What are the key components of chromosomes?A. DNA -heterochromatin -euchromatinB. ProteinsC. Found in nucleusD. You should understand the relationship between DNA and proteins (chromatin packing and histones)
Key termsA. Eukaryotic chromosomes-made of DNA and proteins (histones)B. Gene-heritable factor that controls specific characteristics -made up of a length of DNA, found on a specific chromosome location (a locus)C. Allele-one specific form of a gene (all found at the same locus) -Example: Everyone has the gene for eye color. The possible alleles are blue, brown, green, etc.
More Key TermsD. Genome-total genetic material of an organism or species (Example: The Human Genome)E. Gene pool-total of all genes carried by individuals in a population
MutationsA. Chromosome mutations-involve large sections of chromosomes (or the whole thing) -Ex: Down’s syndrome, Turner’s syndrome
MutationsB. Gene mutation-involves changes in single base pairs -Some mutations may not have any effect on the cell and may involve: 1. part of the sense strand of DNA which is not transcribed 2. part of the DNA that a cell does not use 3. changes in second or third bases of a codon (since the genetic code is degenerate the same base may still be coded for)
MutationsB. Gene mutation-involves changes in single base pairs Example: Insertion or deletion of single organic bases -changes the DNA sequence that will be transcribed and translated original DNA sequence: ATG-TCG-AAG-CCC transcribed: UAC-AGC-UUC-GGG translated: tyr-ser-phe-gly addition of base A: ATA-GTC-GAA-GCC-C transcribed: UAU-CAG-CUU-CGG translated: thy-glu-leu-arg
Mutations: Base substitutions and sickle-cell anemiaA. Hemoglobin-protein that helps RBC carry oxygenB. Hb is a gene that codes for hemoglobin -made of 146 amino acidsC. In some cases one base is substituted for another normal: (HbA) base substitution: (HbS) CTC CAC GAG GUG-after transcription and translation HbA produces glutamic acid and HbS produces valine
Mutations: base substitutions and sickle-cell anemiaD. The altered hemoglobin HbS is crystalline at low oxygen levels causing the RBC to become sickled and less efficient at oxygen transportE. Symptoms of sickle cell anemia -physical weakness -heart or/and kidney damage -death
Mutations: base substitutions and sickle-cell anemiaF. In heterozygous people (one normal allele and one sickle cell allele) -the alleles are codominant, but the normal allele is expressed more strongly -in codominance both alleles are expressed (one is not dominant to the other) -some sickled cells present, but most are normal -some people show mild anemia (deficiency of the hemoglobin, often accompanied by a reduced number of red blood cells and causing paleness, weakness, and breathlessness)
Mutations: base substitutions and sickle-cell anemiaG. Advantages of being heterozygous -In areas where malaria is infested: -Plasmodium cannot live in erythrocytes with HbS -Heterozygous individuals have a reduced chance of contracting the protist that is carried by mosquitoes
KaryotypingA. Karyotyping-process of finding the chromosomal characteristics of a cell -chromosomes are stained to show banding and arranged in pairs according to size and structure -You should be able to look at a karyotype and determine the sex of the individual and if non-disjunction has occurred
Amniocentesis and karyotypingA. Amniocentesis -performed at around 16 wks -sample of amniotic fluid is taken and cultured -when there are enough cells, a karyotype can be performed -chromosomes are arranged into pairs to detect any abnormalities -can be used to detect Down’s syndrome (a.k.a. trisomy 21) -can be used to recognize sex or non-disjunction
Karyotyping and Chorionic villus samplingSampling is performed around 11 weeks of pregnancyCells are gathered from chorionic villi (cells from the zygote)Cells are cultured, DNA is extracted and a karyotype is made
Meiosis Key TermsA. Diploid-having two sets of chromosomesB. Homologous chromosomes-matching pairs of chromosomes -have the same genes -are not identical (one chromosome comes from each parent, thus alleles may be different) -found in diploid cellsC. Haploid-having only one set of chromosomes
More Meiosis TermsD. Chromatids -two parts of a chromosomeE. Centromere -part of a chromosome that connects the chromatidsF. Reduction division-in organisms that reproduce sexually -reduction of the number of chromosomes by half (from a diploid nucleus to a haploid nucleus) -think of eggs and sperm; both are haploid (when they unite the diploid number is restored)
