How to Clear RAM on the GDC

Study Tips for Math

STUDY TIPS FOR MATH

1. Always read math problems completely before beginning any calculations. If you "glance" too quickly at a problem, you may misunderstand what really needs to be done to complete the problem.

2. Whenever possible, draw a diagram. Even though you may be able to visualize the situation mentally, a hand drawn diagram will allow you to label the picture, to add auxiliary lines, and to view the situation from different perspectives.

3. Know your calculator! If you must borrow a calculator from your peers, be sure that you have used that "brand" of calculator on previous occasions. If you are not familiar with how a particular calculator works, your calculations may be incorrect. 

4. If you know that your answer to a question is incorrect, and you cannot find your mistake, start over on a clean piece of paper.  Oftentimes when you try to correct a problem, you continually overlook the mistake.  Starting over on a clean piece of paper will let you focus on the question, not on trying to find the error.

5. Do not feel that you must use every number in a problem when doing your calculations. Some mathematics problems have "extra" information.  These questions are testing your ability to recognise the needed information, as well as your mathematical skills. 

6. Be sure that you are working in the same units of measure when performing calculations.  If a problem involves inches, feet AND yards, be sure to make the appropriate conversions so that all of your values are in the same unit of measure (for example, change all values to feet). 

7. Be sure that your answer "makes sense" (or is logical).  For example, if a question asks you to find the number of feet in a drawing and your answer comes out to be a negative number, know that this answer is incorrect.  (Distance is a positive concept - we cannot measure negative feet.) 

8. Remember, that it may be necessary to "solve" for additional information in a problem before being able to arrive at the final answer.  These questions are called "two-step" problems and are testing your ability to recognise what information is needed to arrive at an answer. 

9. If time permits, go back and resolve the more difficult problems on the test on a separate piece of paper.  If these "new" answers are the same as your previous answers, chances are good that your solution is correct. 

10. Remain confident!  Do not get flustered!  Focus on what you DO know, not on what you do not know.  You know a LOT of math!! 

11. When asked to "show work" or "justify your answer", don't be lazy.  Write down EVERYTHING about the problem, including the work you did on your calculator. Include diagrams, calculations, equations, and explanations written in complete sentences.  Now is the time to "show off" what you really can do with this problem. 

12. If you are "stuck" on a particular problem, go on with the rest of the test.  Oftentimes, while solving a new problem, you will get an idea as to how to attack that difficult problem. 

13. If you simply cannot determine the answer to a question, make a guess.  Think about the problem and the information you know to be true.  Make a guess that will be logical based upon the conditions of the problem. 

14. In certain problems, you may be able to "guess" at an approximate (or reasonable) answer.  After you perform your calculations, see if your final answer is close to your guess. 

A PROBLEM A DAY KEEPS FEAR OF MATH AWAY!

Jessica To

IB Biology SL - Unit 3 - Chemistry of Life

Unit 3

3.1  Chemistry of life

3.1.1
State that the most frequently occurring chemical elements in living things are carbon, hydrogen, oxygen and nitrogen.

3.1.2

State that a variety of other elements are needed by living organisms, including sulfur, calcium, phosphorus, iron and sodium.

3.1.3

State one role for each of the elements mentioned in 3.1.2.

Nitrogen:

- Required by proteins (included in amino acid structure).
- Essential to enzymes required for plant function

Calcium:

- Strengthens bone and teeth
- Regulates cell wall construction in plants

Phosphorus:
- Part of the phosphate groups in ATP and DNA molecules
- Make up part of the ‘backbone’ of DNA
-  Formation of the phospholipid bilayer
(Hydrophilic phosphate heads ATTRACTED + Hydrophobic hydrocarbon tails)

Iron:
- Found in the structure of hemoglobin and is essential for the production of red blood cells.
- Involved in light energy transferring compounds involved in photosynthesis in plants.

Sodium:
- major ion associated with the propagation of a nerve impulse.
- Can replace potassium in some plants

Sulfur:

- Found in some amino acids (proteins)

3.1.4  Draw and label a diagram showing the structure of water molecules to show their polarity and hydrogen bond formation.

