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Chapter 10&11 DNA, Genes and Genomics 1.The Discovery of DNA EL: To revise DNA structure and learn about the discovery of DNA Discovery of DNA • DNA interactive interviews • Page 377-78 of NOB Prokaryotic DNA • The prokaryotes usually have only one chromosome, and it bears little morphological resemblance to eukaryotic chromosomes. • Consists of single, circular DNA molecule located in the nucleoid region of cell. Referred to as being “naked” • Bacterial cells may also contain multiple plasmids - small circular fragment of DNA separate from the main chromosome. Eukaryotic DNA Structure DNA consists of two molecules that are arranged into a ladder-like structure called a Double Helix. A molecule of DNA is made up of millions of tiny subunits called Nucleotides. Each nucleotide consists of: 1. Phosphate group 2. Ribose sugar 3. Nitrogenous base Nucleotides Phosphate Nitrogenous Base Ribose Sugar Nucleotides The phosphate and sugar form the backbone of the DNA molecule, whereas the bases form the “rungs”. There are four types of nitrogenous bases. Nucleotides A Adenine C Cytosine T Thymine G Guanine Nucleotides Each base will only bond with one other specific base. Adenine (A) Thymine (T) Form a base pair. Cytosine (C) Guanine (G) Form a base pair. A DNA Structure T G C T C A G T Because of this complementary base pairing, the order of the bases in one strand determines the order of the bases in the other strand. A G T C A DNA replication • DNA interactive animation DNA Replication 1. unfolding and unwinding of the DNA double helix at hundreds of points, known as replication origins, along the chromosome. 2. The enzyme helicase separates the two DNA strands, separating them like opening a zipper, with the point of opening being termed the replication fork. 3. Where the DNA strands are separated, a short length of RNA binds to each DNA strand under the control of the enzyme, DNA primase. This RNA acts as a primer (see figure 11.26a page 405). DNA Replication 4. A DNA polymerase enzyme can then proceed to build new DNA strands using each of the old strands as a template (see figure 11.26b). 5. Replication of DNA can occur only in the 5´ to 3´ direction. This is no problem with the so-called leading strand because its new complementary strand can be built continuously in the 5´ to 3´ direction. The other strand, known as the lagging strand, can be built only backwards and in short discontinuous pieces (Okasaki fragments - see figure 11.26b). 6. When finished, the RNA primers are removed, the gaps are filled by another DNA polymerase and the pieces are joined by the enzyme, DNA ligase. DNA Replication Watch DNAi clip Mitochondrial DNA • Mitochondria contain mtDNA, a double stranded circular molecule comprising: – 16 568 base pairs and code for 37 genes: – 13 genes code for proteins that are involved in cellular respiration – 2 genes code for ribosomal RNA (rRNA) – 22 genes code for transfer RNAs (tRNAs). Activity • Activity 7.1 – Simulation of DNA replication (handed in please) • Ch 10 quick check qu 9-11 (pg 353), 12-14 (pg 356), 18-19 (pg 406), • Chapter 10 review qu 2, 6, 7, 8 Reflection • After listening to some of the interviews with scientists, do you think you would have the patience to see out a major scientific discovery like that of DNA? Why or why not? • What learning was new today? • What learning was revision or built on what I already know? • What did I find most challenging and what strategies will I put in place to help me? • What percentage of the class did I spend on task and how can I improve this if needed? 2. DNA to proteins EL: To explore how protein synthesis occurs PROTEIN SYNTHESIS An individuals characteristics are determined by their DNA. The DNA determines which proteins are made. The most important proteins are enzymes. The sequence of bases in the DNA determines the sequence of amino acids in the protein. This is known as the GENETIC CODE. THE GENETIC CODE TRIPLET CODE 3 bases in the DNA code for one amino acid in the protein. Each triplet is known as a CODON. UNIVERSAL Found in all organisms. DEGENERATE More than one codon for each amino acid. NON-OVERLAPPING START AND STOP CODONS PROTEIN SYNTHESIS TRANSCRIPTION AMINO ACID ACTIVATION TRANSLATION Protein Synthesis Summary TRANSCRIPTION U U A C U U C A G C A G U G A A U A TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U U A C U U C A G C A G U G A A U A TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U U A C U U C A G C A G G A A U AU TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U U A C U U C A G C A G A A U AU G TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C U U A U A G C A G A A U A U GC TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C U U A U A G A G A A U A U GC C TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C U U A U A G A A A U A U GC C G TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C U U A U A G A A U A U GC C GA TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C U A U A G A A U A U GC C GA U TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C A U A G A A U A U GC C GA U U TA C GG CT AA CA TA C AATC G U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C A U A G A A U A U GC C GA U U G TA C GG CT AA CA TA C AATC U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C A G A A A U A U GC C GA U U GU TA C GG CT AA CA TA C AATC U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G U C A G A A U A U GC C GA U U GU A TA C GG CT AA CA TA C AATC U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G C A G A A U A U GC C GA U U GU A U TA C GG CT AA CA TA C AATC U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G C A A A U A U GC C GA U U GU A U G TA C GG CT AA CA TA C AATC U HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). G C A A A U A U GC C GA U U GU A U GU TA C GG CT AA CA TA C AATC G HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). C A A A A U GC C GA U U GU A U GUU TA C GG CT AA CA TA C AATC G HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). C A A A U GC C GA U U GU A U GUU A TA C GG CT AA CA TA C AATC G HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). C A A A U GC C GA U U GU A U GUU A G TA C GG CT AA CA TA C AATC HELICASE unwinds and unzips the relevant part of the DNA helix. RNA POLYMERASE attaches to the DNA. One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE STRAND). As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise order due to complementary base pairing (A <-> T/U ; G <-> C). Fully formed mRNA peels off the DNA and leaves the nucleus via a nuclear pore. The DNA rewinds. Fully formed mRNA peels off the DNA and leaves the nucleus via a nuclear pore. The DNA rewinds. AMINO ACID ACTIVATION ACTIVATION OCCURS WHEN THE tRNA COMBINES WITH A SPECIFIC AMINO ACID U A C THE ANTICODON DETERMINES WHICH SPECIFIC AMINO ACID IS ATTACHED TRANSLATION U A C G G C A U G C C G A U U G U A U G U UA G A ribosome binds to the mRNA near the START CODON. G G C U A C A U G C C G A U U G U A U G U UA G tRNA with the complementary ANTICODON (UAC) binds to the start codon (AUG) held in place by the large subunit of the ribosome. It brings with it the amino acid methione. G G C U A C A U G C C G A U U G U A U G U UA G The ribosome now slides along the mRNA to “read” the next codon. U A C G G C A U G C C G A U U G U A U G U UA G A second tRNA now bind to this codon, bringing a second amino acid with it. U A C G G C A U G C C G A U U G U A U G U UA G A peptide bond is formed between the two amino acids. G G C A U G C C G A U U G U A U G U UA G The tRNA which carried the first amino acid is released but leaves its amino acid behind as a DIPEPTIDE. G G C A U G C C G A U U G U A U G U UA G The ribosome now slides along the mRNA to “read” the next codon. G G C A U G C C G A U U G U A U G U UA G One by one each codon is read as the ribosome moves along the mRNA. G G C A U G C C G A U U G U A U G U UA G Each time the growing polypeptide is linked to the amino acid on the incoming tRNA. A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G A U G C C G A U U G U A U G U UA G STOP ! A U G C C G A U U G U A U G U UA G The polypeptide is complete when the ribosome reaches the STOP codon. A U G C C G A U U G U A U G U UA G The polypeptide is released. A U G C C G A U U G U A U G U UA G The polypeptide is released. The polypeptide may combine with other polypeptides and will become variously coiled/folded to produce a protein. The tRNAs are recycled. A U G C C G A U U G U A U G U UA G The mRNA may be used again in this form, or it may be broken down into nucleotides which can be reassembled to produce a different polypeptide. The ribosome is free to move along another mRNA. Ribosomes work in groups so that many slide along a mRNA molecule simultaneously. These groups are called POLYRIBOSOMES. Each ribosome takes about 1 minute to travel along a mRNA molecule. Transcription and translation • DNAi animations QUIZ What sort of chemical is helicase? Why is DNA double stranded if one strand is redundant? Where in the cell are the