DNA, Genetics, Evolution

Question 1

Differences between DNA and RNA

Answer 1

Question 2

DNA structure, Crick and Watson, Rosalind Franklin

Answer 2

1. DNA molecular structure correctly proposed by James Watson & Francis Crick.
2. They constructed models to quickly visualise & assess viability of potential structures
3. 1^st model was triple helix with bases on outside & sugar-phosphate residues in centre, with Mg cross-links between strands.
4. Guided by:
   
   1. Molecular distances & bond angles
      – Linus Pauling
   2. DNA composed of nucleotides made up of sugar + phosphate + base.
      – Phoebus Levene
   3. DNA X-ray crystallography data showed it’s organised into helical structure – RF
      (without permission):
      
      1. DNA purified, then fibres were stretched in thin glass tube (to make most strands parallel)
      2. DNA targeted by X-ray beam, which diffracts when it contacted an atom
      3. Scattering pattern of X-ray distinctive
         so recorded on film & used to find
         DNA molecular structure:
      4. Structure:
         1. ​X in centre of diffraction pattern indicated DNA = double helix.
         2. DNA molecule shown to twist at regular intervals to form helix.
         3. X-shape angle showed pitch of helix.
      5. Orientation: N-bases closely packed together on inside & phosphates form an outer backbone
5. Rosalind Franklin rejected model as not enough Mg & didn’t support Erwin Chargaff’s findings (# of A + G = # of T + C).
6. Using trial & error, Watson & Crick assembled DNA model that showed:
   
   1. DNA strands antiparallel & form double helix.
   2. DNA strands pair via comp. base pairing (A = T ; C Ξ G)
   3. Outer edges of bases remain exposed (allows access to proteins involved in transcription & replication)
   4. Potential DNA replication mechansims:
      
      1. Replication occurs via comp. base pairing (A pairs with T, G pairs with C)
      2. Replication is bi-directional due to antiparallel nature of strands

Question 3

DNA Replication

Answer 3

Context:

1. Occurs during S phase of Interphase.
2. Replication is a semi-conservative process as new DNA contains 1 old & 1 new strand.
3. Parent DNA strands act as templates for new
   strands.
4. Comp. bases from opp. strands form H-bonds (A = T & C ≡ G)
5. Occurs in a 5’ to 3’ direction.
6. Consequently, when DNA replicated:
   1. Each new strand formed identical to original strand separated from template
   2. Both DNA molecules formed have equal
      base seq. to original molecule.

DNA replication Process

1. Helicase: Unwinds double helix by breaking H-bonds between 2 strands. → forms replic. fork with strands running in antiparallel directions.
2. DNA Gyrase: ↓ strain created by helicase by relaxing + supercoils (via (–) supercoiling).
3. Single Stranded Binding (SSB) Proteins:
   
   1. Split DNA strands & prevent strand from re-annealing
   2. Prevent nucleases from digesting SS-DNA
   3. Dislodged from strand when new comp. strand synth. by DNA pol III.
4. RNA Primase: Synthesises RNA primers on each template strand, RNA primers are binding spots for DNA-PIII.
5. DNA-PIII: Once bound to primer:
   
   1. Cleaves PO₄’s from nearby DNsTP, to form nucleotides which are added to 3’ end of primer; according to comp. base pairing
      (A = T + C ≡ G)
   2. Cleaving PO₄’s releases nrg used to form covalent bond between nucleotides.
   3. Thus, DNA-PIII synthesises new DNA between RNA primers (Okazaki frags).
6. Leading Strand: Con. synth. where DNA-PIII moves towards replic. fork.
7. Lagging Strand: Discon. synth. where DNA-PIII moves away from fork → create Okazaki frags as fork exposes more temp. strand.
8. DNA-PI: Excises RNA primers & replaces them with DNA;
9. DNA ligase: Joins gaps between Okazaki frags by making cov. bonds between nucleotides to form a continuous strand.

Question 4

Transcription

Answer 4

Sections of Gene:

1. Promoter: Non-coding seq. that acts as binding site for RNA pol, thus starting transcription.
2. Sense Strand: DNA strand with same base sequence as RNA. Not transcribed.
3. Antisense Strand: Temp. strand with comp. base seq. to RNA & sense strand. Transcribed.
4. Terminator: Non-coding seq. that signals RNA pol to detach from DNA, thus ends transcript.

Process:

1. Transcription: RNA seq. synthesis using DNA template, by RNA polymerase. Occurs in nuc.
2. Initiation:
   1. RNA pol binds to promoter & unwinds + separates DNA strands by breaking H-bonds between comp. base pairs (using NRG).
   2. Nucleoside triphosphates (NsTP) line up opp. their comp. base partner.
3. Elongation:
   
   1. RNA pol excises 2 phosphates → NRG
   2. NRG used to bind (now) free nucleotides together (thus, synth. RNA) in 5’ → 3’ dir.
   3. (Forms coding seq.)
4. Termination:
   1. ​RNA pol reaches terminator → both RNA pol & synth. RNA strand detach &
      DNA double helix re-forms.

Question 5

Nucleosomes

Answer 5

1. In eukaryotic organisms, DNA packaged with histones to form (nucleosome).
2. Histones: Proteins used by cell to package DNA into nucleosomes.
3. Nucleosomes: Molecules consisting of 8 histones (octamer) with DNA coiled around.
   
   1. Help to supercoil DNA, resulting in greatly compacted structure that allows for more efficient storage.
   2. Histone tails are + charged, so associate to DNA & determine tightness of packing
      
      1. ​​Acetylation: Adding acetyl group to histone tail, dec. + charge → DNA less tightly coiled & inc. transcript.
         (Euchromatin)
      2. Methylation: Adding methyl group to histone tail, inc. + charge → DNA more tightly coiled & dec. transcript.
         (Heterochromatin)
4. Supercoiling helps:
   
   1. Protect DNA from damage
   2. Mobilise chrom’s during mitosis & meiosis.

Organisation of Eukaryotic DNA

1. DNA packaged with 8 histone proteins (an octamer) → complex (nucleosome)
2. H1 histone binds to linker DNA, which binds nucleosomes together, to form
   * *chromatosomes**
3. These coil to form more condensed solenoid fibre structure, which then form loops.
4. Loops compressed & folded around protein scaffold to form chromatin (eu- or hetero-).
5. Chromatin then supercoils during cell division to form chromosomes visible (when stained) under microscope

Question 6

Epigenetics

Answer 6

Epigenetics: Study of phenotypic changes caused by variations in gene expression levels.

1. DNA Methylation prevents TF binding, so ↓ gene expression/transcription.
2. Thus, transcription of gene ind. prop. to DNA methyl.
3. Epigenetic analysis shows that DNA methylation patterns/gene expression may change over course of a lifetime
4. Diff cell types in same organism may have markedly diff DNA methylation patterns.
5. Env. factors (e.g. diet, pathogen exposure, etc.) also affect DNA methyl. within cells.
6. Also influenced by heritability but not genetically pre-determined, so identical twins may have diff DNA methyl. patterns.

