Macromolecules and Key Biological principles

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Macromolecules and Key Biological principles

TAQ 1

1

In order to form a strand of messenger RNA, a process of transcription from DNA takes place. The enzyme RNA polymerase attaches ahead of the region coding for this protein on the DNA, and splits the hydrogen bonds between the base pairs of the DNA, exposing the bases, in this example, assuming the 3' direction is at the leftmost end, TAC-GCA-TTT-CGT-CTC-CCT-GTT-ATT. RNA polymerase then attaches bases in the 5' to 3', direction, having read the DNA in the 3' to 5' direction, giving the RNA sequence of AUG-CGU-AAA-GCA-GAG-GGA-CAA-UAA, before releasing the RNA strand at a predetermined point. RNA contains uracil as the complimentary base to adenine, hence the U in the RNA as opposed to the T which would be present in the DNA. The strand of DNA not used for transcription is the coding strand, so called as it contains the same sequence of bases as the RNA strand, once the thiamine/uracil substitution is accounted for.

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Word Count: 157

2

Once the messenger RNA is released into the cytoplasm, it will attach to a ribosome for the process of translation. Ribosomal RNA forms an active site for translation and works as an enzyme to catalyse the reaction forming peptide bonds. Transfer RNA contains the complimentary base pair to each triplet in the sequence, attached to a particular amino acid. There are 20 amino acids, but 64 possible base codons, so some amino acids have multiple possible codons. Our RNA strand AUG-CGU-AAA-GCA-GAG-GGA-CAA-UAA, will translate to:

Methionine – Arginine – Lysine – Alanine – Glutamic acid – Glycine – Glutamine – STOP.

AUG is the most common start codon coding for methionine, and even if a different start codon is used, it will still translate to methionine in eukaryotes. UAA is one of three stop codons, used to signal that the protein is complete and to begin the release of the protein into the cell for it to fulfil its biological purpose.

Word Count: 160

References

Www2.le.ac.uk, (2015). 34 codon table.jpg — University of Leicester. [online] Available at: https://www2.le.ac.uk/departments/genetics/vgec/diagrams/34 codon table.jpg/view [Accessed 26 Jan. 2015].

Seeley, R., Stephens, T., Tate, P. and Seeley, R. (1992). Anatomy & physiology. St. Louis: Mosby Year Book.

TAQ 2

Part 1

Mendel studied inheritance of 7 distinct characteristics of the pea plant (Pisum sativum). Mendel discovered that cross-breeding of true breeding plants with violet flowers and white flowers gave only purple flowered offspring. Self-fertilisation of the offspring gave 929 plants in the F2 generation, with a purple:white ratio of 705:224, or 3.15:1.

All 7 characteristics studied fell within ratio of 2.82-3.15:1, and statistical analysis showed Mendel's hypothesised 3:1 ratio could not be rejected at the 99% confidence level (Table 1)

Dominant Characteristic

Recessive Characteristic

Dominant expression in F2

Recessive expression in F2

Dominant :

Recessive

Chi-squared value (Pearsons goodness of fit)

P value

Purple Flowers

White Flowers

705

224

3.15:1

0.368

0.544

Axial Flowers

Terminal Flowers

651

207

3.14:1

0.305

0.581

Yellow Seeds

Green Seeds

6022

2001

3.01:1

0.017

0.897

Round Seeds

Wrinkled Seeds

5474

1850

2.96:1

0.263

0.608

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Inflated Pod

Constricted Pod

882

299

2.95:1

0.072

0.788

Green Pod

Yellow Pod

428

152

2.82:1

0.451

0.502

Normal stem

Dwarf Stem

787

277

2.84:1

0.607

0.436

Table 1 – Results and statistical analysis of Mendels experiments

Mendel promulgated three laws of inheritance.

The law of segregation: During gamete creation, two copies of each allele separate, and so each parent only passes on one allele, and the combination of the alleles from each parent give rise to the characteristics.

