Question:
Part A
1 Hexanamine & triethylamine both have 101.190 amu molecular masses.
Despite this similarity in molecular weight, the boiling points for 1 Hexanamine & triethylamine respectively are 131.5deg C & 89.5deg C.
Let’s discuss the reason why these molecular weight compounds have different boiling points.
Propanol is boiling at 97°C while propanal boils at 48°C.
Explain the difference in boiling points for these two compounds that have similar molecular mass.
Discuss intermolecular attitudinal attraction between molecules to support the answer.
Linoleic acid and stearic acids are both long-chain carboxylic acids that have 18 carbon atoms.
The melting points of these fatty oils are vastly different.
Stearic acid (18.0) melts around 70°C while linoleic (18.2) melts about?5°C.
Explain why the melting points of these two fatty acids are so different.
Some amides may have antibacterial properties.
Discuss the commonality of all amides with antibacterial activities, as well as their mechanisms and potential applications.
Part B
Human body functions well when it has a healthy balance of carbohydrate, protein and lipid.
These biomolecules are essential for understanding the functions they play in the body.
Provide an overview of the structural classifications of carbohydrates and the levels and organization of protein structure. Explain the role of lipids for maintaining the structure in the cell membrane.
Part C
Cyanide can inhibit a key enzyme of cellular metabolism.
Discuss the mechanism, and how it affects enzyme inhibition.
Sequence 1 (or sequence 2) are short sequences with DNA that carry a message.
First you must translate and transcribe the sequences.
Use the single letter code for each amino acids obtained from translating the sequence to crack the message.
It is important to note that there are only twenty amino acids.
For completeness, you may need all of the letters J, O, U or X, to complete the message.
Below are parts of the normal and sickle hemoglobin DNA sequences.
With your knowledge of transcription, translation, and other aspects of haemoglobin genetics, you can determine the mRNA or amino acid sequences for both normal and sickle cells.
Indicate whether the following compounds are in DNA or RNA, or both.a) Adenine
b. Guaninec) 2?deoxy?D?ribose
d. Cytosinee) Thyminef) D?riboseg) Uracil
In the following table is a partial aminoacid sequence of Cytochrome?C protein derived from 6 species.
Sort the organisms into five groups based on their similarities to human.
It is generally agreed that the more amino acid (or nucleotide), differences between organisms makes them more important in evolution.
If there are very few differences between two organisms, they are likely closely related.
Do you think that the ranking, based on the information in the table and your ranking, gives an indication of how much further apart organisms have gone in evolution? Why/why not?
Answer the following questions using double-stranded genetic DNA.
a. Identify the coding and the template strands of the double-stranded DNA.
b. Determine sequence of mRNA determined from transcription of DNA.
c. Determine the anticodons in tRNA that are necessary to translate mRNA codons.
d. Using genetic code, identify sequence of amino acid obtained from translation of mRNA.
Answer to Question: BHS105A Biochemistry 1
Part A
1 Hexanamine & triethylamine both have 101.190 amu molecular masses.
Despite this similarity in molecular weight, the boiling points for 1 Hexanamine (and triethylamine) are 131.5deg C versus 89.5deg C.
These boiling points are very different.Molecular Weight Compounds.
Hydrogen bonding is possible in both types of amines because the hydrogen atoms are bonded with the nitrogen atom.
This bonding is not as strong, however, as it is in alcohols.
More Activeintermolecular force of 1?Hexanamine.
The lack of hydrogen atom bonds with nitrogen means that there is no intermolecular hydro bonding in Tertiary Amines.
The result is a lower boiling temperature due to decreased N-H bonding
Propanol is boiling at 97°C while propanal boils at 48°C.
Explain why these two compounds have different boiling points.
Discuss intermolecular interactions between molecules to help you understand.
The difference in boiling temperature between propanol, propanalis and propanol is due to their molecular weights. These determine the London’s dispersion (molecular attraction forces).
It could also be due to the difference in the polarization o the carbonyl group.
You can see this in aldehydes that are less polarized as ketones.
Ketones exhibit a stronger interaction between their molecules than aldehydes’ molecular interactions.