ribosomes? What is the start codon? Give the three alternative stop codons. Give the primary structure (sequence of amino acids) of the polypeptide made in this animation. What is the difference between a polypeptide and a protein? What is the advantage of ribosomes operating as polyribosomes? What are the similarities and differences between DNA replication and protein synthesis? ANSWERS Helicase is an enzyme and therefore also a protein. DNA is double stranded to permit replication. Ribosomes are located in the cytoplasm. The start codon is AUG. The three stop codons are UGA, UAG and UAA. Give the primary structure of the polypeptide is methionine, proline, isoleucine, valine, cysteine. Polypeptides have less than 100 amino acids, protein have more. A protein may consist of several polypeptides. Polysomes increase efficiency, they enable one mRNA molecule to produce many polypetides simultaneously. HINT - Think about the enzymes and nucleotides used, the end product and the location of the process. Really good website • http://www.le.ac.uk/ge/genie/vgec/he/index. html Activity • In pairs, complete activity 11.1 • Quick check qu 4-8 (pg 391) • Ch 11 ch review qu 2, 3, 4, 9 (pg 414-416) Reflection • Did the hands-on activity help cement your understanding of transcription and translation? If not, how else do you think you could ensure your understanding? • What learning was new today? • What learning was revision or built on what I already know? • What did I find most challenging and what strategies will I put in place to help me? • What percentage of the class did I spend on task and how can I improve this if needed? 3.Genomics EL: What a genome is and how gene expression is regulated What is a genome? • All of the genetic material (the base pairs) found in one complete set of an organism’s chromosomes. • The study of genomes is called genomics. Does genome size matter? COMMON NAME SPECIES NAME Approx GENOME SIZE (millions of base pairs) Fruit fly Drosophila melanogaster 180 Snake Boa constrictor 2100 Human Homo sapiens 3100 Onion Allium cepa 18000 Lungfish Protopterus aethiopicus 140000 Amoeba Amoeba dubia 670000 Why would a single celled animal like the amoeba need a genome that is about 200 times larger than the human genome? Ans: They carry a lot of junk DNA! What is a gene? • Segment of DNA that codes for formation of a protein – Structural genes express structural and/or functional proteins. – Regulatory genes are short nucleotide sequences that express proteins that control the activity of structural genes by feedback mechanisms. Number of genes COMMON NAME SPECIES NAME No. GENES Human Homo sapiens 25000 Mustard plant Arabidopsis thaliana 27000 Fruit fly Drosophila melanogaster 14000 Baker’s yeast Saccharomyces cerevisiae 6000 Gut bacterium Escherichia coli 4000 Should we be offended that a mustard plant has as many genes as a human? An Overview of Gene Structure Coding Region DNA sequence that will be transcribed from the template strand. 5’ 3’ 3’ Regulatory region S T A R T Promoter region S T O P Terminator region 5’ Gene Expression • The expression of genetic information is one of the fundamental activities of all cells. • Instructions stored in DNA are transcribed and translated into various RNA molecules. Introns and Exons • The coding region in eukaryotes contain: – introns - non-coding regions of DNA – exons - coding regions of DNA • Prokaryotes do not have introns – why? – They don’t carry “junk DNA” due to short replication cycles RNA Processing in Eukaryotic Cells DNA Template Strand INTRON EXON INTRON EXON EXON Spliceosome Pre-mRNA transcript of DNA template strand EXON EXON Spliceosome EXON Introns are spliced out by spliceosomes leaving only the sequences that will be expressed. This is an example of RNA processing. The introns usually are degraded. The result is a mature mRNA strand that will leave the nucleus to be translated. Genome to proteome • The human genome has about 25,000 genes but our proteome (the total number of different proteins) is much larger (~100,000) – How can this occur? • Many genes can produce more than one protein because the mRNA transcript contains different combinations of exons. This process is called alternative splicing. Alternative splicing INTRONS Pre-mRNA transcript EXON 1 EXON 2 EXON 3 EXON 4 Possible mRNAs using different combinations of exons EXON 1 EXON 1 EXON 2 EXON 2 EXON 2 EXON 3 EXON 4 EXON 3 EXON 4 EXON 4 PROTEIN 1 PROTEIN 2 PROTEIN 3 Result: when each mRNA is translated, a different protein is produced. Activity: • In pairs, complete activity 10.2 “Sequencing a genome” • Quick check qu 15-17 (pg 362), 18-23 (pg 364), 24-27 (pg 370), 16&17 (pg 403) • Chapter review qu 3, 5, 9 (pg 381-382) Reflection • What learning was new today? • What learning was revision or built on what I already know? • What did I find most challenging and what strategies will I put in place to help me? • What percentage of the class did I spend on task and how can I improve this if needed? 