Pre 2016

• (IMAGE) NAME THE PROCESS SHOWN

  translation

  
• (IMAGE) IDENTIFY MOLECULE Q

  tRNA

  
• THE FIRST CODON IS AUG. GIVE THE BASE SEQUENCE OF:

  THE COMPLEMENTARY DNA BASE SEQUENCE

  THE MISSING ANTICODON

  the complementary DNA base sequence: TAC

  the missing anticodon: UAC
• ASPARTIC ACID AND PROLINE ARE BOTH AMINO ACIDS. DESCRIBE HOW 2 AMINO ACISS DIFFER FROM ONE ANOTHER.

  Have different R group

  
• DELETION OF THE SIXTH BASE (G) IN THE SEQUENCE SHOWN IN THE DIAGRAM ABOVE WOULD CHANGE THE NATURE OF THE PROTEIN PRODUCED BUT SUBSTITUTION OF THE SAME BASE WOULD NOT. USE THE INFORMATION IN THE TABLE AND YOUR OWN KNOWLEDGE TO EXPLAIN WHY.

  1. Substitution would result in CCA/CCC/CCU
  2. which all code for the same amino acid proline
  3. deletion would cause a frame shift which would change the next codon from UAC to ACC

  
• MRNA IS USED DURING TRANSLATION TO FORM POLYPEPTIDES. DESCRIBE HOW MRNA IS PRODUCED IN THE NUCLEUS OF A CELL.

  1. Helicase
  2. breaks hydrogen bonds
  3. one strand of the DNA acts as a template
  4. RNA nucleotides attracted to exposed bases
  5. which bind to their complementary base pairs
  6. RNA polymerase joins RNA nucleotides together
  7. the Pre-mRNA formed is spliced to remove introns forming mRNA
• DESCRIBE THE STRUCTURE OF PROTEINS

  1. Polymer of amino acids
  2. joined by peptide bonds
  3. formed by condensation
  4. primary structure is the sequence of amino acids
  5. secondary structure is the folding of a polypeptide chain due to hydrogen bonding
  6. tertiary structure is the 3D folding due to hydrogen and ionic and disulfide bonds
  7. quaternary structure is 2 or more polypeptide chains
• DESCRIBE HOW PROTEINS ARE DIGESTED IN THE HUMAN GUT

  1. Hydrolysis of peptide bonds
  2. endopeptidases break polypeptides into smaller peptide chains
  3. exopeptidases remove terminal amino acids
  4. dipeptidases hydrolyse dipeptides into amino acids
• (IMAGE) MITOCHONDRIAL DISEASE (MD) OFTEN CAUSES MUSCLE WEAKNESS. USE YOUR KNOWLEDGE OF RESPIRATION AND MUSCLE CONTRACTION TO SUGGEST EXPLANATIONS FOR THIS EFFECT OF MD.

  1. Reduction in ATP production by aerobic respiration
  2. less force generated because fewer actin and myosin interactions in muscle
  3. fatigue caused by lactate from anaerobic respiration

  
• (IMAGE)

  COUPLE A:

  1. Mutation in mitochondrial DNA affected
  2. all children got affected mitochondria from the mother
  3. possibly mutation during formation of mothers eggs

  COUPLE B:

  1. mutation in DNA in nucleus affected
  2. parents are heterozygous
  3. so only 1 in 4 homozygous affected

  
• (IMAGE) SUGGEST HOW THE CHANGE IN THE ANTICODON OF A TRNA LEADS TO MD

  1. change to tRNA leads to wrong amino acid being incorporated into protein
  2. tertiary structure of the protein is changed
  3. protein required for oxidative phosphorylation so less ATP made

  
• (IMAGE) IF SOMEONE HAS MD, THE CONCENTRATION OF LACTATE IN THEIR BLOOD AFTER EXERCISE IS USUALLY MUCH HIGHER THAN NORMAL. SUGGEST WHY.

  1. Mitochondria not producing a lot of ATP
  2. with MD there’s increased use of ATP supplied by increased aerobic respiration
  3. more lactate is produced leaves muscle by facilitated diffusion

  
• (IMAGE)

  1. Enough DNA using PCR
  2. compare DNA sequence with normal DNA

  
• THE FAULTY ALLELE THAT CAUSES ELLIS-VAN CREVELD SYNDROME IS THE RESULT OF A MUTATION OF A GENE CALLED EVC. THIS MUTATION LEADS TO THE PRODUCTION OF A PROTEIN THAT HAS ONE AMINO ACID MISSING.