Question 7

Transcription Regulation

Answer 7

1. Transcription regulated by 2 groups of proteins that mediate binding of RNA pol to promoter:
   
   1. Transcription factors (TF) form complex with RNA pol at promoter & don’t allow initiation without factors, hence their levels regulate gene expression.
   2. Regulatory proteins bind to non-coding DNA seq. outside of promoter & interact with transcription factors:
      
      1. Activator proteins bind to enhancer
         seq. & ↑ transcript. rate
         (by mediating complex formation)
      2. Repressor proteins bind to silencer
         seq. & ↓ transcription rate
         (by preventing complex formation)
2. Control Elements: Exist in large amounts to further tighten control & coordination.
   1. Distal control elements bind to regulatory proteins.
   2. Proximal control elements bind to transcription factors.
3. Presence of certain transcription factors or regulatory proteins may be tissue-specific
4. Intracellular chem signals may also trigger change in [reg. prots] or [TF] in response to stimuli → gene expression changes in response to changes in conditions in/out cell:
   
   1. Humans produce different amounts of melanin depending on light exposure
   2. Morphogens: Uneven distr. in embryo & contribute to diff. gene express. patterns depending on their conc.
   3. ​Mutant allele “c^s” in “C” gene in Siamese cats only produces tyosinase (pigment production) at < body temp.

Answer 8

1. In eukaryotes, post-transcriptional mod. of transcript mRNA needed to form mature mRNA
2. Ribosomes also separated from genetic material (DNA & RNA) by nucleus, so gen. needs to be moved.

3 post-transcriptional events:

1. Capping: Involves addition of methyl group to 5’-end of transcribed RNA
   
   1. Methylated cap provides protection against degradation by exonucleases
   2. Allows transcript to be recognised by ribosome.
2. Polyadenylation: Addition of poly-A tail to 3’-end of mRNA. (NOTE: A stands for adenine)
   
   1. Poly-A tail improves RNA transcript stability
   2. Facilitates its export from nucleus.
3. Thus, mRNA mods allow ribosome to it out of nuc. (via nuclear pores) before transl.
4. Splicing: Removing introns from mRNA transcript.
   
   1. Within eukaryotic genes exist:
      
      1. Introns: Non-coding seq. which must be removed prior to forming mature mRNA.
      2. Exons: Coding regions, which fuse
         together when introns removed to form continuous seq.
   2. ​Alternative splicing: Removing specific
      exons → Gene seq. make diff polypep’s.
5. Polysome: Group of ribosomes translating an mRNA seq. simul.
   • In prok. they couple transl. + transc. due to no comp. & both occuring in 5’ → 3’.

Question 9

Translation

Answer 9

Context:

1. Translation: Polypeptide synthesis using base sequence of mRNA molecules (in ribosomes).
2. mRNA seq. read by ribosome in base triplets (codons). Each codon codes for 1 AA.
   
   • Gen code degenerate as >1 codon can code for same AA. Also allows silent mutations to occur, whereby a change in DNA seq. doesn’t alter polypeptide seq.
3. Thus, order of codons in mRNA seq. determine AA order in polypeptide chain.
4. 64 codon possibilities (4 bases ^{3 bases/codon})
5. 3 components work together to synthesise polypeptides by translation:
   
   1. mRNA has sequence of codons that determines AA sequence of polypeptide.
   2. tRNAs have anticodons that bind to comp. codon on mRNA; they carry AA corresponding to that codon.
   3. Ribosomes are mRNA & tRNA binding sites; also catalyse polypeptide assembly.

Process:

• Initiation: Assembly of components that carry out translation (mRNA, tRNA, ribosome).

1. Next, appropriate tRNA molecule binds to
   codon via its anticodon (according to comp. base pairing)
2. Finally, large ribosomal subunit aligns itself to the tRNA molecule at the P site and forms a complex with the small subunit
3. Ribosome composed of 2 sub-units:
   
   • Small subunit binds to mRNA & moves along it until reaching start codon (AUG).
   • Large Subunit containing 3 tRNA binding sites: A, P, and E binds to small subunit.
     * Elongation:
4. Initiator tRNA (with methionine) binds to start codon “AUG” in P site of large sub.
5. 2^nd tRNA (with anticodon comp. to 2^nd codon) binds to 2^nd codon in A site of large subunit.
   (max 2 tRNAs bound at once).
6. Ribosome catalyses pep bond between AAs in A & P site via condensation reactions → dipep.
7. tRNA in P site now deacylated (no AA), whilst
   tRNA in A site carries dipeptide.
8. Ribosome translocates 3 bases along mRNA in 5’ to 3’ dir, so 1^st tRNA moves from P to E site, releasing it.
9. 2^nd tRNA takes place of 1^st. so moves from A to P site, freeing A site.
10. 3^rd tRNA binds with anticodon comp. to 3^rd
    codon on mRNA in vacant A site.
11. Process repeated until stop codon is reached.

• Termination: Disassembly of components & release of polypep chain.

1. Non-coding stop codon reached, which release factor signalling transl. end; polypep released.
2. Ribosome disassembles back into its 2 indep. subunits.

Question 10

Activating tRNA

Answer 10

Context:

1. tRNA activation occurs in cytoplasm via tRNA-activating enzyme, tRNA, AA and ATP.
2. Each AA recognised by specific enzyme
3. But multiple tRNA molecules recognised by enzyme due to degeneracy.

Process:

1. tAE binds to specific AA & ATP.
2. Enzyme catalyses ATP hydrolysis → AMP + 2P
3. AA binds to AMP → AA-AMP complex, linked by high energy bond, 2P is released.
4. Bond act as energy store to provide most of
   nrg used to make pep. bond during transl.
5. tRNA binds to tAE.
6. AA cov. bonded to 3’ terminal of tRNA, releasing AMP attached to enzyme.
7. tRNA molecule now “activated” & released.

Question 11

Insulin production in bacteria

Answer 11

Context:

1. Diabetes type II due to destruction of ß-cells that secrete insulin (hormone).
2. Used to be treated with insulin produced from other animals (e.g. pigs) as they bind to human insulin receptor, but found to cause allergies.
3. Genetic code = universal as same codons code for same AAs in all living things, gen. info
   transferrable between species
4. Ability to transfer genes between species used to produce human insulin in bacteria (for mass production), with exactly same AA seq. as gene
   transcribed & translated in human cells.