MeiosisB. Produces gametes (sperm and egg)C. Overview 1. homologous chromosomes pair (diploid) 2. two divisions (meiosis I and meiosis II) 3. result=4 haploid cellsD. When gametes unite (to produce a diploid cell) a cell with homologous pairs is created -one set of chromosomes is from the mom and one set of chromosome is from the dad
chiasmata Meiosis (details) TetradA. Interphase -DNA replicationB. Prophase I -chromosomes condense -nucleolus becomes visible -spindle formation -synapsis-homologous chromosomes are side by side (they become a tetrad and are intersected at the chiasmata) -nuclear membrane begins to disappear
Meiosis (details)C. Metaphase I: Bivalents move to equatorD. Anaphase I: -Homologous pairs split -One chromosome from each pair moves to opposite poleE. Telophase I: -Chromosomes arrive at poles -Spindle disappears
Meiosis (details)F. Prophase II – new spindle is formed at right angles to the previous spindleG. Metaphase II – Chromosomes move to equatorH. Anaphase II – -Chromosomes separate -Chromatids move to opposite poles
Meiosis (details)I. Telophase II -Chromosomes arrive at poles -Spindle disappears -Nuclear membrane reappears -Nucleolus becomes visible -Chromosomes become chromatin -Cytokinesis
Crossing over of homologous chromosomesA. An important source of variation -creates new combinationsB. Happens during prophase IC. Called a synapsisD. Recombination-reassortment of genes into different combinations from those of the parents Chiasmata formed during a synapsis
Crossover H E StartA A genotype HEB B genotype HEC C genotype he D genotype heD h e H E A Crossover B and C become B recombinants C D h e H E ResultsA A genotype HEB B genotype HeC C genotype hE D genotype heD h e
Meiosis and genetic variationA. The number of possible gametes produced by random orientation of chromosomes is 2n (where n is = to the haploid number of chromosomes)B. In humans the production of 1 gamete has over 8 million possible combinations (223)C. Recombination (in prophase I) + Random orientation of chromosomes (in metaphase I) = an infinite number of variations
Meiosis and Non-disjunctionA. Disjunction - when the homologous chromosomes separate in anaphase IB. Aneuploidy -happens when chromosomes do not separate (in anaphase I or II) -caused by non-disjunction -result: one cell missing a chromosome and one cell having an extra chromosome -Total number of chromosomes = 2n 1C. Polyploidy- having more than two complete sets of chromosomes (common in plants)
Non-disjunction and Down’s syndromeA. One of the parent gametes contains two copies of chromosome 21B. The zygote then ends up with 3 copies -2 from one parent -1 from the otherC. Down’s syndrome = trisomy 21D. Chances of non-disjunction of chromosomes increases with age in females (in males too, but less of an effect)
Non-disjunction and Down’s syndromeE. Female age has a greater effect because: -gametes are produced before birth -more exposure to chemicals, radiation, etc.F. Male age has less effect because they do not produce gametes until pubertyG. Genetics may also increase the likelihood of having a child with Down’s syndrome
Theoretical Genetics Key TermsA. Dominant allele-the allele that always shows in the heterozygous state (Example: Bb=brown)B. Recessive allele-the allele that only shows in the homozygous recessive state (Example: bb=blue)C. Codominant alleles-pairs of alleles where two differing alleles are shown in the phenotype in a heterozygoteD. Homozygous -having two identical alleles of a gene (Example: BB or bb)E. Heterozygous -having two different alleles of a gene (Example: Bb)
More VocabularyF. Carrier- a person who has a recessive allele, but does not express it (they are generally heterozygous, Bb)G. Genotype-alleles that a person has (the letters) Ex: BbH. Phenotype- the physical characteristics the a person shows (caused by the genotype) Ex: brown hair or blue eyesI. Test cross- crossing two or more genotypes to find the possible genetic outcomes
Mendel’s Monohybrid CrossesA. Punnett square-shows possible outcomes from a test crossB. Mendel studied characteristics of pea plants Wrinkled and round peas (round peas are dominant)
Gregor Mendel’s Findings Dominant Recessive Trait Expression Expression1. Form of ripe seed Smooth Wrinkled2. Color of seed albumen Yellow Green3. Color of seed coat Grey White4. Form of ripe pods Inflated Constricted5. Color of unripe pods Green Yellow6. Position of flowers Axial Terminal7. Length of stem Tall Dwarf
Mendel’s Monohybrid CrossesC. Mendel found tall is dominant over shortD. His procedures were: 1. Start with 2 pure breeding homozygous plants (This is the P generation.) -Plant 1=tall (TT) -Plant 2= short (tt) 2. Cross-breed the plants -Place pollen from the tall plant in the short plant and vice versa.