4.Gene Regulation and Genetic Mutations EL: How genes are switched on and off The effects genetic mutations can have on organisms Gene Regulation • Each cell contains an entire organism’s genome. • All cells of an organism have the same genome, but can have different phenotypes. • For example, cells in your eye have the gene for producing fingernail protein (keratin) but this gene is not expressed. • How do genes get switched on or switched off? Why regulate gene expression? • Cells conserve energy and materials by blocking unneeded gene expression. • If a substrate is absent in the environment why produce the enzyme for that substrate! • Repressor molecules keep the cell from wasting energy by not transcribing mRNA or making enzyme molecules that have no use. • The cell can control its metabolism – resources are used only when there is a metabolic need and can be redirected to other metabolic pathways. Gene regulation in prokaryotes • Bacteria have groups of genes that are controlled together and are turned on/off as required. • E.g. the lac operon is a set of genes in bacteria used for lactose metabolism. • Bacteria produce the enzymes to break down lactose to glucose and galactose only when lactose is present. • http://www.sumanasinc.com/webcontent/animations/content/lacop eron.html • http://pages.csam.montclair.edu/~smalley/LacOperon.mov Gene regulation in Eukaryotes • Still being investigated • Proteins involved (like prokaryotes) – but more complicated – Enhancers: act as binding sites for activator proteins and increase number of DNA polymerase molecules transcribing genes – Chemical modification: eg presence or absence of histone proteins Changing the Genetic Information Mutations • A mutation refers to any permanent change in the DNA nucleotide base sequence of an organism. • Mutations occur spontaneously and randomly throughout the lifetime of all organisms This red delicious apple illustrates a somatic mutation. A mutation to the ovarian wall gives rise to a sector of yellow colored fruit. The mutation does not affect the seeds (germline) which give rise to the standard red delicious type. • The effects of mutations vary depending on their location both within the chromosome (or gene) and the body of the organism. Acquiring Mutations • Mutations in the DNA of an organism can be caused by: - Mistakes in DNA replication. • This is a natural process • 1 mistake in 1,000,000,000 bases • Proof reading enzymes correct most mistakes - Environmental factors that increase the rate of mutations are called mutagens. • Radiation • Various chemicals • High temperatures Can you inherit a mutation? • Yes! If a mutation occurs in the cells that produce gametes (germ-line cells) the change will be passed onto the offspring. • If a mutation occurs in any other cell of the body (somatic cells) it will not be inherited, but it may affect the individual during their lifetime. The biological consequences of mutations • Mutations may be beneficial, neutral or harmful! • Mutations are a source of genetic variation – new alleles in a population – that may be selected for by environmental factors and confer an advantage on the organism. • Mutations are a source of biological novelty for evolution. Types of Mutations • Point Mutations (i.e. spelling mistakes) - Changes in a single DNA nucleotide - These can occur within a gene’s coding region or within regulatory regions of genes • Block Mutations - Changes in a segment of a chromosome These changes usually involve the rearrangement of a number of genes • Chromosome Number Mutations - Changes in the number of chromosomes Point Mutations • Single nucleotide substitutions may result in: 1. Changed amino acid sequence 2. No amino acid change because genetic code is degenerate 3. Results in “stop” instruction and formation of a new allele. Mutation: Substitute T instead of C Original DNA Mutant DNA Point Mutations As a reference for the following screens, the diagram below illustrates the transcription and translation of DNA without a point mutation. Original Unaltered Code Original DNA