Process

1. Desired gene seq. obtained by either:
   
   • DNA isolated from cells & nuclei by centrifugation (heavy cell organelles sink).
   • Using rev. transcriptase to convert mRNA → dDNA.
2. Interest gene specifically amplified via PCR
3. Plasmids used as vectors as they can auton. self-replicate & express genes
4. Gene + plasmid cut with same restriction endo.
   at specific recognition sites by cleaving sugar-phosphate backbone to create “sticky ends”
5. Gene now binds to plasmid as sticky ends of gene & vector overlap via comp. base pairing.
6. Gene & vector spliced together by DNA ligase (which fuses their backbones together with phosphodiester bond) to form recomb DNA.
7. Recomb. DNA introduced into host cell/org.
   (transfection if prok or transformation if euk).
8. Antibiotic selection commonly to ID which cells have successfully incorporated recomb. DNA.
9. Transgenic cells, once isolated & purified, express desired trait encoded by int. gene, so placed in fermenter to reproduce lots.
10. (e.g. Insulin) produced, purified and sold for use (e.g. in diabetics).

Question 12

Hershey-Chase Experiment

Answer 12

1. Proteins & nucleic acids believed to be involved in composition of genetic material.
2. Alfred Hershey & Martha Chase conducted a series of experiments to prove DNA was gen material (not protein).
3. Known that viruses (E.g. T2 Bacteriophage)
   consisted solely of DNA & protein coat and could transfer their genetic material into hosts.
4. T2’s were grown in 1 of 2 isotopic mediums in order to radioactively label specific viral part.
5. Viruses grown in radioactive S (³⁵S) had radiolabelled proteins (S present in proteins but not DNA)
6. Viruses grown in radioactive P (³²P) had radiolabelled DNA (P present in DNA but not proteins)
7. Viruses allowed to infect bacterium (E. coli).
8. Virus & bacteria separated via centrifugation
9. Larger bacteria forms solid pellet whilst smaller viruses remains in supernatant
10. So when pellet found to be radioactive when infected by ³²P–viruses (DNA) but not ³⁵S–viruses (protein), showed DNA passed on.
11. Showed DNA, not protein, was gen material as DNA was transferred to bacteria.

Question 13

Meselson and Stahl

Answer 13

1. Prior to experiment, 3 hypotheses had been proposed for the method of replication of DNA:
   
   1. Conservative Model: Entirely new molecule is synthesised from a DNA template (which remains unaltered)
   2. Semi-Conservative Model: Each new molecule consists of 1 newly synthesised strand & 1 template strand
   3. Dispersive Model: New molecules made of segments of new & old DNA
2. Meselson & Stahl experimentally tested validity of these models using N¹⁵ (heavier radioactive isotope of N¹⁴), an element present in bases).
3. DNA cultured in N¹⁵ for many gens to ensure N¹⁵ was only N source in DNA, then transferred to, & induced to replicate in N¹⁴-only medium.
4. DNA samples separated via centrifugation to find DNA composition in replica. molecules.
5. DNA detected as it absorbs UV, hence creating dark band when tubes illuminated with UV:
   
   • Single band in 1^st gen falsifies cons. replica. (shows mix of old & new DNA/
     N¹⁵ & N¹⁴).
   • 2 bands in 2^nd gen falsifies dispersive replication (New-only & mixed DNA /
     Only-N¹⁴ and N¹⁵ & N¹⁴).
6. Hence, showing DNA Replication is semi-conservative.

Question 14

DNA Sequencing

Answer 14

1. DNA sequencing: Process by which base order of a nucleotide sequence is elucidated
2. Dideoxynucleotides (ddNT): Lack 3’-OH
   group needed for making PPD bond.
   
   1. Thus, ddNTs stop further elongation of nucleotide chain & effectively end replica.
   2. Resulting DNA seq. reflects specific nucleotide pos. ddNT was added.
3. Sanger Method:
   1. PCR mixes set up, each containing stocks of deoxyribonucleotides + ddNT + fluoresc. primers + enzymes for replica.
   2. PCR makes lots of DNA molecules quickly, so PCR mixes should have all possible terminating frags for that spec. base.
   3. Frags separated using gel electroph & base seq. determined by ordering frags according to length.
   4. Fuorescently labelled primer included in each mix allow frags to be detected by automated seq. machines.
   5. If Sanger method conducted on coding strand, resulting seq. elucidated will be identical to template strand.

Question 15

Non-coding DNA

Answer 15

1. Vast majority of human genome is comprised of non-coding DNA, which serve other functions (table).

Context:

1. DNA profiling: Technique by which individuals identified & compared via resp. DNA profiles
2. Within non-coding regions of individual’s genome exists satellite DNA: DNA seq. made up of repeating elements (STRs).

Process:

1. DNA sample collected (e.g. from blood, semen, saliva, etc.), then amplified using PCR
2. Sat. DNA (with STR seq.) cut with specific restriction endo. to create frags, which differ between indivs due to # of STRs in frags.

3. Frags separated using gel electrophoresis &
 resulting profiles (composed of bands) are compared to see if bands match.

4. For family tests:
   
   1. Paternal lineage determined by analysing VNTR from Y-chromosome.
   2. Maternal lineage deduced by analysing mitochondrial DNA variations in single nucleotides at hyper-variable regions.

Question 16

Genes, Locii, Alleles, Chromosomes, Homologous Chromosomes, SNPs, Mutations,

Answer 16

1. Polygenic traits: Traits influenced by multiple genes.
2. Gene: Heritable factor consisting of a length of DNA and influences a specific characteristic.
3. Locus: Specific position on chrom occupied by gene.
4. Allele: Alt. forms of same gene, they have same locus, but only one can occupy it.
   
   • Code for diff. variations of specific trait
   • Alleles have very similar gene seq. & only diff. by 1 or few bases (SNP’s).
   • Single Nucleotide Polymorphisms (SNP): Positions in gene where >1 base may be present.
5. Chromosomes: Groups of linked genes.
   23 types of chromosomes in humans.
   
   • Homologous: Chromosomes that carry same sequence of genes but not necessarily same alleles of those genes, allowing species members to interbreed.
6. Mutations: Random changes in base seq. of gene. May lead to new alleles forming from other alleles.
   
   • Base Substitution: 1 base in seq. of gene replaced by different base.
   • Almost all mutations either neutral or harmful as random change to allele selected for by NS over time unlikely to be beneficial.
   • Mutations in body cells eliminated when individual dies, but mutations in sex cells can be passed on to offspring & cause genetic disease.

Question 17

Mutations

Answer 17

1. Gene mutation: Change in nucleotide/base seq. of gene coding for a specific trait.
2. New alleles are formed by mutation
   (alleles only differ by few bases (SNP’s)).
3. Either spontaneous or induced.

Gene mutations can be:

1. Beneficial: Change gene seq. (missense mutations) to create new variations of a trait
2. Detrimental: Truncate gene seq. (nonsense
   mutations) to STOP normal function of trait.
3. Neutral: Have no effect on functioning of specific trait (silent mutations).