3.1.5  Outline the thermal, cohesive and solvent properties of water.

Thermal properties

• Hydrogen bonds between polar water molecules cause water to resist change
• High specific heat (energy required to change water temperature)
• High heat of vaporization (energy required to boil water)
• High heat of fusion (loss of energy required to freeze water)
• Thus, water produces a stable environment for aquatic organisms

Cohesion

• Hydrogen bonds between polar water molecules cause them to cohere
• Allowing for transpiration in plants moving water against gravity
• Surface tension between cohering water molecules
• Allowing for animals such as water striders to walk over the surface of ponds even though they are denser than water

Solvent properties

• The polarity of water attracts, or dissolves, any other polar or charged particles by forming hydrogen bonds with them
• Proteins, glucose, or ions, such as sodium or calcium are all soluble
  
  

3.1.6. Explain the relationship between the properties of water and its uses in living organisms as a coolant, medium for metabolic reactions and transport medium.

Coolant

• Hydrogen bonds between polar water molecules cause water to resist change
• High heat of vaporization (energy required to change liquid water to vapor) because hydrogen bonds must be broken
• Thus, evaporation of water from plant leaves (transpiration) or from human skin (sweat) removes heat, acting as a coolant

Medium for metabolic reactions

• Cytoplasm is primarily water, providing a polar medium in which other polar or charged molecules dissolve
• Many enzymes are globular proteins that are water soluble so they dissolve in cytoplasm where they control metabolic reactions

Transport medium

• Hydrogen bonds between polar water molecules cause them to cohere
• as water is lost by transpiration/evaporation from plant leaves, hydrogen bonds between adjacent water molecules pull water up columns of xylem
• Thus, plants move water against gravity, by as much as 100 meters

3.2 Carbohydrates, lipids and proteins

3.2.1 Distinguish between organic and inorganic compounds.

Organic: Compounds containing carbon that are found in living organisms (except hydrogencarbonates, carbonates and oxides of carbon) are regarded as organic.

• exceptions: carbonates (e.g. CaCO3), hydrogen carbonates (e.g. HCO3), and oxides of carbon (e.g. CO, CO2)

• all other compounds are regarded as inorganic

3.2.2  Identify amino acids, glucose, ribose and fatty acids from diagrams showing their structure

Amino Acid.

Glucose

Ribose

Fatty acids

3.2.3 List three examples each of monosaccharide, disaccharides and polysaccharides

Monosaccharide:
- Glucose
- Fructose
- Galactose

(ribose)

Disaccharides:

- Maltose
- Lactose

- sucrose

Polysaccharides:

- starch
- glycogen
- cellulose

3.2.4 State one function of glucose, lactose and glycogen in animals, and of fructose, sucrose and cellulose in plants.

glucose, lactose and glycogen in animals

• glucose = monosaccharide: major energy source for fueling cellular respiration

• lactose = disaccharide: energy source found in mammalian milk

• glycogen = polysaccharide: energy storage in liver

and fructose, sucrose and cellulose in plants

• fructose = monosaccharide: energy component in flower nectar

• sucrose = disaccharide: energy molecule transported via phloem

• cellulose = polysaccharide: major structural component of plant cell walls

3.2.5 Outline the role of condensation and hydrolysis in the relationships between monosaccharides, disaccharides and polysaccharides; between fatty acids, glycerol and triglycerides; and between amino acids and polypeptides.

condensation synthesis

• condensation synthesis reactions link two monosaccharide monomers

• forming one disaccharide molecule and one H2O molecule

• repeated additions of monosaccharides produces a polysaccharide

hydrolysis

• The addition of water to split polysaccharides into smaller fragments. –OH and –H are added to the fragments. 

• H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

fatty acids, glycerol and triglycerides:

condensation synthesis

• three separate condensation synthesis reactions

• link three fatty acid monomers to a single glycerol monomer

• forming one triglyceride molecule and three H2O molecules

hydrolysis

• a triglyceride is broken down into one glycerol and three fatty acid molecules

• with three H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

amino acids and polypeptides:

condensation synthesis

• two amino acid monomers are linked to form a dipeptide

• releasing one H2O molecule

• repeated condensation synthesis reactions produce polypeptides (=proteins)

hydrolysis

• a polypeptide is broken down into separate amino acid molecules

• with H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

3.2.6 State three functions of lipids.
- Energy storage

- Thermal insulation

- Hormones – used in messenger molecules (used as insulators to speed up the transfer of nerve impulses from nerves)