  SUGGEST HOW A MUTATION CAN LEAD TO THE PRODUCTION OF A PROTEIN THAT HAS ONE AMINO ACID MISSING.

  loss of three bases
• SUGGEST HOW THE PRODUCTION OF A PROTEIN WITH ONE AMINO ACID MISSING MAY LEAD TO A GENETIC DISORDER SUCH AS ELLIS-VAN CREVELD SYNDROME

  1. Change in active site
  2. so faulty enzyme so fewer enzyme-substrate complexes formed
• A MUTATION OF A TUMOUR SUPPRESSOR GENE CAN RESULT IN THE FORMATION OF A TUMOUR. EXPLAIN HOW.

  1. The tumour suppressor gene isn’t able to control cell division
  2. so rate of cell division is out of control
• NOT ALL MUTATIONS RESUOT IN A CHANGE TO THE AMINO ACID SEQUENCE OF THE ENCODED POLYPEPTIDE. EXPLAIN WHY.

  1. The genetic code is degenerate
  2. mutation in intron
• A MUTATION CAN LEAD TO THE PRODUCTION OF A NON-FUNCTIONAL ENZYME. EXPLAIN HOW.

  1. mutation in nucleotide sequence of DNA
  2. change in amino acid sequence
  3. change in hydrogen, ionic, disulfide bonds
  4. change in tertiary structure
  5. change in active site
  6. substrate isn’t complementary so no enzyme-substrate complexes form

  
• ONE WAY IN WHICH THE STRINGENT RESPONSE GIVES RESISTANCE TO THIS ANTIBIOTIC IS BY STOPPING CELL DIVISION.

  THE SCIENTISTS CONCLUDED THAT STOPPING CELL DIVISION IS NOT THE ONLY WAY IN WHICH THE STRINGENT RESPONSE GIVES RESISTANCE TO THIS ANTIBIOTIC. EXPLAIN HOW FIGURE 1 SUPPORTS THIS CONCLUSION.

  1. Division stopped of Both strains
  2. SR strain still more resistant at higher concentrations of antibiotic

  
• (IMAGE)

  1. non-SR strain falls more up to 10ug/cm-3
  2. above 10 SR strain levels out and non-SR strain continues to decrease
  3. greater difference between strains with increasing concentration of antibiotic

  
• THE STRINGENT RESPONSE INVOLVES A NUMBER OF ENZYME-CATALYSED REACTIONS.

  EXPLAIN HOW SCIENTISTS COULD USE THIS KNOWLEDGE TO DESIGN DRUGS THAT MAKE THE TREATMENT OF INFECTIONS CAUSED BY THE SR STRAIN MORE SUCCESSFUL.

  1. Maid a competitive or non-competitive inhibitor
  2. the competitive inhibitor competes with non-competitive inhibitor which changes the active site

  OR

  1. make a drug that inhibits stringent response
  2. give Before an antibiotic

  
• (IMAGE)

  1. fewer radicals than non-SR
  2. produces more catalase than non-SR
  3. catalase may be linked to production of fewer free radicals

  
• (IMAGE) DESCRIBE AND EXPLAIN THE APPEARANCE OF ONE OF THE CHROMOSOMES IN CELL X

  1. chromosome is formed of two chromatids
  2. because DNA replication has occured
  3. sister chromatids held together by centromere

  
• (IMAGE) DESCRIBE WHAT HAS HAPPENED DURING DIVISION 1 IN THE FIGURE ABOVE

  1. chromosomes in homologous pair
  2. one of each goes into daughter cells

  
• IDENTIFY ONE EVENT THAT OCCURRED DURING DIVISION 2 BUT NOT DURING DIVISION 1

  separation of sister chromatids or division of centromere

  
• NAME TWO WAYS IN WHICH MEIOSIS PRODUCES GENETIC VARIATION

  1. Independent segregation of homologous chromosomes
  2. crossing over

  
• (IMAGE) CAN YOU CONCLUDE THAT THE INSECT PEST RESISTANT TO BT TOXIN FOUND IN THE YEARS 2002 TO 2005 WAS THE SAME INSECT SPECIES? EXPLAIN.

  graph only shows number of species, not the name of the species.