Induced by:

1. Ionising Radiation ↑ mutation rate if enough nrg to cause chemical changes in DNA.
2. Chemical Mutagens ↑ mutation rate by causing chemical changes in DNA.
   (e. g. mustard gas, benzene, ROS, tar)
3. Biological: Viruses (HPV) and Bacteria

Question 18

Hiroshima + Nagasaki and Chernobyl

Answer 18

Question 19

Sickle Cell Anaemia

Answer 19

1. Disorder caused by single base sub. mutation to a gene (Hb), which codes for haemog. prod.
2. Most humans have co-dom. allele Hb⁴.
3. DNA: CTC → CAC on 6^th codon of gene
   → new co-dom. allele formed: Hb⁵.
4. mRNA: GAG → GUG.
5. Polypeptide: 6^th AA in haemog: Glu → Val when GUG transcribed.
6. Mutation → changes haemog. structure →
   stick to form insoluble fibrous strands → rigid enough to distort RBCs into sickle shape.
7. Consequences:
   
   1. ​Damage to tissues by becoming trapped in & blocking blood capills → ↓ bloodflow.
   2. Both haemog & plasma memb. dmg
   3. RBC Life ↓ → ↓ RBC count → Body can’t replace RBC at same rate → anaemia.

Question 20

Genome + HGP (PM MS MEGI)

Answer 20

Genome: Whole of gen. info of organelle, cell, or organism. Includes genes + non-coding DNA seq.

• In animals, genome = DNA molecules that form chromosomes in nucleus + DNA molecule in mitochondrion.
• 46 chromosomes in humans.
• In plants, genome = DNA molecules that form chromosomes in nucleus + DNA molecules in mitochondrion + chloroplast.
• In prokaryotes, genome = DNA in circular chromosome + any plasmids that are present. Much smaller as a result.
• Genome size generally ∝ size:
  • Viruses & bacteria tend to have smallest genomes
  • Plant genome size varies dramatically due to capacity for plant species to self-fertilise & become polyploid.
  • Size may also change due to chromosomes fusing or splitting, but rare.

HGP: International cooperative venture established to sequence human genome

1. HGP showed that humans share majority of their seq, with SNP’s contributing diversity.
2. HGP aided by improvements in tech that rapidly inc. speed of gene sequencing.
3. Completion of HGP led to many outcomes:
   
   1. Mapping: #, location, size & seq. of human genes now established
   2. Screening: Allowed for prod. of specific gene probes to detect sufferers & carriers of genetic diseases.
   3. Medicine: Discovery of new proteins & causes of gen. diseases → ↑ treatments.
   4. Evolution: Comparing with other genomes → ↑ knowledge of origins, evolution & migratory patterns of man
   5. Gene transfer (genetic engineering)
   6. Promote International co-operation.
   7. Understanding that genome > proteome & that most genome not transcribed.
   8. Mutations discovered.

(e.g. introns, promoters, STR’s, etc.)

Question 21

Euk + Prok DNA + Plasmids

Answer 21

1. Plasmids:
   
   1. Contain few genes
   2. Capable of self-replication
   3. Exchanged between bacteria via pili (conjugation) → bacteria evolves new features within gen (horiz. gene transfer).
   4. Plasmids may also cross species barriers if plasmid released when prok. absorbed by cell of different species.
2. Plasmid’s ability to self-replicate & synth. proteins → vectors for genetic engineering.

Question 22

Homologous Pairs

Answer 22

1. Sexually reproducing organisms inherit their genetic seq. from both parents
2. So organisms possess 2 copies of each chrom (homologous chrom), which share:
   
   1. Same structural features (e.g. same size, same banding patterns, same centromere positions)
   2. Same genes at same loci positions (whilst
      genes are same, alleles may be different)
   3. Homo. chrom. separated in gametes (via meiosis) prior to reproduction, in order to prevent chrom. numbers continually doubling with each generation.
3. Organisms with diff. diploid #’s unlikely to be able to interbreed (can’t form homo. pairs in zygotes).
   
   • In cases where diff. species do interbreed, offspring usually infertile (can’t form functional gametes). (e.g. horse + donkey).

Question 23

Karyograms

Answer 23

1. Karyotypes: # & types of chrom. in euk:
   
   1. Harvest cells (usually from foetus or adult WBC’s).
   2. Cell div chemically induced, then mitosis arrested whilst chrom. are condensed
      (so they’re visible).
   3. Stage during which mitosis halted determines whether chrom. appear with sister chromatids or not.
   4. Chroms stained & photo taken, then
      arranged into homo. pairs by size
      (sex chrom. shown last) → karyogram.
2. ​Karyotyping usually done prenatally to:
   
   1. ​Determine gender of unborn child (via ID of sex chrom.)
   2. Test for chromosomal abnormalities (e.g. Down)
3. ​Down syndrome:
   
   1. ​Due to non-disjunction in 1 of parental gametes: Failure of chrom separation resulting in 1 xtra/1 less chrom.
   2. Non-disjunction may occur via:
      
      1. Sis chromatids in Anaphase II
         → 2 affected daughter cells
      2. Bivs failing to separate in Anaphase I → 4 affected daughter cells.
   3. Down = Trisomy due to: 1 parental gamete having 2 chrom 21 copies (due to non-disj) + 1 normal parental gamete with 1 copy fusing → 3-copy zygote.
   4. ​Studies show that:
      
      1. Non-disj ∝ parental (esp. mom) age ↑
         
         1. May be due to developing oocytes being arrested in prophase I until ovulation as part of oogenesis.
         2. ​Higher incidence of chrom. errors in offspring due to anaphase I non-disjunction.
         3. Mean maternal age ↑, → ↑
            # of Down syndrome offspring.
4. ​Karyotyping sources:
   
   1. ​Amniocentesis: Extraction of amniotic fluid (has fetal cells) with needle inserted through abdomen.
   2. Chorionic Villus Sampling (CVS):
      Extraction of CV (placental tissue) with suction tube inserted through cervix.
   3. CVS can be done earlier in preg. than
      Amniocentesis, but risk = 2% (opp. to 1%).

Question 24

Autoradiography

Answer 24

John Cairn 1^st to create image of chroms before condensation, which made measuring inaccurate.

1. Incubate cells in radioactive ³H-T solution.
2. ³H-T incorporated into chromosomal DNA of cell as T not present in RNA.
3. Chromosomes isolated by gently lysing cells & fixing chrom. to photographic surface
4. Surface then immersed in radioactively-sensitive emulsion containing AgBr.
5. Radiation released from ³H-T converts Ag⁺ ions in AgBr into insoluble Ag grains.
6. Following period of exposure, excess AgBr
   washed away, leaving Ag grains = small black •
7. When photographic film developed, chrom. DNA visualised with an electron microscope.