Ketones have higher boiling point than aldehydes, due to the above difference.
The reason the boiling points differ is because of the incompatibility in the electronic distribution of hydrogen and carbon.
Ketones are more electronic than aldehydes because they have a lower electronic distribution. This is due to the interplay between the polarity and the hydrogen-carbon bond in the carbonyl.
Ketones have a decreased enol characteristic, which is due to the complexity in the bonds between the hydrogen and carbon in the carbonyl group. This gives them the higher boiling points.
Ketones are more polar due to their presence of two electron-donating Alkyl groups.
A higher dipole moment indicates a greater degree of polarization and thus high boiling points.
Propanol contains a hydroxyl (C-OH), which is the covalent link between C and H. This makes it less dispersive, Van Dawaal force of attraction, and has a lower boiling points.
Propanal contains a double bond at the end between the carbon- and oxygen atoms, making it more polar. It also has a dipole/dipole interaction which results in a higher boiling temperature (Feng, et al. 2017).
Linoleic acid and stearic acids are both long-chain carboxylic acids that have 18 carbon atoms.
The melting points of these fatty oils are vastly different.
Stearic acid (18.0) melts around 70°C while linoleic (18.2) melts about?5°C.
Explain why these fatty acid melt at different melting points.
This is because an increased molecularweight increases the boilingpoint as in the case Stearic acids.
Lauric acid melts at a liquid temperature between -50 and 250 degrees Celsius. The differences in melting points of the two acids are determined by their molecular arrangement in the solid state.
Saturated fatty oils (stearic) melt at higher temperatures than unsaturated ones (linoleic).
This means that the molecular structure is stacked.
Unsaturated fatty acids, due to their double bond structure, have bends which result in weaker intermolecular interactions.
Because of this, unsaturated oils have melting points that are lower than saturated ones.
Some amides may have antibacterial properties.
Discuss the common feature among all amides possessing antibacterial activity, including their mechanism of action and possible applications.
Action: AMPs possess cationic amphipathic capabilities that enable them to penetrate the membranes microorganisms, such as bacteria, protozoa or fungi.
They are attracted the the negative charges of outer microbial cells, which support highly-selective interaction.
AMPs can selectively attack microbial membranes due to the cholesterol-zwitterionic and sphingomyelin around mammalian skins. Secondary structures are formed when they come in contact with pathogens membranes.
Permeability occurs when the electrostatic force is applied to the bacteria membrane.
They can also be inhibited cell-wall synthesis, activate relevant enzymes, or cause DNA-protein synthesis to be affected.
They are typically coated to reduce the formation of biofilms and microbial contamination, and to protect polymers from microbial contamination (Reinhardt &Neundorf 2016).
Because there are no electrons to form hydrogen bonds, they do not have basic properties like amines. (Electrons are pulled by a more electronegative atom within the carbonyl group.
Polyamides are stronger and more durable because of the presence of additional aromatic rings.
Part B
Human body functions well when it has a healthy balance of carbohydrate, protein and lipids.
These biomolecules are essential for understanding the functions they play in the body.
Explain the roles of lipids and carbohydrates in maintaining the structure and integrity of the cell membrane.
The body’s molecules that perform a significant physiological function include carbohydrates, proteins, and cholesterol.
All three components, proteins, carbohydrate, as well as lipids, have some similarities.
The building blocks of all elements form long chains. This digestion reduces the size and structure of the macromolecules, allowing your body to absorb the part Part efficiently.
But they do have some differences.
Overview Of The Structural Classification Of Carbohydrates
Carbohydrates are also known as saccharides. This term is derived from a Greek word sakkron that means sugar.
It is often divided into three building blocks: starches; sugars; and fiber.
Sugars are simple carbohydrates, such as glucose, lactose and fructose.
They stimulate rapid blood glucose increases.
Starches are complex carbohydrates, which contain multiple molecules of glucose.
Because they slow down the rise of blood glucose, they can help to lower it (Barr1991).
Importantly, fiber and resistant starsch are not digestible or broken down by the small intestinal tract.
Fiber is another complex carbohydrate and is not easily digestible.