  
• (IMAGE) ONE FARMER STATED THAT THE INCREAE IN THE USE OF BT CROP HAD CAUSED A MUTATION IN ONE OF THE INSECT SPECIES AND THAT THIS MUTATION HAD SPREAD TO OTHER SPECIES OF INSECT. WAS HE CORRECT? EXPLAIN YOUR ANSWER.

  NO

  1. mutations are spontaneous
  2. only the rate of mutation is affected by environment
  3. different species don’t interbreed / don’t produce fertile offspring
  4. so mutation can’t be passed from one species to another

  
• (IMAGE) THERE WAS A TIME LAG BETWEEN THE INTRODUCTION OF BT CROPS AND THE APPEARANCE OF THE FIRST INSECT SPECIES THAT WAS RESISTANT TO THE BT TOXIN.

  EXPLAIN WHY THERE WAS A TIME LAG.

  1. initially there are a few insects with favourable alleles
  2. individuals with favourable alleles will have more offspring
  3. takes many generations for favourable alleles to become the most common allele of this gene

  
• (IMAGE) SUGGEST AN EXPLANATION FOR THE RESULTS IN THE FIGURE.

  1. mutation produced KDR minus / resistance allele
  2. DDT use provides selection pressure
  3. mosquitoes with KDR minus allele mode likely to survive and reproduce
  4. leading to increase in KDR minus Allele in population
• (IMAGE) THE KDR PLUS ALLELE CODES FOR THE SODIUM ION CHANNELS FOUND IN NEURONES.

  WHEN DDT BINDS TO A SODIUM ION CHANNEL, THE CHANNEL REMAINS OPEN ALL THE TIME. USE THIS INFORMATION TO SUGGEST HOW DDT KILLS INSECTS.

  1. neurones remain depolarised
  2. so no action potentials

  
• (IMAGE) SUGGEST THE THE KDR MINUS ALLELE GIVES RESISTANCE TO DDT

  1. Mutation changes shape of sodium ion channel protein
  2. DDT no longer complementary so can’t bind

  
• GUVE THREE WAYS IN WHICH COURTSHIP BEHAVIOUR INCREASES THE PROBABILITY OF SUCCESSFUL MATING

  1. recognise same species
  2. synchronises mating
  3. attraction of mate / opposite sex
  4. indication of sexual maturity
  5. formation of a bond between two organisms to raise young
• (IMAGE) THE SCIENTIST WANTED TO KNOW IF THE RECORDED NATURAL SONG WAS LESS EFFECTIVE THAN THE NATURAL SONG IN STIMULATING COURTSHIP BEHAVIOUR.

  SUGGEST HOW THE SCIENTISTS COULD DETERMINE IF THE RECORDED NATURAL SONG (L) WAS LESS EFFECTIVE THAN THE NATURAL SONG.

  1. use a male with intact wings
  2. determine percentage response of females compared with L

  
• (IMAGE) A STUDENT CONCLUDED FROM THE DATA TAYT THE NUMBER OF CHIRPS AND TICKS IS ESSENTIAL FOR SUCCESSFULLY STIMULATING COURTSHIP BEHAVIOUR.

  DOES THIS DATA SUPPORT THIS CONCLUSION? EXPLAIN.

  1. only 30% courtship with no song - so still courtship occurred when no song played - K
  2. reduced courtship when no ticks - M
  3. Reduced courtship when no chirps - N
  4. so courtship must involve a visual stimulus
  5. Chirps more important as lowest courtship when none - N. Ticks less important as similar courtship when changed - M
  6. data only shows presence and absence of chirps

  
• (IMAGE) TO WHICH CLASS DOES THE WHITE FRONTED GROUND DOVE BELONG?

  Aves

  
• (IMAGE) GIVE THE SCIENTIFIC NAME FOR THE WHITE-FRONTED GROUND DOVE

  Gallicolumba kubaryi

  
• (IMAGE) THIS CLASSIFICATION SYSTEM CONSISTS OF A HIERARCHY AS THERE ARE SMALL GROUPS WITHIN LARGER GROUPS. GIVE ONE OTHER FEATURE OF A HIERARCHY THAT IS SHOWN.

  no overlap

  
• WHAT IS MEANT BY A HIERARCHY?

  1. groups within groups
  2. with no overlap between groups
• (IMAGE) HOW MANY DIFFERENT FAMILIES ARE SHOWN IN THE FIGURE

  3