The law of independent assortment: As long as genes are not linked to each other, they will not influence the inheritance of other genes. For example, a dihybrid cross where both parents are heterozygous for both alleles; shows a 9:3:3:1 ratio of genotypes, but the genes do not affect each others expression.

The law of dominance: Should an organism inherit differing alleles for an individual characteristic, only one will show (dominant allele) and one will be suppressed, but may be passed on to offspring, and show. This does not take into account incomplete dominance and co-dominance.

Word Count: 217 disregarding table

Part 2

R

r

r

Rr

rr

r

Rr

rr

  • R shows the dominant allele, and r the recessive allele for tongue rolling,
  • All children from this cross inherit a recessive allele from one parent, and will therefore either show the phenotype of a non-tongue roller or be a carrier for this gene.
  • The homozygous parents genes have no bearing on the expression of the gene.
  • The heterozygous parent will have an equal chance of passing on either allele therefore half of the children have one dominant allele (blue boxes) and are able to roll their tongue, and half inherit two recessive alleles and being unable to do so (red boxes).

Word count: 102 disregarding Punnett square.

Part 3

T

t

T

TT

Tt

t

Tt

tt

  • T show the dominant allele and t the recessive allele for phenylthiocarbamate tasting.
  • Both parents have an equal chance of passing on a dominant or recessive allele.
  • The punnet square shows three genotypes: TT, Tt and tt in the ratio 1:2:1 respectively.
  • With four children, three will show the phenotype of being phenylthiocarbamate tasters, and one will be a non-taster, being homozygous for the recessive alleles (green box in square).
  • Of the three children who can taste phenylthiocarbamate, one will be homozygous for the dominant allele (blue box in square), the other two will be heterozygous for these alleles (red boxes in table)

Word count: 103 disregarding Punnett square.

Part 4

Rt

Rt

Rt

Rt

RT

RRtT

RRtT

RRtT

RRtT

Rt

RRtt

RRtt

RRtt

RRtt

rT

RrtT

RrtT

RrtT

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RrtT

rt

Rrtt

Rrtt

Rrtt

Rrtt

  • There are 4 genotypes, but only 2 phenotypes.
  • All sixteen children can roll their tongue, as they will all have inherited the dominant allele from their mother.
  • All children will receive a recessive phenylthiocarbamate tasting allele from the mother.
  • Half the children will receive a dominant phenylthiocarbamate tasting allele from their father, the other half a second recessive allele.

In total:

  • Four children will be homozygous tongue rollers and heterozygous phenylthiocarbamate tasters (Blue).
  • Four will be homozygous tongue rollers and homozygous phenylthiocarbamate non-tasters (Red).
  • Four will be heterozygous tongue rollers and heterozygous tasters (Green).
  • The final four will be heterozygous tongue rollers and homozygous non-tasters. (Yellow)

Word count: 108 disregarding Punnett square.

References

Mendel, G. (1865). Experiments in Plant Hybridisation.

English translation from http://www.mendelweb.org/Mendel.html

https://www.khanacademy.org/math/probability/statistics-inferential/chi-square/v/pearson-s-chi-square-test-goodness-of-fit

TAQ 3

Structure

Roles in Nature

Lipids

Fatty acids: carbon chains with an ketone group at one end. May be saturated, or unsaturated chains, and usually up to 25 carbon atoms in length.

Triglycerides: Hydrophillic glycerol group with each hydroxyl group forming an ester with a hydrophobic fatty acid

fig 1 – triglyceride with saturated, monounsaturated and polyunsaturated fatty acids

Phospholipids: Hydrophillic phosphate groups attached to hydrophobic tails.

Steroids: Formed around four carbon rings e.g. Cholesterol

Membranes: Formed due to the amphiphyllic nature of phospholipids. Can be a bilayer, or a spherical structure known as a micelle which can transport non-fat soluble nutrients across a membrane.