Question 25

Meiosis

Answer 25

Meiosis: Reduction div of diploid germline cells in reproductive organs into 4 gen. distinct hap nuclei (gametes)

1. 1^st meiotic division separates homo. pairs to halve chrom. # (diploid → haploid) (red. div).
2. 2^nd meiotic division separates sister chromatids
3. Interphase:
   
   1. ​DNA replic. during S phase so red. div. → ½ diploid # of chroms in gametes.
   2. Thus, when gametes fuse during fertilisation, they form diploid zygote.
   3. If not, polyploidy occurs.
   4. As meiosis results in gen distinct gametes, random fert. by egg & sperm always generates diff. zygotes
4. Prophase I:
   
   1. Synapsis: Homo. chromosomes pair up at points (chiasmata) to form bivs (synapsis)
   2. Crossing-over: DNA exchanged between non-sister chromatids across chiasmata → chromatid alleles recomb.
   3. Chiasmata: Inc. stability of bivalent, so likely occcur at random pos &/or >1 in each, so cross-over can occur anywhere.
5. Metaphase I:
   
   • Random Orientation: Orientation of bivs is random.
6. Anaphase I:
   
   • Disjunction: Splitting of homologous chroms to opp. poles.
   • Random & ind. orientation → random & ind. assortment → Gamete combos = 2ⁿ
     (n = haploid number).
   • If crossing over also occurs, gamete combos becomes immeasurable

Question 26

Gregor Mendel’s Laws

Answer 26

1. Gregor Mendel came up with inheritance laws by performing experiments on pea plants
2. He crossed diff. purebred pea plant varieties.
3. Collected & grew their seeds to determine their traits.
4. Next, he self-fertilised offspring & grew their seeds to similarly determine their traits.
5. Crosses were performed many times to establish reliable data trends.
6. Mendel discovered the following things:
   
   1. When crossing 2 diff. purebred varieties together, results weren’t a blend – only 1 feature expressed
   2. When Mendel self-fertilised offspring, 2 diff traits expressed in 3:1 ratio.
7. From these findings, Mendel drew the following conclusions:
   
   1. Organisms have discrete factors (genes) that determine its features.
   2. Furthermore, organisms possess 2 versions (alleles) of each factor (gene).
   3. Each gamete contains only 1 allele (hap).
   4. Parents contribute equally to inheritance of offspring due to fert. of randomly selected egg & sperm
8. Certain laws were derived:
   
   1. Law of Segregation: When gametes form, alleles separated so that each gamete carries only 1 allele for each gene.
   2. Law of Independent Assortment: The segregation of alleles for 1 gene occurs independently to that of any other gene*
      
      • * Not true for linked genes.
   3. Principle of Dominance: Recessive alleles will be masked by dominant alleles.*
      
      • * Some genes show co-dom.

Question 27

Genotype, Phenotype, Dominant, Recessive, Co-dominant

Answer 27

1. Genotype: Allele combo for specific trait.
2. Phenotype: Observable traits of a specific trait.
   (determined by genotype & env. influences).
3. Dominant: Expresses trait that is always present in phenotype when present.
4. Recessive: Only expressed in phenotype when in a homozygous state.
5. Co-dom: Allele pairs that affect phenotype equally when present in heterozygote.

Question 28

ABO Blood groups

Answer 28

1. Human RBCs categorised into blood groups based on structure of its antigen.
2. I^A, I^B & i alleles all produce basic antigen on surface of RBCs:
   
   1. I^A & I^B alleles co-dom & each modify antigen to produce different variants.
   2. i allele is recessive & doesn’t modify the basic antigenic.
3. Incorrect blood transfusions lead to agglutination and lysis of RBCs.

Question 29

Genetic Disorders

Answer 29

1. Sex Linkage: Gene controlling trait located on sex chrom. Most X-linked as ↓ genes exist in Y.
2. Sex Rules:
   1. Only ♀ carriers as ♂ can’t be hetero.
      (only have 1X).
   2. ♂ always inherit X-linked trait from mom
      (as Y received from dad).
   3. ♀ can’t inherit X-linked recessive trait from unaffected dad (must receive dom allele).
3. Many genetic diseases identified, but most are rare because:
   
   1. Alleles that ↓ survival + reprod. unlikely to be passed onto offspring.
   2. Recessive conditions more common, as faulty allele can be present in carriers without causing disease/harm.
   3. Dominant conditions may have late onset, as this doesn’t prevent reprod. & the transfer of faulty allele.

Question 30

Pedigree Charts

Answer 30

Question 31

PCR

Answer 31

1. PCR: Artificial DNA replic. technique used to
   amplify specific DNA seq.
2. Useful when only small amount of DNA available for testing. E.g. crime scene samples of blood, semen, tissue, hair, or from fossils
3. Reaction occurs in thermal cycler & uses variations in temp to control replic:
   
   1. Denaturation: DNA sample heated (90ºC) to break H-bonds holding strands tog. →
      separates them into single DNA strands.
   2. Annealing: Sample cooled (50ºC) to allow primers to anneal to target DNA seq. + re-form double strand apart from seq. with primers (target seq).
   3. Elongation: Sample heated to opt. temp.
      for heat-tolerant Taq polymerase (doesn’t denature at high temp) to function (75ºC), isolated from thermophilic bacteria.
4. Repeat procedure. For n cycles, PCR produces 2ⁿ copies of DNA sample.

Question 32

Gel Electrophoresis

Answer 32

1. Gel Electrophoresis: Used to separate charged molecules, like proteins or DNA fragments, according to their size and charge:
2. Fragmented DNA samples placed in gel block & current applied → samples move through gel
3. Small fragments less impeded by gel so move faster than large fragments, so orders by size.

DNA Separation:

1. DNA fragmented with restriction endonuclease - diff. DNA samples generate diff frag lengths.
2. DNA samples placed into agarose gel &(DNA)¯ due to (PO₄³¯) so moves to anode, but at diff. rates.
3. Frag sizes calculated by comparing against known industry standards.
4. Separated seq. moved to membrane &
   specific seq. ID’d by adding comp hybridisation probe, which appears in autoradiograph.

Protein Separation:

1. Proteins may fold into diff shapes/ diff. sizes & have + & – regions, so treated with anionic detergent (SDS) to induce uniform – charge.
2. Protein samples placed into polyacrylamide gel & move to anode at diff. rates dep. on size.
3. Protein sizes calculated by comparing against known industry standards.
4. Separated proteins moved to membrane &
   target proteins identified by staining with specific monoclonal antibodies.

Question 33

GMO’s

Answer 33

Question 34

Bt corn

Answer 34

1. Bt corn: GM maize with insecticide-prod. gene (Bt)
2. Bt-toxin kills corn borers, which eat crop
3. Monarch butterflies also feed on milkweed with Bt-corn pollen moved by wind.