Mendel’s Monohybrid CrossesD. His procedures were: 3. The F1 generation is the 1st group of offspring. -All were tall, and had a heterozygous genotype. (Tt) -This is an application of the law of segregation. -All offspring had a ‘T’ from one parent and a ‘t’ from the other parent
Mendel’s Monohybrid CrossesD. His procedures were: 4. The F1 offspring were then crossed. (Tt x Tt) -Possible outcomes can be found using a Punnett square GENOTYPES T t -25% TT T TT Tt -50% Tt -25% tt t Tt tt PHENOTYPES -75% Tall -25% Short
Multiple allelesA. When genes have more than two allelesB. Example: Blood type has 4 phenotypes based on three alleles (IA, IB and i)C. IA and IB are codominant and i is recessiveD. This is why parents can have kids with different blood types Phenotypes Genotypes A IAIA or IAi B IBIB or IBi AB IAIB O ii
Multiple allelesD. Perform a test cross for P: mother with O blood type and father with AB blood type. What are the possible phenotypes? Phenotypes Genotypes A IAIA or IAi B IBIB or IBi AB IAIB O ii
Multiple allelesPerform a test cross for P: mother with O blood type and father with AB blood type. What are the possible phenotypes for F1? P = ii x IAIB i i IA IA i IA i IB IB i IB i None of the children can have the same blood type as either of the parents.
CodominanceA. When neither allele for a gene is recessiveB. Example: Blood typeC. Alleles A and B are both dominant (both are expressed)D. i is recessive to alleles A and BE. One letter is chosen and the possible alleles are written in upper case letters to illustrate codominance Phenotypes Genotypes A IAIA or IAi B IBIB or IBi AB IAIB O ii
Sex chromosomes and genderOnly possibility for P generation = XX and XY X X The sex of all babies is determined by the X XX XX chromosomes in the sperm from the Y XY XY man.-The X chromosome is larger and carries more genes than the Y chromosome -Examples of genes on X, but not Y = color blindness and hemophilia -Many sex-linked traits are related to the X chromosome.
Sex linkageA. Genes carried on sex chromosomes (usually X)B. Example: Hemophilia-a blood disorder that prevents clotting -patients do not produce clotting factors that allow coagulation of blood, and thus torn blood vessels are prevented from closing -most common in boys (they get it from their mother’s X chromosome, as they only get one X, which means only one chance to get the dominant allele)
Sex linkageC. Two parents without hemophilia: XHXh and XHY XH Xh XH XHXH XHXh Y XHY XhY*The XhXh does is very rare*Males cannot be heterozygous carriers because they only have one X.*Females can be carriers and pass on the trait to the next generation. They can be heterozyg. or homozyg.* XH -Normal and Xh –Hemophilia
Predict the genotypic and phenotypic ratios of monohybrid crosses for each of the following.1. Sickle cell anemia: HbA=normal and HbS=sickle cell HbAHbS x HbAHbS2. Colorblindness: XB=normal and Xb=colorblind XBXb x XbY3. Hemophilia: XH=normal and Xh=hemophilia XhXh x XHY4. Blood type: IAi x IBi**You should be able to determine if alleles are codominant because both alleles will be represented by capital letters. You should also know if the inherited traits are sex-linked.