3.2.7  Compare the use of carbohydrates in energy storage:

A. carbohydrates

• Glucose, when in excess, linked together by condensation synthesis to form polysaccharides

• Such as starch, glycogen, and chitin

starch

• the primary energy storage compound in plants, with å-glucose

• forming either as amylose (unbranched chains)

• amylopectin (branched chains)

glycogen

• a branched polysaccharide formed from å-glucose

• primary energy storage compound in animals

• found mainly in liver and muscle cells

B. lipids

triglycerides

• a primary long-term energy storage compound

• containing twice the amount of energy per unit mass or volume as do either carbohydrates or proteins  (more energy compared to proteins/carbohydrates)

C-H bonds

• along the length of fatty acids are very stable

• because the electrons in the C-H bond are equally shared

• thus, when these bonds are broken, they yield large sums of energy

3.3 DNA structure

3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate.

- A DNA is composed of a deoxyribose, a phosphate group and a nitrogen base (Adenine, Cytosine, Thymine, and Guanine).

- The phosphate group is covalently bonded to the carbon of the deoxyribose and then nitrogenous base is attached to the deoxyribose.

3.3.2 State the names of the four bases in DNA.
Cytosine
Guanine
Thymine
Adenine

 3.3. 3 Outline how DNA nucleotides are linked together by covalent bonds into a single strand.

Two DNA nucleotides can be linked together by a covalent bond between the sugar of one nucleotide and the phosphate of another. More nucleotides can be added to form a single strand.

3.3.4 Explain how a DNA double helix is formed using complementary base pairing and hydrogen bonds.
DNA molecules consist of two strands of nucleotides which are then wound together to form a double helix. These are formed between the bases of two strands. However, it is formed by complementary base pairing because Adenine only forms hydrogen bonds with Thymine and Cytosine only forms hydrogen bonds with Guanine.

3.4 DNA replication

3.4.1 Explain DNA replication in terms of unwinding the double helix and separation of the strands by helicase, followed by formation of the new complementary strands by DNA polymerase.

1) Production of new DNA à semi conservative à one old and one new strand

2) Helicase unwinds the DNA/double helix to allow the strands to be separated
by breaking hydrogen bonds between bases.
3) Single strand act as a template.
4) Free bases form H-bonds with complementary bases.
5) Nucleotides are linked by DNA polymerase.
6) Daughter DNA molecules rewind.
7) Daughter strand identical to parent strand.

3.4.2 Explain the significance of complementary base pairing in the conservation of the base sequence of DNA.

Because the nitrogenous bases that compose DNA can only pair with complementary bases, any two linked strands of DNA are complementary. This ensures that the old base sequence is conserved.

A bonds with T and G bonds with C. Complementary base pairing ensures proper base incorporated into DNA strand thus making identical copies of the DNA strand (ensuring conservation not the base sequence).

How the base sequence of DNA is conserved during replication.

DNA replication is semi-conservative.
DNA is split into two single strands.
Nucleotides are assembled on to each single strand by complementary base pairing
AT, CG
Strand newly formed on each single strand is identical to other single strand.

DNA polymerase used.

3.4.3 State that DNA replication is semi-conservative.

DNA replication copies DNA to produce new molecules with the same base sequence. It is semi-conservative because each new molecule formed by replication uses one new strand and one old strand which is conserved from the parent DNA molecule.

3.5 Transcirption and translation

3.5.1 Compare the structure of RNA and DNA.

1. The number of strands.

·       DNA has two strands forming a double helix.

·       RNA has one strand.

2. The type of sugar.

·       DNA has a deoxyribose.

·       RNA has a Ribose.

3. The Nucleotides.

·       DNA has A,C,G,T

·       RNA has A,C,G,U, (Uracil replaces Thymine)

3.5.2 Outline DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase.

Transcription: The copying of the base sequence of a gene by making an RNA molecule.

Initiation:

- The DNA double helix uncoils and the two strands separate.

- RNA Polymerase attaches to promoter regions of the DNA strand (starting point)

Elongation:
- RNA polymerase makes a mRNA molecule that is complementary to the DNA.
- (Uses DNA anti-sense strand as a template, synthesizes a complementary RNA strand using base pairing rules)

Termination:

- The mRNA separates from the DNA.                  