Experiments:

1. 1^st Experiment conducted comparing monarch caterpillars DR’s & Bt pollen-based diets.
   1. Monarchs fed milkweed leaves with Bt
      pollen (sim. spread via wind).
   2. Growth & DR’s compared in monarchs fed on non-dusted, non-GM & dusted diets.
   3. Caterpillars exposed to Bt pollen ate less, grew more slowly & exhibited higher DR.
2. But, Bt pollen on leaves > found naturally (rain could wash) & Larva restricted in diet (in field, larva could avoid eating pollen dusted leaves).
3. 2^nd experiment conducted comparing monarch butterfly DR’s & proximity to Bt corn fields:
   
   1. No sig. ↑ DR when monarch larva placed in or near actual Bt corn field
   2. Exposure to Bt pollen poses no sig. risk to monarch butterfly pops.

Question 35

Clones

Answer 35

1. Clones: Groups of gen. identical organisms, derived from single original parent cell.

Natural Methods of Cloning:

1. Bacteria & fungi reproduce asex. to produce genetic clones via binary fission (mainly bacteria) or by producing spores (mainly fungi)
2. Vegetative propagation: Small pieces induced to grow indep. due to totipotent meristematic tissue in adult plants differentiating:
   
   • Onion/garlic bulbs = modified plant leaves – all bulbs in group are gen. identical
   • Underground stems (e.g. potato tubers) can form new plants gen. identical to parent plant
3. Some animal species also reproduce asexually:
   
   1. Binary Fission: Parent organism divides equally → 2 daughter organisms
      (e.g. flatworms)
   2. Budding: Cells split off parent organism, creating smaller daughter organism that
      eventually separates from parent (Hydra)
   3. Fragmentation: New organisms grow from separated frag of parent org
      (e.g. starfish)
   4. Parthenogenesis: ♀ prod. diploid egg cells instead of haploid (e.g. ♀ aphids)
4. Human Twins:
   1. Monozygotic (ID) twins created when fertilised egg splits into 2 identical cells, each forming an embryo. Gen. identical.
   2. Dizygotic (non-ID) twins created when unfertilised egg splits into 2 cells & each fertilised by diff. sperm.

Artificial Methods of Cloning

Embryonic Division:

1. Embryonic cells retain pluripotency, so
   differentiate to form all tissues comprising org.
2. Embryonic cells separated artificially in lab, early in developmental cycle.
3. Separated groups of cells implanted into surrogate uterus to develop into clones.
4. Limited by fact that embryo used still formed randomly via sexual reproduction & so specific genetic features of resulting clones unknown.

​(SCNT): Artificial method by which cloned embryos prod. using differentiated adult cells.

1. Genetic features of resulting clone known.
2. Somatic cells taken from adult donor & their cultured (for their diploid nuclei).
3. Enucleated egg cell made by taking unfertilised egg & removing its nucleus.
4. Egg fused with somatic nucleus → diploid egg.
5. Electric current used to stimulate egg to divide & develop into embryo.
6. SCNT cloning split into 2 purposes:
   
   1. Reproductive cloning: If embryo is implanted into surrogate uterus, new cloned organism of donor will develop.
   2. Therapeutic cloning: Embryonic cells
      induced to differentiate to create specific tissues or organs for transplantation

Answer 36

Question 37

Stem Cuttings (MELT HAP)

Answer 37

1. Stem cutting: Separated portion of plant stem that can regrow into new indep. clone via vegetative propagation.
2. All stems possess nodes, from which leaf, branch or aerial root may grow.
3. Stem cuttings typically placed in soil with lower nodes covered & upper nodes exposed
4. Stem cutting: Common method employed to rapidly propagate plant species (including sugar cane, grapes & roses)
5. Factors affecting rooting of stem cutting:
   
   1. Cutting pos: Cutting stem above/below
      node & relative proximity of cut to node)
   2. Cutting length: How many nodes remain on cutting.
   3. Growth medium: Soil, H₂O, potting mix, compost, or open air
   4. Use & conc. of growth hormones
   5. Temp: Most cuttings grow optimally at temps common to spring & summer
   6. H₂O availability: Groundwater or humidity
   7. Soil pH:
   8. Light Exposure:

Question 38

Thomas Morgan

Answer 38

1. Thomas Morgan discovered non-Mendelian ratios in fruit flies, which aided in understanding gene linkage.
2. Cross-breeding R-eyed wild types with W-eyed mutants → clear sex bias in phenotypic distr.
3. All ♀ offspring of R-eyed ♂ were R-eyed,
4. All ♂ offspring of W-eyed ♀ were W-eyed.
5. Morgan inferred that eye colour gene dep. on X, as it was found on X-chromosome.
6. Morgan then investigated other traits & found that certain phenotypic combos occurred in much lower freq. than expected.
7. Based on this data, Morgan proposed:
   1. Alleles for Link-traits located on same
      chrom (link-genes), so didn’t indep. assort.
   2. Linked alleles could be uncoupled via cross-over to create alt pheno, but new
      phenotypes would occur at lower freq.
   3. Morgan also observed that cross-over freq. between linked genes differed depending on distance between 2 genes on chrom (cross-over freq. ∝ distance between genes). Which he used to dev 1^st gene linkage maps, showing relative positions of genes on chrom.

Question 39

Chi-Squared Test

Answer 39

1. Offspring with unlinked genes inherit any potential phenotypic combo equally
   (due to random assortment).
   of alleles (due to independent assortment).
2. Offspring with linked genes only express the phenotypic combos present in either parent (unless crossing over occurs)
3. Thus, unlinked recomb. phenotypes occur less freq. than ‘linked’ parental phenotypes.

Chi-squared: Statistical measure used to determine whether diff. between obs. & exp. freq. distribution is statistically significant.

1. Idenify Hypotheses:
   1. H₀: No sig. diff. between obs. & exp. freqs
      (i. e. genes are unlinked)
   2. H₁: Sig. diff. between obs. & exp. freqs
      (i. e. genes are linked)
2. Construct freq. table:
   
   1. ​Draw a dihybrid cross to find exp. ratios.
   2. Total x exp. ratio = exp. freq
3. Apply Chi-Squared: ​(O - E)² ÷ E
4. Degrees of Freedom: (m - 1)(n - 1) = 3 for dihybrid crosses.
5. ID p value:
   
   1. When df = 3, Chi-squared > 7.8 to be considered stat. sig. (p < 0.05).
   2. If > 7.8 → p < 0.05 → Sig. diff
      → genes are linked.
   3. If < 7.8 → p > 0.05 → Sig. diff.
      → genes are unlinked.

Question 40

Polygenic Inheritance

Answer 40

1. Monogenic traits (controlled by 1 gene loci) exhibit discrete variation, with individuals expressing 1 of several distinct phenos.
2. Polygenic traits (controlled by >2 gene loci) exhibit continuous variation, with individuals
   expressing pheno existing in a bell-shaped cont. spectrum of potential phenos.
   