High density, slow release energy storage as fats

Carbohydrates

General formula of Cx(H2O)y Exceptions include RNA

Reversible between straight carbon chain and oxygen bridged ring forms, with several isomers. Ring form more stable.

fig 2 – straight chain glucose

fig 3 - α-glucose ring

Monosaccharides are usually 5 to 6 carbon atoms in size.

Polysaccharides are multiples of monosaccarides joined by a dehydration reaction.

Multiple polysaccharides possible, depending on location of dehydration between two given monosaccharides, and will have differing chemical properties.

Monosaccarides:

Most commonly an energy source for an organism.

Disaccarides: usually formed for transport such as sucrose in plants.

Polysaccarides: Used for easily accessed storage of energy, e.g. glycogen, or structurally like cellulose.

Proteins

Primary structure: The specific order of the amino acids, as coded for by the RNA.

Fig 4 – primary structure of lysozyme

Secondary structure: Structure taken on by the polypeptide chain, can be an α-helix (right handed helix) or pleated β-sheet.

May change type and direction within one polypeptide chain.

Caused by hydrogen bonding.

fig 5 – secondary structure

Tertiary Structure:

The folded structure of a protein, created by amino acid side chain reactions, salt bridges, and the hydrophobic/phillic nature of the varying groups to form a specific structure.

Fig 6 – tertiary structure of yeast 5-aminolaevulinic acid dehydratase

Quarternary structure: Two or more tertiary structures interact to form a protein complex.

Enzymes: Reliant on a specific conformation of the active site created by a proteins' tertiary and quaternary structure.

Structural: Proteins such as collagen used in ligaments.

Motor proteins: used to transfer chemical energy to kinetic energy.

Transport of molecules, such as haemoglobin.

Word Count: 330 (image labels excluded)

References

Open chain glucose structure. (2015). [image] Available at: http://fog.ccsf.edu/~mmalacho/physio/oll/Lesson1/images/glucose.gif [Accessed 19 Jan. 2015].

RCN, (2015). Primary structure of Lysozyme. [image] Available at: http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/L/lysozyme.gif [Accessed 30 Jan. 2015].

Protein secondary structure. (2015). [image] Available at: http://www.drzolghadri.com/wp-content/uploads/2012/01/Protein_-_Secondary_Structure.jpg [Accessed 20 Jan. 2015].

Tertiary protein structure. (2015). [image] Available at: http://www.biochemj.org/bj/373/0733/bj3730733f01.htm [Accessed 20 Jan. 2015].

Structure of triglyeride. (2015). [image] Available at: http://upload.wikimedia.org/wikipedia/commons/b/be/Fat_triglyceride_shorthand_formula.PNG [Accessed 30 Jan. 2015].

TAQ 4

An enzyme is a protein based biological catalyst used in the body to make a specific chemical reaction possible, or to speed up a reaction to the point it is viable for life to continue.

The active site of an enzyme is a precisely shaped region where the substrate fits the enzyme in the lock and key model (Fischer, 1958)

Koshland (1958) noted the flexibility of the active site and that the substrate may also change slightly in shape as it binds, proposed the induced fit theory. As well as shape; the electrical charge, and hydrophillic/phobic regions also aid enzyme selectivity.

Figure 1 – Lock and key vs. induced fit models

Mechanisms of enzyme activity include:

Lowering of the activation energy, meaning less free energy is needed to reach the transition state. Transition state stabilisation, in the enzyme-substrate complex, making the reaction more likely. Substrates are also placed in the correct orientation, lowering the entropy, and providing a pathway unlikely to occur naturally at a sufficient rate.

Fig. 2 – Reduced activation energy by enzme activity

Enzymes can couple to use a favourable reaction to provide the energy for an unfavourable one.

Enzymes can only speed up the reaching of the equilibrium, and are equally effective in both the catabolic and anabolic directions of a reaction should the concentration of substrate and product change, however the bodies use of the products will drive the equilibrium.