- The DNA strands reforms into a double helix.

1.    

3.5.3 Describe the genetic code in terms of codons composed of triplets of bases

triplet code = 3 nucleotide bases code for one amino acid

codon = a group of 3 nucleotide bases à codon

there are 64 different codons (4 x 4 x4 = 64)

3.5.4 Explain the process of translation, leading to polypeptide formation

1) mRNA attaches to ribosome (small unit)

2) Many ribosome bind to the same mRNA, which carries codons.
3) Transfer tRNA each have specific anticodon that binds to ribosomes to corresponding triplet base/codon.
4) A second tRNA binds to next codon.
5) Two amino acids bind together in a peptide linkage.
6) First tRNA detaches, ribosomes move along mRNA.
7) Another tRNA binds to next codon and it continues until polypeptide/protein is formed to stop codon (UAG, UAA and UGA.)
8) Stop codon has no corresponding tRNA, causes release of polypeptide

3.5.5 Discuss the relationship between one gene and one polypeptide.

Polypeptides are long chains of amino acids.

Amino acids must be linked up in a precise sequence to make a polypeptide.

Genes store the information needed to make a polypeptide in a coded form.

The sequence of bases in a gene codes for the sequence of amino acids in a polypeptide.

The information in the gene is decoded during the making of the polypeptide.

This process is conducted in two stages known as transcription and translation.

DNA (sequence) is copied to (m)RNA.
DNA is separated.
RNA polymerase separate the strands, only one strand is copied.
RNA polymerase binds to promoter of template strand.
ATP provides energy for attachment.
RNA polymerase catalyzes the formation of the polymer
RNA lengthens in 5’ to 3’ direction.
RNA polymerase reaches terminator on DNA to stop transcription process.
mRNA is separated from DNA.

3.5.5 Discuss the relationship between one gene nad one polypeptide.

3.6 Enzymes

3.6.1 Define enzymes and active site.

Enzyme: A globular protein functioning as a biological catalyst, speeding up reaction rates by lowering activation energy
Active site: Site on surface of the enzyme to which the substrate binds.

3.6.2 Explain enzyme–substrate specificity.

Enzymes fit together with substrates similar to a lock and key.
Active site has shape that gives specificity.
Enzymes catalyze a reaction with a specific substrate.

3.6.3 Explain the effects of temperature, pH and substrate concentration on enzyme activity.

pH:
Enzymes have an optimum pH, active site works best at this pH.
Activity decreases above and below the optimum by interfering with active site structure, denaturing by extremes of pH so enzyme activity stops.

Temperature:

Each enzyme has an optimal temperature for it’s maximum activity
Small temperature increases result in increased enzyme activity to a point.
Increase activity due to increased movement of molecules (more KE).
Temperature increases above the optimum causes loss of activity due to denaturation.

Substrate concentration:
Low – As increase concentration get increase in rate of reaction.
More chance of collision between substrate and enzyme/active site.
At high substrate concentration- have no change in rate as increase concentration; all active sites occupied.

3.6.4 Define denaturation
Denaturation is a structural change in a protein that results in the loss of it’s biological properties.

3.6.5 Explain the use of lactase in the production of lactose-free milk.

-  Used to improve the digestion of milk by some people.

- Breaks down lactose into glucose and galactose.
- Increase sweetness of milk (glucose and galactose are sweeter)
- No need to add extra sugar in manufacture of flavoured milk drinks/frozen desserts,

IB Biology SL - Unit 3 - Chemistry of Life

Unit 3

3.1  Chemistry of life

3.1.1
State that the most frequently occurring chemical elements in living things are carbon, hydrogen, oxygen and nitrogen.

3.1.2

State that a variety of other elements are needed by living organisms, including sulfur, calcium, phosphorus, iron and sodium.

3.1.3

State one role for each of the elements mentioned in 3.1.2.

Nitrogen:

- Required by proteins (included in amino acid structure).
- Essential to enzymes required for plant function

Calcium:

- Strengthens bone and teeth
- Regulates cell wall construction in plants

Phosphorus:
- Part of the phosphate groups in ATP and DNA molecules
- Make up part of the ‘backbone’ of DNA
-  Formation of the phospholipid bilayer
(Hydrophilic phosphate heads ATTRACTED + Hydrophobic hydrocarbon tails)

Iron:
- Found in the structure of hemoglobin and is essential for the production of red blood cells.
- Involved in light energy transferring compounds involved in photosynthesis in plants.