   • # of loci responsible for particular trait ∝ # of possible phenos.
3. Maize grain colour: Controlled by 3 gene loci:
   1. Grain colour ranges from W to dark R, dep. on amount of pigment expressed
   2. Each gene has 2 alleles, which either code for R or W pigment.
   3. Most freq. combos have equal # of both.
   4. Conversely, combos of extremes are rare
   5. Overall pattern of inheritance shows continuous variation.
4. Height + Skin Colour also affected by env:
   
   1. Added effect of env pressures functions to ↑ variation seen for particular trait
   2. Human height controlled by mutliple genes, but also affected by diet & health (disease).
   3. Skin colour controlled by multiple melanin producing genes, but also affected by sun exposure.

Question 41

Evolution by Natural Selection (ICE AGE)

Answer 41

1. Evolution: Cumulative change in allele freq. of a pop’s gene pool over successive gens.
2. Gene pool: Sum total of alleles for all genes present in a sexually reproducing population
3. Gene pool size ∝ amounts of gen diversity → ↑ chances of biological fitness & survival
4. Natural Selection: Freq. of alleles that adapt
   indivs to env. ↑, + vice-versa to bad alleles.

Process (ICE CAGE)

1. Inherited variation exists within population
2. Competition ← offspring > env. capacity.
3. Env pressures → differential reprod. within pop
4. Adaptations (traits that make indiv. suited to its env & way of life) that benefit survival select for and passed onto offspring.
5. Genotype/Allele freq. changes cumulatively within pop gene pool across gens (Evolution).

Question 42

Types of selection

Answer 42

1. Stabilising Selection: Natural sel. favours intermediate pheno over both ends of the range of variation.
   
   1. Results in removal of extreme phenotypes (phenotypic distribution becomes centrally clustered to reflect homogeneity)
   2. Operates when env conditions are stable & competition is low.
   3. E.g. Human birth weights (too large = birthing complications ; too small = risk of infant mortality)
2. Directional Selection: Natural sel. favours 1 end of the range of variation over another.
   
   1. Operates in response to gradual or sustained changes in env conditions
   2. Causes progressive change in pop in that direction.
   3. Causes species to change enough over time to be regarded as different species (speciation).
   4. E.g. Dev of antibiotic resist in bacteria pop
3. Disruptive Selection: Natural sel. favours both ends of the range of variation at cost of
   intermediate phenos
4. Causes pheno distr to deviate from centre & results in bimodal spread
5. Occurs when fluctuating env conditions (e.g. seasons) favour presence of 2 diff
   phenos, which are adapted to diff. niches.
6. Extreme types adapted to diff. niches
   (e. g. seasons).
7. Reproductive barriers become established between extreme types (e.g. plants grow in diff. seasons)
8. E.g. proliferation of black or white moths in regions, but not grey-

Question 43

Change in allele freq.

Answer 43

1. Population bottlenecks & founder effect will exacerbate genetic diffs between geographically isolated pops.
2. Pop bottlenecks: Natural or anthropomorphic event that ↓ pop size by an order of magnitude (~ >50%).
   1. Surviving pop has less genetic variability than before, so subject to higher lvl of genetic drift.
   2. As surviving members begin to repop, newly deving gene pool div to original.
3. Founder effect: Occurs when small pop breaks away from larger pop to colonise new territory
   
   1. As this pop subset doesn’t have same degree of diversity as larger pop, it’s subject to more genetic drift
   2. Consequently, as this new colony ↑ size, its gene pool diverges from original.
   3. Original pop stays largely intact as opposed to pop bottlenecks.

Question 44

Evidence for Evolution

Answer 44

Fossils:

1. Fossils show changes over time in organisms; as fossilised orgs are diff. from existing ones; yet share homo structures with existing orgs;
2. Seq. of fossils at various geological eras matches expectations of evo:
   
   1. Bacteria
   2. Simple invertebrates
   3. Complex vertebrates.
3. They also show intermediate stages in evolution of groups.
4. Thus suggest orgs share common ancestors.
5. Some have not changed much, due to little selection pressure. There are also gaps in fossil record, which make conc. difficult.

Selective Breeding: Form of artificial selection whereby humans select for desirable traits to be passed onto future gens (rather than env).

1. Modern varieties of wheat & rice prod. higher yields & more pest-resist. than wild ancestors.
2. Dog breeds are numerous and sig. different from wild wolf ancestor.
3. Artificial selection shows changes in domestic species can be achived in relatively short time.
4. Does not prove that evolution of new species has occured.

Homo. structures: Same ancestor that had this structure, but has become diff. as they perform different functions. Implying common ancestry.

1. The more similar the homo. structures between 2 species are, the more closely related they are likely to be.
2. E.g. pentadactyl limbs in all vertebrae:
   
   1. Human hands adapted for tool manipulation.
   2. Wings adapted for flying
   3. Hooves adapted for galloping
   4. Fins adapted for swimming
3. AA seq. of diff. species also show how closely related species are.
4. ​Speciation:
   
   1. Provides evidence for evolution of species & origin of new species by evolution due to continuous range in variation between populations.
   2. But doesn’t match either belief that:
      
      1. Species created as distinct types of organism, so should be constant across their geographic range
      2. Species are unchanging.

Question 45

Fossils

Answer 45

1. Fossil: Preserved remains or traces of any organism from the past.
   
   1. Reserved Remains: Provide direct
      evidence of ancestral forms.
      (e.g. bones, teeth, shells, leaves, etc.)
   2. Traces: Provide indirect evidence of ancestral forms.
      (e.g. footprints, tooth marks, burrows)
2. Fossil Record: Totality of fossils, both discovered & undiscovered.
3. Fossil record provides evidence for evo. by revealing features of ancestor for comparison against living descendants.

​Law of Fossil Succession: Chron. seq. by which traits appear to develop.

1. Fossils dated by determining age of rock layer in which fossil is found (rock layers develop in chronological order so oldest = bottom).
2. Different kinds of organisms found in rocks of particular ages in consistent order, indicating a seq. of dev:
   
   1. Prokaryotes appear in fossil record before eukaryotes
   2. Ferns appear in fossil record before angiospermophytes.
   3. Invertebrates appear in fossil record before vertebrates
3. Ordered succession of fossils suggests that newer species likely evolved as a result of changes to ancestral species

Question 46

Industrial Melanism

Answer 46

1. Inherited variation exists within pop (mel. + pep.)
2. Comp. ← offspring > env. capacity.
3. Env press (poll due to ind. revo.) → SO₂ kills pale lichen + C dark trees → differential reprod. in pop.
4. Adaptations (black colour) benefit from camo provided by black trees → pass on to offspring.
5. Genotype/Allele freq. (mel. variety) cum. ↑
   within pop gene pool across gens (Evolution).

Question 47

Antibiotic Resistance in Bacteria

Answer 47

1. Antibiotics: Chemicals produced by microbes that either kill or inhibit growth of bacteria
2. Antibiotics are commonly used as treatment for bacterial infections (but not viral infections).