Transcription mRNA Translation Amino acids Amino acid sequence forms a normal polypeptide chain 1. Changed amino acid sequence • A single base is substituted by another. • Usually results in coding for a new amino acid in the polypeptide chain. Mutation: Substitute T instead of C Original DNA Mutant DNA mRNA Amino acids Polypeptide chain with wrong amino acid 2. No amino acid change A single base is Normal DNA substituted by another. mRNA Called silent or neutral Amino acids mutations and produce little or no change in the phenotype. A change in the third base of a codon still codes for the same amino acid. Amino acid sequence from the non-mutated DNA forms a normal polypeptide chain Mutation: Substitute C instead of T Mutant DNA mRNA Amino acids Despite the change in the last base of a triplet, the amino acid sequence is unchanged 3. Results in “stop” instruction and formation of a new allele • A single base is substituted by another. • This results in a new triplet that does not code for an amino acid. • The resulting triplet may be an instruction to terminate the synthesis of the polypeptide chain. Mutation: Substitute A instead of C Original DNA Mutant DNA mRNA Amino acids Mutated DNA creates a STOP codon which prematurely ends synthesis of the polypeptide chain Reading Frame Shift by Insertion or Deletion A single base is inserted, upsetting the reading sequence for all those after it. • A reading frame shift results in new amino acids in the polypeptide chain from the point of insertion onwards. • The resulting protein will be grossly different from the one originally encoded (it is most likely to be non-functional). Mutation: Insertion of C Original DNA Mutant DNA mRNA Amino acids Large scale frame shift results in a new amino acid sequence. The resulting protein is unlikely to have any function. Trinucleotide Repeat Expansions • Many normal human genes contain multiple copies of a three base sequence called a trinucleotide. • These repeating sequences can expand in number. This mutation gives rise to several inherited conditions. • The mutant allele that causes “fragile X syndrome” has 200 to 2000 repeats of the trinucleotide CGG, in contrast to 6 to 50 repeats in a normal person, in the untranslated region of the FMR1 gene. Fragile X Syndrome • Occurs 1 out of every 4000 males and 1 out of every 6000 females. • Mutation of the FMR1 gene on the X chromosome leads to loss of the fragile Xmetal retardation protein, FMRP. This FMRP protein is involved in the translation of a number of essential neuronal mRNA’s. • Characteristics of this disease include: - Metal retardation Shyness and limited eye contact Elongated face Large or protruding ears Large testicles (macroorchidism) Low muscle tone Activity • Quick check qu 28&29 (pg 373), 30&31 (pg 376), 20-24 (pg 412) • Ch 10 ch review qu 11, 12, 13 pg 382-383 Reflection • What learning was new today? • What learning was revision or built on what I already know? • What did I find most challenging and what strategies will I put in place to help me? • What percentage of the class did I spend on task and how can I improve this if needed? 5. Genetic Mutations EL: Chromosomal mutations and defects Block Mutations • The rearrangement of blocks of genes within a chromosome. Can occur during crossing over in meiosis • The rearrangement of blocks of genes between non-homologous chromosomes (translocation). A piece of one chromosome is broken off and joined to another chromosome. • Block mutations result in a new gene order along a chromosome. They can be highly disruptive! translocations Normal Deletion: Pieces of chromosome are lost. Duplication: Pieces of chromosome are repeated so there are duplicate segments Inversion: Pieces of chromosome are flipped so the genes appear in reverse order. Translocation: Pieces of chromosome are moved from one chromosome onto another . Changes to the Chromosome Number Examples of Polyploid Plants Name Number • Aneuploidy – changes to the number of specific chromosomes Common wheat 6N = 42 Tobacco 4N = 48 • Polyploidy- changes to the number of whole sets of chromosomes Potato 4N = 48 Banana 3N = 27 Boysenberry 7N = 49 Strawberry 8N = 56 Aneuploidy • Change to the number of specific chromosomes. • The extra or missing chromosome may be an autosomal or a sex chromosome. • Such changes are due to nondisjunctions. These events are due to errors in chromosome segregation in meiosis. • Pairs of homologous chromosomes may fail to separate in meiosis I or the centromere may fail to separate the sister