according to the similarity of the organism to human: based on the similarity between the cytochrome C aminoacid sequences, 1 being the closest to human. It is largely agreed that the greater the number of amino acid (or nucleotide) differences between a given pair of organisms, the further apart they are in evolution. On the other hand, if two organisms show very few differences, they are likely to be closely related. With the given information in the table and your ranking of organisms, do you think this ranking gives definite information on how further apart the organisms are in evolution? Why/why not?

6.Complete the following questions using the given double stranded DNA.

Double stranded DNA:

5' - ATGGACGGTTGA - 3'

3' – TACCTGCCAACT - 5'

a) Identify the coding strand and the template strand in the given double stranded DNA.

b) Determine the sequence of mRNA obtained from the transcription of the DNA strand.

c) Determine the anticodons of tRNA required to translate the mRNA codons.

d) Using the genetic code, identify the sequence of amino acids obtained upon translation of the mRNA.

Part A

The report involves questions and answers that discuss the structure and function of biomolecules and identifies the basic features of amines, esters, and amides. The report also discusses genetic expression and the mechanisms of DNA and RNA process in an organism.

1.Both hexamine and triethylamine have the same number of electrons and molecular weight but their boiling point varies because of their differences in intermolecular hydrogen bonding. Hexamine has higher boiling points than triethylamine because it has hydrogen bonds as well as van der Waals dispersion forces whereas triethylamine has not.

2.The boiling point of propanol is higher than that of propanal.  The main reasons for this variation are as a result of their differences in intermolecular forces and polarization. Propanal is an aldehyde which has Oxygen which is highly polar and has no hydrogen bonds (Nelson, Cox, & Lehninge, 2005). It has only dipole-dipole interactions which are weak intermolecular force whereas propanol which is an alcohol has hydrogen bonds which have the strongest intermolecular forces acting between hydrogen molecules thus their variation in their boiling points.

3.Both stearic and linoleic acid are all carboxylic acids with 18 carbon atoms but have differences in their melting points. Their large differences in melting point are as a result of the saturation. Stearic fatty acid is saturated fatty acid whereas linoleic acid is unsaturated fatty acid. The higher the unsaturation, this is why linoleic has a lower melting point than stearic( Ko, et al., 2010).

4.The common characteristic that all amides with antibacterial activity have is nitro and Bromo substituent on the phenyl ring. The nitro and Bromo substituents are the one that causes antimicrobial activity. Their mechanism of action is that amide peptide causes membrane permeation and metabolism disruption as well as interruption of protein and DNA functions of bacteria.  Amides are applied in pharmaceutical industries to make antibacterial and antifungal agents against yeast, gram positive and gram negative bacteria (Nugent, 2010).

Carbohydrates are a molecular compound that is made up of carbon, hydrogen, and oxygen elements. Carbohydrates are classified into three groups according to their structural features.  They include monosaccharides, disaccharides, and polysaccharides.  Monosaccharides are the simplest sugars and are made up of one polyhydroxy aldehyde or ketonic unit. Their number of carbon mostly ranges from three to seven and are classified into their functional groups. An aldose sugar is monosaccharides that has aldehyde group whereas a ketose sugar is the one that has ketonic unit. Also, mono sugars can be classified depending on the number of carbons in the sugar. For examples sugars with three carbons can be referred to as trioses, with five carbon pentoses or with six carbon hexoses. Some examples of monosaccharides include glucose, glyceraldehyde, ribose and dihydroxyacetone (Devlin, 2011).

Disaccharides involve more than one monosaccharide unit. The mono sugars are linked together by glycosidic linkages. Some of the common disaccharides include fructose, maltose, and lactose. On hydrolysis, disaccharides sugars yield monosaccharides units which may be similar or dissimilar. On the other hand, polysaccharides consist of a large number of monosaccharides units which may be branched or unbranched. When polysaccharides are hydrolyzed they mostly yield similar monosaccharide units.  Some examples of polysaccharides include chitin, gum, mucilage, and pectic substances.