Sodium:
- major ion associated with the propagation of a nerve impulse.
- Can replace potassium in some plants

Sulfur:

- Found in some amino acids (proteins)

3.1.4  Draw and label a diagram showing the structure of water molecules to show their polarity and hydrogen bond formation.

3.1.5  Outline the thermal, cohesive and solvent properties of water.

Thermal properties

• Hydrogen bonds between polar water molecules cause water to resist change
• High specific heat (energy required to change water temperature)
• High heat of vaporization (energy required to boil water)
• High heat of fusion (loss of energy required to freeze water)
• Thus, water produces a stable environment for aquatic organisms

Cohesion

• Hydrogen bonds between polar water molecules cause them to cohere
• Allowing for transpiration in plants moving water against gravity
• Surface tension between cohering water molecules
• Allowing for animals such as water striders to walk over the surface of ponds even though they are denser than water

Solvent properties

• The polarity of water attracts, or dissolves, any other polar or charged particles by forming hydrogen bonds with them
• Proteins, glucose, or ions, such as sodium or calcium are all soluble
  
  

3.1.6. Explain the relationship between the properties of water and its uses in living organisms as a coolant, medium for metabolic reactions and transport medium.

Coolant

• Hydrogen bonds between polar water molecules cause water to resist change
• High heat of vaporization (energy required to change liquid water to vapor) because hydrogen bonds must be broken
• Thus, evaporation of water from plant leaves (transpiration) or from human skin (sweat) removes heat, acting as a coolant

Medium for metabolic reactions

• Cytoplasm is primarily water, providing a polar medium in which other polar or charged molecules dissolve
• Many enzymes are globular proteins that are water soluble so they dissolve in cytoplasm where they control metabolic reactions

Transport medium

• Hydrogen bonds between polar water molecules cause them to cohere
• as water is lost by transpiration/evaporation from plant leaves, hydrogen bonds between adjacent water molecules pull water up columns of xylem
• Thus, plants move water against gravity, by as much as 100 meters

3.2 Carbohydrates, lipids and proteins

3.2.1 Distinguish between organic and inorganic compounds.

Organic: Compounds containing carbon that are found in living organisms (except hydrogencarbonates, carbonates and oxides of carbon) are regarded as organic.

• exceptions: carbonates (e.g. CaCO3), hydrogen carbonates (e.g. HCO3), and oxides of carbon (e.g. CO, CO2)

• all other compounds are regarded as inorganic

3.2.2  Identify amino acids, glucose, ribose and fatty acids from diagrams showing their structure.

Amino acid

Glucose

Ribose

Fatty acids

3.2.3 List three examples each of monosaccharide, disaccharides and polysaccharides

Monosaccharide:
- Glucose
- Fructose
- Galactose

(ribose)

Disaccharides:

- Maltose
- Lactose

- sucrose

Polysaccharides:

- starch
- glycogen
- cellulose

3.2.4 State one function of glucose, lactose and glycogen in animals, and of fructose, sucrose and cellulose in plants.

glucose, lactose and glycogen in animals

• glucose = monosaccharide: major energy source for fueling cellular respiration

• lactose = disaccharide: energy source found in mammalian milk

• glycogen = polysaccharide: energy storage in liver

and fructose, sucrose and cellulose in plants

• fructose = monosaccharide: energy component in flower nectar

• sucrose = disaccharide: energy molecule transported via phloem

• cellulose = polysaccharide: major structural component of plant cell walls

3.2.5 Outline the role of condensation and hydrolysis in the relationships between monosaccharides, disaccharides and polysaccharides; between fatty acids, glycerol and triglycerides; and between amino acids and polypeptides.

condensation synthesis

• condensation synthesis reactions link two monosaccharide monomers

• forming one disaccharide molecule and one H2O molecule

• repeated additions of monosaccharides produces a polysaccharide

hydrolysis

• The addition of water to split polysaccharides into smaller fragments. –OH and –H are added to the fragments. 