chromatids in meiosis II. Down’s Syndrome An extra chromosome 21 Non-Disjunction: Meiosis I Non-disjunction • • Meiosis I A non-disjunction in meiosis I occurs when homologous chromosomes fail to separate properly during anaphase I. Meiosis II One gamete receives two of the same sort of chromosome and the other gamete receives no copy. n+1 Gametes n+1 n–1 n–1 Non-Disjunction: Meiosis II • • • Meiosis I A non-disjunction in meiosis II occurs when sister chromatids fail to separate properly during anaphase II. One gamete receives two of the same sort of chromosome and the other gamete receives no copy. Some gametes are unaffected. Meiosis II Nondisjunction n+1 Gametes n–1 n n Turner’s Syndrome 44 + X • Occurs in 1 out of every 25,000 births • Only one X chromosome is present and is fully functional. • Common symptoms include: – – – – – Short stature Swelling of hands an feet Broad chest Low hairline Webbed neck Klinefelter’s Syndrome 44 + XXY Mildly impaired IQ (intelligence) Chest hair is sparse Poor beard growth Frequently some breast development (low levels of testosterone) Osteoporosis Female type pubic hair pattern Penis and testes underdeveloped, low levels of testosterone. Always infertile. Limbs tend to be longer than average Sex chromosomes: XXY • • • • • • • • Occurs in about 1 in 500 to 1000 births. Characteristics vary widely. Males are normally sterile Some degree of language impairment Youthful build Rounded body type Some degree of gynecomastia Hypogonadism Polyploidy • Organisms that have more than two sets of chromosomes • Very common in plants because they can reproduce asexually, but rarer in animals. • Polyploidy can result in “instantaneous speciation”. • Polyploid plants are usually more robust and sturdy than diploid plants. • In our long history of plant cultivation we have selected out such plants because they produced a higher yield and were less subject to disease. • As a result many of our crops today have been bred to a high level of ploidy. • Wheat was the first crop to be domesticated originating in SW Asia about 10,000 years ago. • Today, bread wheat is a hexaploid. An Expert in Polyploidy • The twenty different species worldwide differ widely in their chromosome number as they exhibit a range of polyploidy. • The haploid number is 7 As a general rule, strawberry species with more sets of chromosomes tend to be more robust and produce larger plants with larger berries. • Strawberry species an be: – – – – – Diploid Tetraploid Hexaploid Octoploid decaploid Fitness of Mutations • The fitness of a mutation describes its value to the survival and reproductive success of the organism. A mutation may turn out to be: • Lethal: Many mutations are lethal and embryos are non-viable. • Harmful: Non-lethal mutations, e.g. Down syndrome and sickle cell disease, may be expressed as effects that lower fitness. • Silent (neutral): Most point mutations are probably harmless, with no noticeable effect on the phenotype. • Beneficial (useful): Occasionally mutations may be useful, particularly in a new environment, e.g. insecticide resistance in insects, antibiotic resistance in bacteria. Mutations: Key Points • All new alleles originate by mutation. • New alleles introduce genetic variation: the raw material on which natural selection can act. • Most mutations occur in somatic cells and are not inherited. • Only mutations in gametes can be inherited. Evolutionary Significance of Mutations • Polyploidy can result in the formation of “instant species” by creating a barrier to chromosome pairing at meiosis (common in plants). • Fusion of chromosomes (a form of translocation) may reduce chromosome number. This can result in reproductive isolation and a new species. Possible fusion of two chromosomes to create the No. 2 chromosome in humans. Note the similar banding patterns of chromosomes from related primate species. 2 12 12 12 13 11 11 Human Chimpanzee Gorilla Orangutan Example: Fusion of chromosomes may have taken place during the course of human evolution. The chromosome number in the great apes is 2N = 48, whereas in humans 2N = 46. Activity: • Complete Activity 11.2 “Changing your genes” • Quick check qu 14&15 (pg 400) • Biochallenge pg 413 • Ch 11 ch review qu 5, 7, 8, 11, 13, 14, 15 Reflection • What learning was new today? • What learning was revision or built on what I already know? • What did I find most challenging and what strategies will I put in place to help me? • What percentage of the class did I spend on task and how can I improve this if needed