BY450 Fundamentals of Genetics & Evolution Lecture 3 Genes & Genomes PDF

BY450 Fundamentals of Genetics & Evolution Lecture 3 Genes & Genomes Andy Overall Coding DNA accounts for just ~ 1.5% of DNA ~ 98.5% is non-coding DNA of which a significant fraction is functionally important (eg, enhancers, promotors) and genes that make RNA...

BY450 Fundamentals of Genetics & Evolution Lecture 3 Genes & Genomes Andy Overall Coding DNA accounts for just ~ 1.5% of DNA ~ 98.5% is non-coding DNA of which a significant fraction is functionally important (eg, enhancers, promotors) and genes that make RNAs (eg, ribosomal and transfer).  A eukaryotic gene. Coding exons separated by non-coding introns. RNA splicing is performed by spliceosomes and small nuclear RNA (snRNA) molecules. ~ 10% of our protein- coding genes have a single exon that does not undergo splicing. The remainder can give rise to splice variants. Strachan & Lucassen, Genetics & Genomics in Medicine, 2 nd ed. Diploid parents Haploid gametes Given that each offspring is generated from half of two parental genomes, it’s no surprise that each cell in the body has the same genome. Diploid offspring The cells differ due to different genes being switched on/off (expressed) To maintain diploid organisms, chromosome pairs need to assort prior to becoming daughter cells (gametes) with a complete haploid component of the genome = meiosis  Chromatids separate Chromosomes duplicate into chromatids To maintain diploid organisms, chromosome pairs need to assort prior to becoming daughter cells (gametes) with a complete haploid component of the genome = meiosis Homologous chromosomes exchange genetic material (crossing over)  C1 C2 D2 (A) Pairs of chromosomes with D1 A1 C2 same genes but different alleles. A2 C1 B1 B2 A1 A2 Each chromosome has sister D1 D2 B1 B2 chromatids. E1 E2 F1 F2 (B) Homologous chromosomes E1 E2 pair. F1 F2 (C) Crossing over (recombination) C1 C2 occurs. There are 2 chiasmata on D1 D2 A1 A2 large chromosome, only 1 on B1 B2 C1 C2 smaller chromosome. A1 A2 D1 D2 B1 B2 (D) Homologous chromosomes E1 E2 E1 E2 begin to separate. F1 F2 F1 F2 Adapted from Strachan & Lucassen, Genetics & Genomics in Medicine, 2 nd ed.  (A) Pairs of chromosomes align on the metaphase plate at the A2 A1 center of the spindle apparatus. B2 B1 A2 A1 A2 A1 Contraction of the spindle fibres B2 B2 B1 B1 draws the chromosomes in the direction of the spindle poles (arrows) C2 C1 C2 C1 D2 D1 C2 C1 (B) Rupture of the chiasmata. D2 D2 D1 D1 E2 E1 E2 E2 E1 E1 F2 F1 F2 F1 F2 F1  (C) Cytokinesis segregates the two chromosome sets, each into a different germ cell (here male C2 C1 C2 C1 spermatozoa; which produces four A2 A1 A2 A1 equivalent germ cells) D2 D2 D1 D1 B2 B2 B1 B1 E2 E2 E1 E1 F2 F1 F2 F1 𝛼1 𝛼2 𝛽1 𝛽2 𝛼3 𝛼4 𝛽3 𝛽4  (D) Meiosis II in each primary spermatozoa gives rise to haploid secondary spermatozoa. This generates unique genetic C2 C2 C1 C1 combinations and hence genetic A2 A1 A1 A2 diversity. D2 B2 D2 B2 D1 B1 D1 B1 E2 E2 E1 E1 Males Females F2 F2 F1 F1 -------Meiosis I------- 𝛼1. 𝛽1 𝛼1 𝛽2 𝛼4 𝛽4. 𝛼4 𝛽3 -------Meiosis II------- Discarded polar bodies C1 C1 C2 C2 A1 A2 A2 A1 D2 B2 D2 B1 D1 B1 D1 B1 E2 E2 E1 E1 F1 F1 F2 F2 𝛼2. 𝛽2 𝛼2 𝛽1 𝛼3 𝛽3. 𝛼3 𝛽4 Model the independent assortment of maternal and paternal homologues during meiosis.  Genes Hereditary “units”: F1 “Genes” for colour phenotype lead to colour being inherited. F2  Genes Hereditary “units”: F1 RR rr “Genes” have alleles R and r, such that “genotypes” lead to F2 colour phenotype inheritance. Rr  Genes Hereditary “units”: “Genes” have alleles R and r, F1 where r is a mutant such that RR rr specific “genotypes” lead to colour phenotype inheritance. * F2 So, “gene” has transformed from an indivisible Mendelian unit of Rr heredity into a continuous nucleotide sequence. * Genes Hereditary “units”: We observe heritable phenotypic variation and assume underlying genetic variation as a cause. However, the root causes of the phenotype take many forms (eg, differences in expression, coding, nature of mutation), so “gene’ remains elusive as a hereditary unit. Genes Hereditary “units”?: Phenotype. Protein mRNA DNA  Gene Expression Example: Bacterium E. coli digests lactose. When lactose is around it produces an enzyme beta-galactosidase, but not when there is no lactose. lacZ gene Upstream (5’) Promotor Termination sequence  Gene Expression Elsewhere in the E. coli genome, a repressor protein is coded for, transcribed and translated that binds to the promotor region of the lacZ gene. No beta- Repressor protein galactosidase lacZ gene Upstream (5’) Promotor Termination sequence This blocks RNA polymerase from transcribing the lacZ gene.  