1.1 - Hexanamine and Triethylamine Boiling Points

Cell membrane serves as a cell wall and cytoskeleton in some organelles. It is composed of a mixture of different classes of molecules. The most common components are lipids and protein. Lipids are a class of molecules that hat shows a wide diversity in terms of biological function and structure. The main role of lipids is in the formation of permeability of barrier cells in form of a lipid bilayer. The percentage of lipids in the cell membrane can be 20 to 80 percent depending on the location and the function of the membrane in the body.

The main component in membrane is phospolipid which forms a lipid bilayer in the cell membrane.  This lipid bilayer has two units namely; hydrophilic and hydrophobic units. The hydrophobic side repels water and points away from intracellular fluids whereas hydrophilic unit is attractive to water thus it face intracellular fluids. The hydrophobic tail is able to repel water and faces away from the intracellular and extracellular fluid. The phospholipids organize themselves such that they expose the hydrophilic regions to water and hydrophobic regions away from water (Yeagle, 2011). The arrangement is spontaneous and does not need any energy. This arrangement of phospholipid forms a layer that represents a wall between the inner and outer side of the cell thus playing a crucial role in maintaining the structure of cell membrane. The lipid bilayer is also semi-permeable which makes it allow specific substances to diffuse across the cell membrane.

The other important lipid component in the cell membrane is cholesterol. It plays a critical role in making the cell membrane to be rigid thus maintaining the structure of cell membrane.  The third lipid component of the cell membrane which helps in maintaining the structure of the cell is glycolipid. Glycolipid consists of the sugar chain and helps the cell to be able to recognize other cells of the body.

A protein structure is organized into four different levels namely; primary, secondary, tertiary and quaternary structures. The primary structure provides the description of the specific order arrangements of amino acids which are bonded together to make a chain of the protein referred to as a polypeptide chain (Alipanahi et al., 2015). The primary structure of this polypeptide chain is identified starting at the amino acid which is located at the N-terminus of the polypeptide chain. In regard to this, each amino acid is identified using its abbreviation. A good example of a protein with primary structure is insulin hormone. Insulin is a relatively small protein which consists of fifty-one amino acids. It is composed of two polypeptide chains where polypeptide A consists of twenty-one amino acids and polypeptide B chain thirty amino acids (Li, et al., 2015).

The second level of protein structure is a secondary structure. It is made up by the spatial rearrangement of amino acid that is adjacent to the primary structure Secondary structure is a common occurring structure in most proteins and is formed via hydrogen linkages. Some examples of secondary structure proteins include Beta pleated sheet and an alpha helix. Long chains of the polypeptide will have some parts that either in form of a twist or pleated sheet or alpha helices shape to form secondary structures.

2 - Boiling Points of Propanol and Propanal

The further folding and twists of secondary structure of proteins forms what is called tertiary structure. It is a compact structure which has hydrophilic groups on the surface of the protein molecule and hydrophobic heads which are in the interior. It is the tertiary structure that determines the functional activity of the protein. Examples of tertiary proteins include fibrous and globular proteins.

On the other hand, quaternary structure is composed of more than one polypeptide chain which is in a specific orientation with respect to each other. Examples of proteins with tertiary structures include myoglobin and heme (Barrett, 2012).

1.Cyanide affects cellular respiration by inhibiting the key enzyme involved in cellular respiration. It reversibly binds to iron iii ions in the hem group found in the cytochrome oxidase within the mitochondria. This inhibits the electron transport chain blocking the reduction of oxygen to water and ATP (Ochiai, 2012).

2.

DNA Sequence                                    3’- TACGAATCAGCTGTA-5’

Complementary DNA Sequence      5’- ATGCTTAGTCGACAT-3’

MRNA seq                                            5-  AUGCUUAGUCGACAU

Codon seq=   AUG –CUU-AGU-CGA-CAU

Amino acid seq= Met-Leu-Ser-Arg-His

3.The normal Haemoglobin DNA consists of a DNA template made up of CAC-GTG-GAC-TGA-GGA-CTC-CTC-TTC. This DNA template undergoes transcription to form a mRNA complementary to the DNA template mRNA Chain;  GUG-CAC-CUG-ACU-CCU-GAG-GAG-AAG

