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.
The two types of amines have hydrogen atoms bonded to thenitrogen atom and thus are capable of hydrogen bonding. However, this bonding is not as strong as that in alcohols. More Activeintermolecular forces of 1?Hexanamine (H-bond). There is no intermolecular hydrogen bonding in tertiary amines due to lack of hydrogen atom bonding with nitrogen. This, therefore, means a lower boiling point resulting from diminished N-H bonding
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
The difference in boiling point between propanol and propanalis because of the molecular weight which determines the London’s dispersion forces (molecular attraction forces). It is also attributed to by the difference in polarization of the carbonyl group. For example, aldehydes which are less polarized than the ketones. Therefore, Ketones have a stronger interaction between its molecules than the molecular interaction in aldehydes. Due to the difference above, Ketones have high boiling points than aldehydes. Subsequently, the difference in boiling points is as a result of lack of similarity in the electronic distribution between carbon and hydrogen. Ketones have a higher electronic distribution than aldehydes which are electronically dense due to the bond between the hydrogen and carbon in the carbonyl group which interferes with the polarity. It is important to note that Ketones have diminished enol character thus the physical property determined by the keto forms resulting from the complexity in the bonds existing in the multiple numbers of carbons giving it the higher boiling point. Additionally, Ketones have a higher boiling point because of the presence of two electron donating alkyl group which makes it more polar. A dipole moment is greater, meaning a greater polarization hence high boiling points. Propanol has a hydroxyl group (C-OH), thecovalent bond between C and H meaning a weaker dispersion, Van Dawaal forces of attraction hence a relatively lower boiling point. Propanal has a double bond between the carbon and oxygen atoms at the end of its molecule which makes it more polar thus a dipole-dipole interaction making it have a higher boiling point (Feng et al., 2017).
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.
This is because an increase in molecular weight increases the boiling point as in the case of Stearic acid. Lauric acid remains solid at room temperature of 250but rather melts at aliquid temperature of -50 0.The difference in melting point between the two acids depends on their molecular structure and arrangement when in thesolid state. Saturated fatty acids (stearic) have higher melting points than the unsaturated fatty acids (linoleic)this is because of the molecular geometrics; tetrahedral bond angles. This means a stacked molecular structure. On the contrary, unsaturated fatty acids due to their double bonds, have bends in their structure thus a weaker intermolecular interaction. Asa result, melting points of unsaturated fatty acids are lower than that of the saturated fatty acids
Some amides are reported to have antibacterial activity. Discuss the feature that is common to all amides with antibacterial activity, mechanism of action and applications.
Action: due to the cationic amphipathic properties of AMPs they can penetrate through the membrane of microorganisms such as fungi, protozoa, and bacteria. They are attracted to the negative charges of the outer microbial membrane, supporting highly-selective interaction. Due to the cholesterol, zwitterionic and sphingomyelin surrounding of the mammalian surfaces, AMPs selectively attacks microbial membranes by developing secondary structures when in contact with pathogen’s membrane. After the electrostatic force binding permeabilization occurs and as a result disrupts bacterial membrane. Also, they may also inhibit cell-wall synthesis, inactivate relevant enzymes or affect DNA-protein synthesis. Usually, they are coated to minimize formation of biofilms infection, reduce microbial contamination on polymers surface (Reinhardt & Neundorf, 2016). They do not exhibit basic properties in asolution like amines due to lack of electrons required for hydrogen bonding and (electrons are pulled by more electronegative atom in the carbonyl group. Polyamidesare more stiff and tough due to additional aromatic rings.
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.
Carbohydrates, proteins, and lipids are molecules within our body that serve a significant physiological function. All the three components proteins, carbohydrate, and lipids exhibit some similarities. For instance, all the elements are made up of small building stocks that form long chains and digestion cuts down the size of the macromolecules definitely so that your body can absorb the element Part smoothly. Nonetheless, they display different distinctions.
Overview of the structural classification of carbohydrates
Carbohydrate is also referred as to as saccharides, which is derived a term from a Greek word sakkron which means sugar. It is also usually categorized into three building blocks: starches, sugars, and fiber.
Sugars are the common and simple carbohydrates like fructose, lactose, and glucose. They trigger rapid increase in blood glucose levels.
Starches: are depicted as complex carbohydrates that contain various molecules of glucose. They aid in blood glucose levels as they cause a less rapid rise (Barr,1991). Most importantly, fiber and resistant starch are not broken down or digestible in the small intestine, thereby these plays a vital role and have positive health effects.
Fiber is another complex carbohydrate that is non-digestible. Human beings don’t possess the special enzymes required to break down the links between the sugars. However, the undigested fiber has a great health benefit as it moves through our gut.
Stages Of Protein Structure
Protein structure is a three-dimensional presentation of molecules in a protein element. Proteins are formed with the monomer (a sequence of amino acids) of the polymer (polypeptides). The process of formation of a protein is quite simple; it is formed by amino acids going through condensation reactions whereby the amino acids lose a water molecule in every reaction to add on another with a peptide bond. There are four levels of protein structure namely ;
Primary protein structure- the primary structure is held affirm with covalent bonds like peptide bonds, which are formed during protein biosynthesis. On its structure, it has two ends of the polypeptide chain which are known as the amino terminus (N-terminus) and carboxyl terminus (C-terminus) by nature of free components on each extremity (Andreeva,2015). The sequence of the protein is specific to any protein and defines the function and the structure of the protein. Such unique sequence of proteins can be either determined by tandem mass spectroscopy or Edman degradation.