• H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

fatty acids, glycerol and triglycerides:

condensation synthesis

• three separate condensation synthesis reactions

• link three fatty acid monomers to a single glycerol monomer

• forming one triglyceride molecule and three H2O molecules

hydrolysis

• a triglyceride is broken down into one glycerol and three fatty acid molecules

• with three H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

amino acids and polypeptides:

condensation synthesis

• two amino acid monomers are linked to form a dipeptide

• releasing one H2O molecule

• repeated condensation synthesis reactions produce polypeptides (=proteins)

hydrolysis

• a polypeptide is broken down into separate amino acid molecules

• with H2O molecules used as a sources of -H and a -OH groups

• catalyzed by enzymes

3.2.6 State three functions of lipids.
- Energy storage

- Thermal insulation

- Hormones – used in messenger molecules (used as insulators to speed up the transfer of nerve impulses from nerves)


3.2.7  Compare the use of carbohydrates in energy storage:

A. carbohydrates

• Glucose, when in excess, linked together by condensation synthesis to form polysaccharides

• Such as starch, glycogen, and chitin

starch

• the primary energy storage compound in plants, with å-glucose

• forming either as amylose (unbranched chains)

• amylopectin (branched chains)

glycogen

• a branched polysaccharide formed from å-glucose

• primary energy storage compound in animals

• found mainly in liver and muscle cells

B. lipids

triglycerides

• a primary long-term energy storage compound

• containing twice the amount of energy per unit mass or volume as do either carbohydrates or proteins  (more energy compared to proteins/carbohydrates)

C-H bonds

• along the length of fatty acids are very stable

• because the electrons in the C-H bond are equally shared

• thus, when these bonds are broken, they yield large sums of energy

3.3 DNA structure

3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate.

- A DNA is composed of a deoxyribose, a phosphate group and a nitrogen base (Adenine, Cytosine, Thymine, and Guanine).

- The phosphate group is covalently bonded to the carbon of the deoxyribose and then nitrogenous base is attached to the deoxyribose.

3.3.2 State the names of the four bases in DNA.
Cytosine
Guanine
Thymine
Adenine

3.3. 3 Outline how DNA nucleotides are linked together by covalent bonds into a single strand.

Two DNA nucleotides can be linked together by a covalent bond between the sugar of one nucleotide and the phosphate of another. More nucleotides can be added to form a single strand.

3.3.4 Explain how a DNA double helix is formed using complementary base pairing and hydrogen bonds.
DNA molecules consist of two strands of nucleotides which are then wound together to form a double helix. These are formed between the bases of two strands. However, it is formed by complementary base pairing because Adenine only forms hydrogen bonds with Thymine and Cytosine only forms hydrogen bonds with Guanine.

3.4 DNA replication

3.4.1 Explain DNA replication in terms of unwinding the double helix and separation of the strands by helicase, followed by formation of the new complementary strands by DNA polymerase.

1) Production of new DNA à semi conservative à one old and one new strand

2) Helicase unwinds the DNA/double helix to allow the strands to be separated
by breaking hydrogen bonds between bases.
3) Single strand act as a template.
4) Free bases form H-bonds with complementary bases.
5) Nucleotides are linked by DNA polymerase.
6) Daughter DNA molecules rewind.
7) Daughter strand identical to parent strand.

3.4.2 Explain the significance of complementary base pairing in the conservation of the base sequence of DNA.

Because the nitrogenous bases that compose DNA can only pair with complementary bases, any two linked strands of DNA are complementary. This ensures that the old base sequence is conserved.

A bonds with T and G bonds with C. Complementary base pairing ensures proper base incorporated into DNA strand thus making identical copies of the DNA strand (ensuring conservation not the base sequence).

How the base sequence of DNA is conserved during replication.

DNA replication is semi-conservative.
DNA is split into two single strands.
Nucleotides are assembled on to each single strand by complementary base pairing
AT, CG
Strand newly formed on each single strand is identical to other single strand.

DNA polymerase used.

3.4.3 State that DNA replication is semi-conservative.

DNA replication copies DNA to produce new molecules with the same base sequence. It is semi-conservative because each new molecule formed by replication uses one new strand and one old strand which is conserved from the parent DNA molecule.

3.5 Transcirption and translation

3.5.1 Compare the structure of RNA and DNA.

1. The number of strands.

·       DNA has two strands forming a double helix.