Mendel’s Law of SegregationA. States: The separation of the pair of parental factors, so that one factor is present in each gamete. (This is how it is written in the IB book.)B. The two alleles for each characteristic segregate during gamete production. This means that each gamete will contain only one allele for each gene. This allows the maternal and paternal alleles to be combined in the offspring, ensuring variation. (This is from wikipedia.)
Mendel’s Law of Segregation and MeiosisA. Mendel looked at genes (on chromosomes)B. Found that each gene appeared twice (in homologous pairs)C. Figured out that when a synapsis occurs in prophase I followed by a separation in anaphase I, homologous chromosomes move to opposite polesD. One chromosome from each pair ends up in a gamete
Mendel’s Law of Independent AssortmentA. States that allele pairs separate independently during the formation of gametesB. Any one pair of characteristics may combine with any one of another pair of characteristicsC. See p. 163 in Green Book
Independent assortment and meiosisA. Any combination of chromosomes is possible in metaphase I (there is no ‘master plan’ for the order that they line up on the metaphase plate before separation)B. Mendel thought all genes were inherited separately and had no relationship -Ex: Pea plants could be green or yellow and wrinkled or round. Shape and color had nothing to do with each other, because the genes are on separate chromosomes. Any combination could have been produced (wrinkled/green, wrinkled/yellow, round/green, round yellow)C. This is demonstrated in Dihybrid crosses.D. Today we realize that there are many exceptions
Possible chromosome alignments A Possible random outcomes B
Mendel’s Law of Independent AssortmentA. Any one of a pair of characteristics may combine with either one of another pairB. Example: gene=shape of pea alleles=round (R) or wrinkled (r) gene=color of pea alleles = yellow (Y) or green (y) *When crossing two plants that are heterozygous for both traits the offspring will show all combinations. This shows that genes for shape and color are independent (unlinked).
Dihybrid crosses Parent genotypes: SsYy x SsYy -S=smooth s=wrinkled -Y=Yellow y=green SY Sy sY syPossibleallele SY SSYY SSYy SsYY SsYycombosfrom one Sy SSYy SSyy SsYy Ssyyparent sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy
F1 Genotypic ratios Possible ratios for 1:SSYY SsYy x SsYy 2:SSYy 1:SSyy SY Sy sY sy 2:SsYY 4:SsYySY SSYY SSYy SsYY SsYy 2:Ssyy 1:ssYYSy SSYy SSyy SsYy Ssyy 2:ssYy 1:ssyysY SsYY SsYy ssYY ssYy F1 Phenotypic ratios 9: smooth-yellow 3:smooth-greensy SsYy Ssyy ssYy ssyy 3: wrinkled-yellow 1: wrinkled-green
Perform a test cross for P: SSYY x ssyy1. What will the genotype and phenotype ratios be for the F1 generation?2. What will the phenotype and genotype ratios be for the F2 generation?3. Determine the recombinants in each generation.
Perform a test cross for P generation: SSYY x ssyy SY SY SY SY sy SsYy SsYy SsYy SsYy sy SsYy SsYy SsYy SsYy Sy SsYy SsYy SsYy SsYy Sy SsYy SsYy SsYy SsYyAll F1 generation offspring are heterozygous (SsYy). What will the outcome be if you cross two individuals from F2?
Autosomal Gene LinkageA. Autosome -all chromosomes that are not sex chromosomesB. Sex chromosomes -determine if an individual is male or femaleC. Linkage group -a group of genes whose loci are on the same chromosomeD. Gene linkage is caused by pairs of genes being inherited together. The presence or absence of one can affect the other.
Autosomal Gene LinkageA. In gene linkage, all of the genes on a chromosome are inherited togetherB. Does not apply to Mendel’s law of independent assortmentC. The closer the loci of the two genes on the chromosome, the smaller the chance that crossing over will occur in a chiasmataD. If no info. is given, assume the genes are not linked
Autosomal Gene LinkageE. These genes do not follow the law of F1 generation independent assortment. PL PL PL PLF. Example: pl P=purple, p=red, L=long, l-round pl P gen.=PPLL x ppll pl pl *all PpLl
Autosomal Gene Linkage F2 generationF. Example continued: P=purple, p=red, PL Pl pL pL L=long, l-round PL F1 gen.=PpLl x PpLl Pl-predicted results would be a ratio of 9:3:3:1 pL-observed results were very pL different than predicted because the genes are for color and pollen shape are linked on the same chromosome. See p. 87 in your review guide.