Denaturing an enzyme is the irreversible process of altering the secondary, tertiary or quaternary structure of the enzyme so it is incapable of performing its role in the body due to the shape change.

High temperature increases reaction rates, due to higher entropy until the point of denaturation is reached, then reactivity falls sharply

A pH change in either direction will affect the hydrogen bonding between the amino acid side chains, causing denaturation.

The optimum conditions are enzyme specific, and body fluids include buffers to maintain pH.

Fig 3 – Enzyme activity graphed against temperature and pH

Enzymes can also be inhibited, either by similarly shaped substrates competitively binding permanently to the active site, or by binding to the enzyme-substrate complex, preventing its release.

Word Count: 330

References

Induced fit Model. (2015). [image] Available at: http://alevelnotes.com/content_images/i68_Induced-fit_model.JPG [Accessed 26 Jan. 2015].

Lock and key model. (2015). [image] Available at: http://alevelnotes.com/content_images/lock-and-key_model.jpg [Accessed 26 Jan. 2015].

Reaction of carbolic anhydrase catalysed an non catalysed. (2015). [image] Available at: http://alevelnotes.com/page_images/504px-Carbonic_anhydrase_reaction_in_tissue.svg.png [Accessed 26 Jan. 2015].

Enzyme activity against pH and Temperature. (2015). [image] Available at: https://biochem4life.files.wordpress.com/2013/04/enzymes.gif [Accessed 26 Jan. 2015].

TAQ 5

Aerobic Respiration.

Most efficient use of glucose, giving a theoretical net gain of 38 molecules ATP for each glucose molecule.

Requires oxygen.

Slower energy release.

Three stages:

Glycolysis: Breakdown of glucose into 2 pyruvate molecules

10 separate reactions.

Two phases; investment, where 2 ATP are consumed, and pay-off where 4 ATP are produced, as well as pyruvate, used in the Krebs cycle.

Overall formula:

C6H12O6 + 2[NAD]+ + 2[ADP] + 2[P]i → 2 CH3COCOOH + 2[NADH] + 2H+ + 2[ATP] + 2H2O

Krebs cycle:

Pyruvate modified to acetyl co-enzyme-A in link reaction.

Starts with citrate, a 6 carbon molecule being from the co-enzyme, and oxoloacetate.

Citrate decomposes to oxoloacetate over multiple reactions, with a net gain of 1 ATP 3 NAD+ and 1 FAD+ per pyruvate. (2 ATP, 6 NAD+ and 2FAD+ per glucose)

NAD+ and FAD+ then enter the electron transport chain for oxidative phosphorylation.

Oxidative Phosphorylation

NAD+ and FAD+ taken to chains on the internal mitochondrial surfaces

Hydrogen removed from reduced co-enzymes and split, with the electrons being taken by a carrier, reducing it, and releasing energy

Hydrogen ions pumped into intra-membrane space using energy released

When H+ returns to matrix, they release enough energy for ATP to form.

Three molecules ATP per NAD+, Two per FAD+.

Final carrier is oxygen, which accepts electrons and hydrogen ions to form water

Anaerobic respiration.

In the absence of Oxygen, the pyruvate from glycolyis is converted into lactic acid which is excreted from the cell into the blood stream.

Causes acidosis in the blood stream, hence becomes painful after short time in anerobic respiration

Simple reaction formula – Pyruvate + NADH ↔ Lactate + NAD+

Occurs via redox reaction with Pyruvate being reduced.

Requires 2 ATP molecules per glucose molecule, and produces 4 for a net gain of 2 per molecule.

Approximately 19 times less efficient than aerobic respiration, but is able to work faster as krebs cycle and oxidative phosphorylation reactions omitted

Excess lactate either converted back to glucose in the liver, or oxidised back to pyruvate for use in Krebs cycle if anaerobic conditions no longer exist.

Word Count: 330

References

Seeley, R., Stephens, T., Tate, P. and Seeley, R. (1992). Anatomy & physiology. St. Louis: Mosby Year Book.

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