Secondary protein Structure it refers to specific sub-structures on the polypeptide backbone chain. There are two primary key components of secondary protein structure, the β-strand and α-helix. The structures are predefined by patterns of hydrogen bonds between primary chain peptide groups. They have unique regular geometry, being forced to specific values on the Ramachandran plot. Both the β-strand and α-helix constitutes a way of saturating all the acceptors and the hydrogen bond donors in the peptide backbone.
Tertiary protein structure- refers to the way whereby the non-helical and helical regions of a polypeptide tied back their self. In a normal protein, it’s the non-helical regions that allow the folding. The polypeptide doesn't fold randomly but appears in a specific fashion, in which some steric properties are imParted to the proteins.
Quaternary protein structure
Most of the proteins contain several polypeptide chains. Specifically, proteins that consist of two or more polypeptide chains, quaternary structure depicts to the classified orientation of the chains in regards to the nature and the environment of the interconnections that stabilize orientation (Web & Sali, 2014).
The role of lipids in maintaining the structure of the cell membrane.
Lipids as a component of molecules show a wide diversity of both biological and structural function. The primary and significant role in the cell membrane is in the formation lipid bilayer by the formation of permeable barrier of cells. There are three major components of lipids present in various cell membranes, this includes;
Phospholipids- is one of the major element of the cell membrane. These elements form a lipid bilayer whereby they have two sections, the head areas which form the Part which is hydrophilic and arrange to face the extracellular fluid and aqueous cytosol whereas the tail areas which form the hydrophobic Part face away from the extracellular fluid and cytosol (Feingold & Elias, 2014).
Cholesterol- is another lipid element of the animal cell membranes. The molecules of the cholesterol are distinctively dispatched between membrane phospholipids. The act aids in keeping the cell membrane from becoming stiff by prohibiting phospholipids from clogging together.
Glycolipids- have an attached carbohydrate sugar chain, and it’s located on cell membrane surfaces (Baenke et al., 2013). Most importantly they aid the cell to acknowledge the existence of other cells of the body.
In conclusion, organic molecules are by far significant in our bodies. All the three organic molecules perform crucial functions starting with, Carbohydrate- it provides energy and structure to other molecules, lipids include hydrophobic organic molecules which aid in the release of sex hormones and lastly, proteins area referred to as the building blocks of the body as plays a vital role in the body.
- Cyanide inhibits a key enzyme in cellular respiration. Discuss the mechanism and effect of enzyme inhibition.
Cyanide is basically a metabolic poison that ultimately interferes with every cell in the body, however due to normalcy of the body to require extremely large amounts of oxygen, it greatly affects the cardiovascular system and the central nervous system extensively. Exposure to high levels of metabolic poisons can lead to respiratory failures, death seizure and even death, but still a person requires medical intervention.
Exposure occurs by ingestion, absorption through skin and inhaling cyanide gas. Any individual who is suffering from cyanide poisoning should remove clothing and wash the individual with running water and soap as quickly as possible.
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.
- 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.
Generally the DNA base pair helically linked
On this hemoglobin TT they are linked sometime and some time linked with CC, hemoglobin is the protein that transport –O2, SO the thr difference between these DNA sequence.
Normal contains- A
Sickle cell contain- AA
State whether the following compounds are present in DNA or RNA or both.
- Adenine- RNA
- Guanine- RNA
- 2?deoxy?D?ribose- DNA
- Cytosine- Both DNA and RNA
- Thymine- DNA
- D?ribose - Both DNA and RNA
- Uracil- RNA
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?
Sometimes proteins are also bonded covalently with carbohydrates. These alterations happen following the reactions of proteins and thus known as post-translational modifications.
These types of alterations give special function to the modified proteins. Proteins that have bonded with carbohydrates are known as glycoproteins. There are two types of glycoproteins, N-linked and O-linked, depending on the position of the covalent bond of the sugar moieties. N-linked sugars are bonded to the amide nitrogen of the R-groups of asparagine. The O-linked sugars are bonded to te hydroxy group which might be serine or threonine. Erythrocytes have glycoproteins on their surface.
The blood group specifities is decided by the degree in the make up of the carbohydrate elements of glycoproteins and glycolipids of erythrocytes. Most blood group determinants are caused by carbohydrate variations. Blood groups A, B, and O are due to the action of Particular genes whose function is to mix specific sugar groups onto RBC membrane.
- Complete the following questions using the given double stranded DNA.
- Identify the coding strand and the template strand in the given double stranded DNA.
- Determine the sequence of mRNA obtained from the transcription of the DNA strand.
- Determine the anticodons of tRNA required to translate the mRNA codons
- Using the genetic code, identify the sequence of amino acids obtained upon translation of the mRNA.
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