Gene linkage and DrosophilaA. When genes are linked actual outcomes do not match expected outcomeB. This would be evident in a dihybrid cross (a cross between two genes)C. Linked genes are represented in vertical pairs with horizontal lines between them Example: b+=tan body b=black body w+=long wings w=short wings b+ w+ *The fly is tan and has long wings. b+ w+ (genotype b+b+w+w+)
Recombinants and gene linkageA. Recombinants are formed as a result of crossing over (during prophase I)B. They are found in combinations that did not exist in the parents
Polygenic InheritanceA. When the inheritance/expression of a characteristic is controlled by more than one geneB. Example: Human skin color -involves at least 3 independent genes -A, B and C represent alleles for dark -a, b and c represent alleles for light skin -P: AABBCC x aabbcc -F1: AaBbCc (all are heterozygous for all alleles) -F2:???
Polygenic InheritanceB. Example: Human skin color -F2:??? -to find out you can make a Punnett square -The more dominant alleles there are, the darker the skin -Dominant alleles are codominant (they are all expressed)
Polygenic inheritance (Other traits)A. HeightB. Eye colorC. Finger prints
Polygenic InheritanceExample: Flower color of beans -Genes A and B control color expression -AP and BP = purple -AW and BW = white -Each color has two alleles -Purple and white are codominant -Because of the codominance the flowers will be shaded depending on their genotypes
P: APAPBP BP x AWAWBWBW F1: APAWBPBW AP BP AP BP AP BP AP BP AWBW AWBW AWBW AWBW
F1: APAWBPBW x APAWBPBWF2: See Punnett square (fill it in)Determine which ones are white, purple andintermediates.) AP BP AP BW AWBP AWBW AP BP AP BW AWBP AWBW
Polygenic inheritance and variation patternsTwo patterns are commonly expressedA. Continuous variation – shows a continuum of variation among a population in a bell curve format - Example: In height expression people can be short, medium or tall (and everything in between)B. Discontinuous variation – does not show a continuum (it is one or the other, there is no in- between expression of phenotype) - Example: Blood type can be A, B, AB or O (nothing in between)
Pedigree ChartsA. Used to show inheritance of traits over several generationsB. Affected individuals-shaded blackC. Unaffected individuals- left blankD. Heterozygous individuals (carriers)- shaded grey or filled in half wayE. Example: Queen Victoria and hemophilia (recessive/sex-linked)
PCRA. PCR=polymerase chain reactionB. PCR=DNA photocopierC. Used to make copies of specific DNA sequences (for study)
How a PCR works1. DNA is heated (H bonds are broken)2. RNA primers are added to start replication3. As the DNA cools primers bind to the single strands (H bonds and complementary base pairing)4. Nucleotides and DNA polymerase are added
How PCR works5. Nucleotides bond with ‘exposed’ bases on the DNA strand6. DNA polymerase joins them7. Complimentary strands are formed8. New strands can be heated, separated and copied9. Animation
Gel electrophoresisA. Technique used to separate molecules base on their rates of movement in an electric field -caused by charge and size of moleculesB. Commonly used in DNA profiling
Gel electrophoresis and DNA profilingA. Scientists cut a mixture of DNA into segments by restriction enzymesB. Place into a special gel with a current running through itC. DNA separates into bands according to sizeD. Mixture is compared to a control groupE. The more similar the DNA strands are the more closely they are relatedF. See p. 30 in review guide or p. 66 in IB textbook for examples
Figure 3. Comparison of known gel results for normalhemoglobin (AA), sickle cell disease (SS) and sickle cell trait (AS). S represents the molecular size marker. What results are present in the lanes marked with a question mark?