·       RNA has one strand.

2. The type of sugar.

·       DNA has a deoxyribose.

·       RNA has a Ribose.

3. The Nucleotides.

·       DNA has A,C,G,T

·       RNA has A,C,G,U, (Uracil replaces Thymine)

3.5.2 Outline DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase.

Transcription: The copying of the base sequence of a gene by making an RNA molecule.

Initiation:

- The DNA double helix uncoils and the two strands separate.

- RNA Polymerase attaches to promoter regions of the DNA strand (starting point)

Elongation:
- RNA polymerase makes a mRNA molecule that is complementary to the DNA.
- (Uses DNA anti-sense strand as a template, synthesizes a complementary RNA strand using base pairing rules)

Termination:

- The mRNA separates from the DNA.                  

- The DNA strands reforms into a double helix.

1.    

3.5.3 Describe the genetic code in terms of codons composed of triplets of bases

triplet code = 3 nucleotide bases code for one amino acid

codon = a group of 3 nucleotide bases à codon

there are 64 different codons (4 x 4 x4 = 64)

3.5.4 Explain the process of translation, leading to polypeptide formation

1) mRNA attaches to ribosome (small unit)

2) Many ribosome bind to the same mRNA, which carries codons.
3) Transfer tRNA each have specific anticodon that binds to ribosomes to corresponding triplet base/codon.
4) A second tRNA binds to next codon.
5) Two amino acids bind together in a peptide linkage.
6) First tRNA detaches, ribosomes move along mRNA.
7) Another tRNA binds to next codon and it continues until polypeptide/protein is formed to stop codon (UAG, UAA and UGA.)
8) Stop codon has no corresponding tRNA, causes release of polypeptide

3.5.5 Discuss the relationship between one gene and one polypeptide.

Polypeptides are long chains of amino acids.

Amino acids must be linked up in a precise sequence to make a polypeptide.

Genes store the information needed to make a polypeptide in a coded form.

The sequence of bases in a gene codes for the sequence of amino acids in a polypeptide.

The information in the gene is decoded during the making of the polypeptide.

This process is conducted in two stages known as transcription and translation.

DNA (sequence) is copied to (m)RNA.
DNA is separated.
RNA polymerase separate the strands, only one strand is copied.
RNA polymerase binds to promoter of template strand.
ATP provides energy for attachment.
RNA polymerase catalyzes the formation of the polymer
RNA lengthens in 5’ to 3’ direction.
RNA polymerase reaches terminator on DNA to stop transcription process.
mRNA is separated from DNA.

3.5.5 Discuss the relationship between one gene nad one polypeptide.

3.6 Enzymes

3.6.1 Define enzymes and active site.

Enzyme: A globular protein functioning as a biological catalyst, speeding up reaction rates by lowering activation energy
Active site: Site on surface of the enzyme to which the substrate binds.

3.6.2 Explain enzyme–substrate specificity.

Enzymes fit together with substrates similar to a lock and key.
Active site has shape that gives specificity.
Enzymes catalyze a reaction with a specific substrate.

3.6.3 Explain the effects of temperature, pH and substrate concentration on enzyme activity.

pH:
Enzymes have an optimum pH, active site works best at this pH.
Activity decreases above and below the optimum by interfering with active site structure, denaturing by extremes of pH so enzyme activity stops.

Temperature:

Each enzyme has an optimal temperature for it’s maximum activity
Small temperature increases result in increased enzyme activity to a point.
Increase activity due to increased movement of molecules (more KE).
Temperature increases above the optimum causes loss of activity due to denaturation.

Substrate concentration:
Low – As increase concentration get increase in rate of reaction.
More chance of collision between substrate and enzyme/active site.
At high substrate concentration- have no change in rate as increase concentration; all active sites occupied.

3.6.4 Define denaturation
Denaturation is a structural change in a protein that results in the loss of it’s biological properties.

3.6.5 Explain the use of lactase in the production of lactose-free milk.

-  Used to improve the digestion of milk by some people.

- Breaks down lactose into glucose and galactose.
- Increase sweetness of milk (glucose and galactose are sweeter)
- No need to add extra sugar in manufacture of flavoured milk drinks/frozen desserts,