S. aureus:

1. S. aureus stim. skin lesions + boils, & pneum.
   + meningitis; treated with antibiotic (methicillin).
2. Initially, only MSSA strands exist.
3. New MRSA strand developed from mutation of antibiotic gene after extensive methicillin use.
4. It reproduces & passes on gene to clones.
5. MRSA survives & reproduces (passing on gene to clones), whilst MSSA die out (don’t reprod.)
6. Gene also transferred to other MSSA in another pop via conjugation (turns into MRSA)
7. Results in cumulative change in allele freq. in S. aureus strands → evolution.

Question 48

Speciation

Answer 48

1. Speciation: Gradual divergence of 2 related pops into diff. species due to geographical separation, which → adaptations.
2. Degree of divergence depends on extent of geographical separation & amount of time since separation occurred:
   
   1. Pops separated recently & are close show less variation (less divergence).
   2. Pops separated long ago & are far show
      more variation (more divergence).
3. As genetic divergence between related pops ↑, their genetic compat. consequently ↓.
4. When 2 pops diverge to point where no longer interbreed & produce fertile, viable offspring
   = separate species (speciation).
5. Endemic Species: Only found in certain geographical area.

Question 49

Isolation Barriers

Answer 49

Question 50

Types of Speciation

Answer 50

Allopatric Speciation: Occurs in diff geographical area. Requires a physical barrier to gene flow.

Sympatric Speciation: Occurs in same geographical area. Requires behavioural or temporal barriers to gene flow.

Both lead to genetically isolated pops; up to the point they can no longer interbreed → new species forms;

Question 51

Pace of Speciation

Answer 51

1. Evolution via speciation may occur via either:
2. Phyletic Gradualism: Speciation generally occurs uniformly, via steady & gradual transformation of whole lineages.
   1. Supported by fossil record of species
      with many intermediate forms connecting ancestral species to modern equivalent.
3. Punctuated Equilib: Speciation is a periodic process that occurs abruptly & rapidly after long periods of stability.
   
   1. In this view, speciation is seen as a periodic process (big changes occur suddenly, followed by long periods of no change)
   2. Supported by general lack of transitional fossils for most species
   3. But absences could also be due to
      unusual & specific events in fossilisation.

Question 52

Darwin’s Finches

Answer 52

1. Charles Darwin’s Theory of Natural Selection used evidence from observing finches in Daphne Major island (in Galapagos).
2. Adaptive Radiation: Growing discrepancy of structures from same ancestral line as they perform diff. functions in diff. species due to diff. env. pressures.
3. “Darwin’s finches” are a bird species endemic to Daphne Major island in Galapagos.
4. Darwin’s finches show adapt. rad. & show beak size & shape (inherited var/adaptation) dep. on
   size of seeds available (env. press):
   1. Drought in 1977: ↓ small seeds / ↑ large
      seeds → adv to have larger beak → ↑ in large beak finches / ↓ in small beak sizes.
   2. Floods in 1983: ↑ small seeds / ↓ large
      seeds → adv. to have small beak → ↓ in
      large beak finches / ↑ in small beak sizes.

Question 53

Clade Reclassification

Answer 53

1. ​Cladograms have shown that Morphology-based class. doesn’t always correspond with evolutionary origins of group of species.
2. 3 Outcomes of reclassification:
   1. Likely closer to truly natural class. (if new classifications based on cladistics); so predictive value = higher.
   2. Unnoticed similarities between groups & diff. between species previously assumed to be similar revealed.
   3. Time-consuming & potentially disruptive for biologists.
3. Reclassification of figwort family:
   1. Until recently, figworts were one of largest family of angiosperms, but problematic as many of figwort plants too diff in structure to function as a meaningful grouping.
   2. Taxonomists compared chloroplast gene base seq in figworts & other sim genera & reclassified fam into diff clades.
   3. Now

Question 54

Binomial System

Answer 54

1. Binomial System: Formal system by which all living species are classified (taxonomy)
2. Periodically assessed + updated at series across international congresses.
3. Good because:
   
   1. International system;
   2. Names agreed at congresses.
   3. All scientists use the same names for species, which prevents miscomm due to language diff.
   4. First name is the genus name and shows which other species are closely related, so traits can be predicted for new species

Question 55

Genus and Taxonomy

Answer 55

1. Drunk Katy Perry Comes Over For Great Sex.
2. Domain, Kingdom, Phylum, Class, Order, Family, Genus, Species
3. All living orgs classified into 3 domains
   
   1. ​Eukarya: Eukaryotes that contain memb-bound nucleus.
   2. Eubacteria: Prokaryotes lacking nucleus & consist of common pathogenic forms (e.g. E. coli, S. aureus, etc.)
   3. Archaea: Prokaryotes lacking nucleus consist of extremophiles (e.g. methanogens, thermophiles, etc.)
4. Genus: A group of species that share certain characteristics.

Question 56

Artificial and Natural Classification

Answer 56

1. Artificial Classification: Grouping species together solely based on physical characteristics.
2. Natural Classification: Grouping species together based on
3. Artificial Disadv is that structures may appear different in similar orgs and sim in diff orgs:
   1. Div-Evo: Growing discrepancy of structures from same ancestral line as they perform different functions in diff species.
   2. Conv-Evo: Growing assimilation of structures with different ancestral line as they perform same/sim functions in diff species.
4. Dichotomous Keys: Consist of pairs of choices;
   whereby each choice in pair leads to another pair of choices or gives the identification.
   1. Requires a good specimen for reliable ID.
   2. Key should only use clear/reliable traits.

Question 57

Plant Phyla

Answer 57

Question 58

Animal Phyla

Answer 58

Question 59

Chordata Classes

Answer 59

Question 60

Clades

Answer 60

1. Cladograms: Tree-like diagrams; used to show evolutionary history.
2. Clade: Group of organisms consisting of all descendants from a common ancestral org.
3. Clade members share inherited characteristics; due to their shared evolutionary history.
4. Nodes represent represent common ancestors
   as well as when speciation occurred, thus represents sequence in which groups diverged
5. Clades based on AA seq differences between organisms; # of diffs ∝ how close orgs are.
6. Cladograms have led to re-classification of some groups (e.g. figworts).
7. Cladistics allow predictions to be made; and estimations of how long ago groups diverged, due to “molecular clock”:
8. Mutation rates generally constant, so used as a “molecular clock” to predict when speciation occurred (indicated by branch length).​
9. Molecular clock limited by:
   
   1. Diff genes/proteins change at diff rates
   2. Over long time, earlier changes may be reversed by later changes → confounding accuracy of predictions.
   3. Rate of change for particular gene may differ between different groups of orgs

Question 61

Speciation by Polyploidy

Answer 61

1. Speciation: Formation of new species;
2. Polyploidy: Form of sympatric speciation due to chrom pairs failing to separate during meiosis; or cell failing to divide in cytokinesis
3. Leads to individuals with multiples of normal chrom number;
4. Polyploid indivs can interbreed with one another, but not with diploid indivs as it would lead to infertile hybrids → reproductive barrier.
5. Common in plants (e.g. Allium genus);