The enzymes that break down the sugars into their components aren’t found in humans.
Undigested fiber can have a tremendous health benefit because it moves through the digestive tract.Stages Of Protein Structure
Protein structure is a three-dimensional representation of molecules in a particular protein element.
The monomer is a sequence (polypeptides) of amino acids that forms proteins.
The process of making a protein involves simple reactions. Amino acids undergo condensation reactions in which they lose a water molecule every time they add to another one with a peptide bonds.
There are four levels of protein structure.
Primary protein structure- The primary structure is held in check by covalent bonds such peptide and amino bonds.
Its structure is composed of two ends of the polypeptide chains, known as the amino terminus, N-terminus, and carboxyl terminal, respectively. Each end has no free components (Andreeva.2015).
The sequence of a protein is unique to each protein. It determines its function as well as its structure.
You can determine this unique sequence by either tandem mass spectroscopy (MS) or Edman degradation.
Secondary protein structure refers only to certain sub-structures in the polypeptide’s backbone chain.
There are two main components to secondary protein structure: the b strand and a helix.
The patterns of hydrogen bonds formed between primary chain groups of peptides are what predetermine the structure.
They exhibit a distinct regular geometry that is forced to certain Ramachandran plot points.
Both the A-helix and the B-strand are used to saturate all the hydrogen bond donors and acceptors within the peptide-backbone.
Tertiary Protein Structure- The way in which the non-helical helical and helical areas of a polypeptide are linked back to each other.
Normal proteins have non-helical regions which allow folding.
The polypeptide folds in a specific way.
Quaternary Protein structure
Most proteins contain multiple polypeptide-chains.
Specifically, proteins with multiple polypeptide-chains are quaternary. The quaternary structure shows the classified orientation of chains in regard to the nature, environment, and interconnections that stabilize orientation. Web & Sali, 2014.
The Role Of Lipids To Maintain The Structure Of The Cell Membrane.
A component of molecules, lipids can have many biological and structural functions.
The main role of lipids is in the formation of lipid bilayers and the formation of permeable barriers between cells.
There are three main components to lipids in cell membranes.
Phospholipids- are one of the main elements of the cell wall.
These elements form a lipid bicell with two sections. The hydrophilic head area faces the extracellular cytosol and aqueous cytosol while the hydrophobic tail areas face away.
Cholesterol – is another lipid element found in the animal cell membranes.
The cholesterol molecules are distinctly distributed among membrane phospholipids.
The act helps prevent the cell membrane becoming stiffen by keeping phospholipids together from clogging.
Glycolipids – have an attached carbohydrate glucose chain and are found on cell surface surfaces (Baenke und al., 2013).
They help the cell recognize other cells.
Organic molecules are fundamentally important in our bodies.
Every one of the three organic molecules have a critical function. Carbohydrate is the main ingredient, providing energy and structure to other compounds. Lipids are hydrophobic organic molecules that aid in the release hormones. Proteins are the building blocks of the human body and play a vital part in it.
Part C
Cyanide can inhibit cellular respiration’s key enzyme.
Discuss the mechanism, and how it affects enzyme inhibition.
Cyanide is a type of metabolic poison that can cause damage to every cell of your body. However, because the body requires extremely high levels of oxygen, it has a significant impact on the cardiovascular system, central nervous system, and other areas.
The exposure to high levels metabolic poisons can result in respiratory failures, death, seizures, or even death. However medical intervention is necessary.
Exposure can happen by ingestion or skin absorption.
Cyanide poisoning patients should immediately remove all clothing and wash their skin with soapy water.
Sequence 1 (or sequence 2) are short sequences with DNA and a message.
To decipher this message, first transcribe and then translate it.
Use the single letter code of every amino acid that you’ve obtained when translating the sequence to crack the message.
It is important to note that there are only twenty amino acids.
For completeness, you may have to insert the letters J, O U, X, and Z.
Below is an example of the DNA sequence of normal and sickle-cell hemoglobin.
Use your knowledge in transcription and translation to determine the mRNA.