By using the messenger RNA chain of normal hemoglobin, the amino acid will be Val-His-Leu-Thr-Pro-Glu

 Sickle cell hemoglobin DNA template is; CAC-GTA-GAC-TGA-GGA-CAC

Then the massager RNA Sickle cell hemoglobin m-RNA = GUG-CAU-CUG-ACU-CCU-GUG

The amino acid sequence of sickle cell hemoglobin will be Val-His-Leu-Thr-Pro-Val

The difference between normal hemoglobin and sickle cell hemoglobin occurs only in a single amino acid at codon six whereby codon six of normal hemoglobin is Glu and Val in sickle cell hemoglobin. Sickle cell hemoglobin and normal hemoglobin vary from each other in just one amino acid. This distinction in one amino acid causes a lot of differences in the properties of these two kinds of hemoglobin. This difference will also affect the role of hemoglobin an oxygen-carrying protein

4.

a) Adenine- Found in both RNA and DNA

b) Guanine-Found in both RNA and in DNA

c) 2-deoxy-D-ribose- Found in DNA

d) Cytosine-Found in both RNA and in DNA

e) Thymine-Found in DNA

f) D-ribose-Found in RNA

g) Uracil-Found in RNA only

5.Depending on the similarity of the sequence, the organism can be ranked as Human, Rhesus monkey, Bullfrog, Tuna, chicken and the furthest organism from the human is silkworm. I think this ranking does not give definite information on how further apart the organism is in evolution. This is because it does not involve crucial information such as features of these organisms.

6.

a)    The template strand is in the bottom one with the direction 3’ to 5’. This is supported by the fact that messenger RNA is transcribed in 5’ to 3’ direction. On the other hand, the coding strand is the first which has the direction of 5’-3’ (Meister, 2012).

b)    The sequence of messanger RNA is AUGGACGGUUGA

c)     AUG= Met

        GAC= Asp

        GGU= Gly

        UGA= Thr

d)  Translation of the mRNA: UAC CUG CCA ACU

UAC= Tyr

CUG= Asp

CCA= Pro

ACU= Thr

Conclusion

In conclusion, the assessment has played a lot in understanding crucial concepts in biochemistry which are related to health science. I can now exhibit my knowledge by applying the principles and concepts in biochemistry which ranges from structure and function of biomolecules to genetic expression, DNA organization, and replication, RNA synthesis, processing, and metabolism.

References

Alipanahi, B., Delong, A., Weirauch, M. T., & Frey, B. J. (2015). Predicting the sequence specificities of DNA-and RNA-binding proteins by deep learning. Nature biotechnology, 33(8), 831.

Barrett, G. (Ed.). (2012). Chemistry and biochemistry of the amino acids. Springer Science & Business Media.

Devlin, T. M. (2011). Textbook of biochemistry. John Wiley & Sons,.

Ko, S. H., Su, M., Zhang, C., Ribbe, A. E., Jiang, W., & Mao, C. (2010). Synergistic self-assembly of RNA and DNA molecules. Nature chemistry, 2(12), 1050.

Li, M., Wang, I. X., Li, Y., Bruzel, A., Richards, A. L., Toung, J. M., & Cheung, V. G. (2011). Widespread RNA and DNA sequence differences in the human transcriptome. science, 1207018.

Meister, A. (2012). Biochemistry of the amino acids. Elsevier.

Nelson, D. L., Cox, M. M., & Lehninger, A. L. (2005). Principles of biochemistry. WH Freeman and Company, New York, fourth edition edition, 1(1.1), 2.

Nugent, T. C. (Ed.). (2010). Chiral amine synthesis: methods, developments and applications. John Wiley & Sons.

Ochiai, E. I. (2012). General principles of biochemistry of the elements (Vol. 7). Springer Science & Business Media.

Rees, D. C., Williams, T. N., & Gladwin, M. T. (2010). Sickle-cell disease. The Lancet, 376(9757), 2018-2031.

Saenger, W. (2013). Principles of nucleic acid structure. Springer Science & Business Media.

Yeagle, P. L. (2011). The structure of biological membranes. CRC press.

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