Gene Expression If the repressor protein comes into contact with lactose, its conformation changes and it detaches from the promotor. lactose Repressor protein beta-galactosidase lacZ gene coded for Upstream (5’) Promotor Termination RNA polymerase sequence RNA polymerase -> transcribes DNA into mRNA  Gene Expression Transcription and translation can then proceed. lactose Beta-galactosidase lacZ gene Upstream (5’) Promotor RNA polymerase Amino acids mRNA translation Ribosomes transcription  Gene Expression Until all the lactose has been metabolized, at which point... Beta-galactosidase Repressor protein lacZ gene Upstream (5’) Promotor A mutation that influences any of these processes will influence the phenotype. Knowing something of how genes function leads to an understanding that in turn can lead to manipulation, for example genetically modifying organisms (GMOs). GMOs contain inserted genes from any other organism. Multiple methods of inserting genes, but using a biological vector is the most common method. What constitutes a gene here? For the coding bit of the gene to function properly in the new organism, it needs a promotor sequence and a terminator sequence. Promotor Transgene Terminator Gene Cassette  The most common promotor element is the CaMV35S promotor derived from the cauliflower mosaic virus. Promotor  Ti plasmid The NOS terminator is derived from the Ti plasmid in Agarobacterium tumefaciens. Promotor Terminator Between these two elements of the gene cassette is ligated the gene whose function we wish to introduce into a new species (“transgene”). One or both of these regulatory sequences are present in most of the genetically modified crops that are approved for distribution in North America, Asia, and Europe. Promotor Transgene Terminator Gene Cassette All life shares DNA as hereditary molecule. DNA molecule must have been present in common ancestor of all known species – microscopic organisms 3.7 billion years ago. Can we identify an organism from a short segment of its DNA? Closely related organisms have more similar DNA molecules that distantly related organisms. DNA molecules are extremely small and can’t be observed in any detail. To ”observe” DNA sequences, it is necessary to amplify the host DNA via a process that emulates DNA replication (PCR). 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Cannot visualize a single strand of a DNA molecule. (1) Place in a microcentrifuge tube with some water. 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Heating up the molecule to around 95°C will break the hydrogen bonds between the complementary bases. HEAT 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCC – 3’ 3’ - ATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ To synthesize new strands of DNA from these two template strands, primers are required that are short sequences of DNA complementary to the base sequences either side of the region of interest. (2) Add primers. Forward primer: 5’ – CGATCTGATATGCC – 3’ Reverse primer: 5’ – AGACAGACAGGATA – 3’ For these to anneal to their complementary bases, the temperature needs to cool down from 95°C to around 56°C. COOL 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAA – 3’ 3’ - TAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Free dinucleotide triphosphate molecules can be ligated to the 3’ (-OH) end of the primers through the catalytic properties of polymerase enzymes. (3) Need to add dATP, dGTP, dCTP and dTTP molecules. (4) Need to add polymerase enzyme (e.g., Taq polymerase) The optimal temperature for the polymerase enzyme to extend the primers is 72°C. WARM 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ At the end of this cycle, two newly synthesised, complementary strands have been added. This procedure is repeated, but now the starting material is 4 template strands, not 2. 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Heat up to 95°C to denature the double helix. HEAT 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCC – 3’ 3’ - ATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCC – 3’ 3’ - ATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Cool to 56°C to allow primers to anneal. COOL 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAA – 3’ 3’ - TAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAA – 3’ 3’ - TAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ Warm up to 72°C to optimize the temperature for the polymerase enzyme to extend the primers. WARM 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ 3’ – GCTAGACTATACGGATAGGACCCATAGACAGATTACAGATGGCAGATTGACATAGTTAAGTTGACAGACGACAGACGTTAAGTAGACAACACAGTTAGATAGGACAGACAGA – 5’................................................................................................................................................................................................................................ 5’ – CGATCTGATATGCCTATCCTGGGTATCTGTCTAATGTCTACCGTCTAACTGTATCAATTCAACTGTCTGCTGTCTGCAATTCATCTGTTGTGTCAATCTATCCTGTCTGTCT – 3’ And so on, with the quantity of DNA sequences between the primers doubling with each cycle of PCR.  One cycle of PCR 95°C 72°C 56°C Room temp https://www.enzo.com/note/what-are-the-differences-between-pcr-rt-pcr-qpcr-and-rt-qpcr/  Once you have this quantity of DNA, it is DNA Template number (millions) possible to visualize it. 30 20 10 https://www.dnasoftware.com/resources/news/amplification-efficiency-is-not-exponential/ The DNA can be dyed and run out on an agarose gel