For both normal and sickle cells, the amino acid sequence is identical for both haemoglobin (DNA)
Normal hemoglobin DNA
C A CT G T T G AT G CTG A CT GA G G O A Ct G a Ct G a Ct Ct T T C
Sickle cell hemoglobin DNA
C A – C G T GT G a C T Ga A G G a C T GA G G a C A c Ct Ct C
Comment on any differences in the sequences, and tell us if it would affect the function of haemoglobin (an oxygen-carrying protein).
Helically linked DNA bases are generally the DNA base pairs
This hemoglobin-TT is linked sometime and sometime linked with C, hemoglobin being the protein that transports -O2, SO or the difference between these DNA sequences.
Normal contains: A
Sickle cell contain-AA
Indicate whether the following compounds are in DNA orRNA, or both.Adenine- RNA
Guanine – RNA2?deoxy?D?ribose- DNA
Cytosine: Both DNA andRNAThymine- DNA
D?ribose is both DNA and RNAUracil- RNA
In the following table is a partial aminoacid sequence of Cytochrome?C protein derived from six species.
The order of the organisms is from 1-5 based on their similarity with human. It is determined by the similarity between the cytochrome C aminocid sequences.
It is generally agreed that the greater number of nucleotide or amino acid differences between two organisms, a more important part of evolution they are.
However, close relatives are most likely to exist if there are very few differences between two organisms.
Do you think that the ranking, based on the information in this table and your ranking, gives an indication of how much further apart organisms are in evolution. Why/why not?
Sometimes, proteins may also be covalently bound with carbohydrates.
These alterations are caused by the reactions of proteins, and are also known as posttranslational modifications.
These types are able to give special functions to modified protein.
Glycoproteins refer to proteins that are able bind with carbohydrates.
There are two types: N-linked and O–linked glycoproteins. These depend on the position of covalent bonds between sugar moieties.
N-linked carbohydrates are bonded the the amide of the R groups of asparagine.
O-linked sugars bind to the hydroxy groups, which can be serine and threonine.
Glycopoproteins can be found on the surfaces of erythrocytes.
The composition of glycoproteins, glycolipids, and carbohydrate elements in erythrocytes will determine the blood group specialties.
Carbohydrate variations account for most of the blood group determinants.
Certain genes play a role in the creation of blood groups O, B, and O. They mix sugar groups onto RBC cell membranes.
Answer the following questions using double-stranded genetic DNA.
Identify the code strand and the templates strand in the given DNA double stranded.
Determine the sequence for mRNA determined from the transcription.
Find out the anticodons of the tRNA necessary to translate the codons of the mRNA
The genetic code allows you to identify the sequence of amino acid obtained during the translation of the MRNA.
Refer toFeng, Y., Su, G., Sun-Waterhouse, D., Cai, Y., Zhao, H., Cui, C., & Zhao, M. (2017).
Optimization of Headspace-solid-phase Micro-extraction, (HS-SPME), for Analyzing Aroma Compounds in Soy Sauce via Coupling and Direct GC/Olfactometry/Gas Chromatography-Mass Spectrometry.
Food Analytical Techniques, 10(3) 713-726.Reinhardt, A., & Neundorf, I. (2016).
Design and Application Antimicrobial Protein Conjugates.
International Journal Of Molecular Sciences. 17(5), 701.Andreeva, A., Howorth, D., Chothia, C., Kulesha, E., & Murzin, A. G. (2015).
SCOP2: Investigating the structure and evolution of proteins
Current protocols in bioinformatics.Baenke, F., Peck, B., Miess, H., & Schulze, A. (2013).
(2013).Barr, J. R., Anumula, K. R., Vettese, M. B., Taylor, P. B., & Carr, S. A. (1991). Structural classification of carbohydrates in glycoproteins by mass spectrometry and high-performance anion-exchange chromatography.Analytical biochemistry, 192(1), 181-192.Feingold, K. R., & Elias, P. M. (2014).
Role played by lipids in maintaining and forming the cutaneous permeability barrier.
Biochimica et BiophysicaActa. Molecular and Cell Biology of Lipids. 1841(3). 280-294.Webb, B., & Sali, A. (2014).
MODELLER. Protein structure modeling. Protein Structure Prediction. 1-15.