Application of DNA profilingA. Criminal investigations -collect blood or semen from suspect -if enough is not present use PCR -compare gel electrophoresis of suspect and victimB. Indentify remains of deceased -take blood samples of living -compare with samples from deadC. Paternal tests
Genetic screeningA. Test individuals in a population for presence of absence of a gene or allele
Genetic ScreeningB. Advantages: 1. pre-natal diagnostics (seek treatment or abortion) Ex: PKU or down’s syndrome 2. Reduce frequency of alleles that cause genetic illnesses (opt to not reproduce or use IVF and select embyros without genetic defect) 3. Genetic diseases that show symptoms in later life can be detected earlier (Ex: Huntington’s disease)
Genetic ScreeningC. Disadvantages: 1. Increased abortion rate 2. Harmful psychological effects -could lead to discrimination (when seeking insurance or medical assistance) -fear of getting older (depression) -creates a genetic ‘underclass’
Human Genome ProjectA. An international cooperative venture established to sequence the complete human genomeB. Hope to determine the location and structure of all genes in the human chromosomesC. Genome-total genetic material of a cellD. Completed in 2000 (about 10 years early)
Advantages of the Human Genome ProjectA. Understand genetic diseasesB. Figure out if any of them can be prevented though screening -could also be negative (in terms of insurance or employers)C. May lead to development of pharmaceuticalsD. Insight into evolution and migration patterns
Genetic ModificationA. Deliberate manipulation of genetic materialB. Genetic code is universal -it can be transferred between organisms because the bases are the sameC. Used to create new combinations of DNAD. Mutation and recombination occur naturally (often are disadvantageous and do not remain in the population)
Genetic ModificationE. Genetic engineers direct the process of recombination -increase chances of favorable combinations
Technique for gene transfer Example-insulin productionA. Must have a vector (bacteriophage), a host (bacteria), restriction enzymes and DNA ligaseB. Plasmids (small circular DNA from bacteria) are cut using restriction enzymesC. Sticky ends are created by adding C nucleotides
Technique for gene transfer Example-insulin productionD. mRNA that codes for insulin is extracted from the pancreasE. Reverse transcriptase makes DNA from the mRNAF. Sticky ends are creating by adding G nucleotidesG. Insulin gene and plamsid are mixedH. They join together at the sticky ends (C pairs with G)
Technique for gene transfer Example-insulin productionI. Plasmid with the human insulin gene is called a recombinantJ. Host cell (usually E. coli) receives the geneK. E. coli are cultured w/ new geneL. Insulin produced by modified E. coli is extracted and eventually given to diabetics
Other examples of genetic modificationA. Insulin production in E. coli (discussed above)B. Bacteria can be modified to produce growth hormone for cows -cows injected with hormone increase milk production by 10-20%C. Breeding of plants to increase disease resistanceD. Dog breeding
Benefits of genetic modificationA. Less disease (possibly) -long term effects are unknownB. More product -Ex: milkC. Some are beneficial with no know side effects -Ex: Insulin production
Negative aspects of genetic modificationA. Introducing plasmids -hospitals fear them -can carry genes for antibiotic resistance -can be passed from one species to anotherB. Don’t know long term effects (Example) -effects of using growth hormone in cows is unknown (some could be in our food) -cows needs extra antibiotics to stay healthy (we get these too)C. Could create super bacteriaD. Less than perfect becomes unacceptable (if anyone can be genetically modified before birth)
Gene therapyA. Treatment of genetic illness by modification of genotype, or base sequence, or allele (dominant for recessive)B. Best if done with stem cellsC. Many attempts have not been successfulD. Read SCID example on p. 28
CloningA. Clone-group of individuals identical in genotype or a group of cells descended from a single parent cell
How Dolly was made1. Udder cells removed from sheep 1 a. cells grown in a culture to turn off their genes2. Embryos removed from sheep 2 a. Nuclei removed from embryos3. Embryos and udder cells fused by electricity to form zygotes4. Zygotes developed to embryos5. Embryos implanted into Sheep 3 (the surrogate mother)6. Dolly develops and becomes first born clone a. Identical to sheep 1
Ethical issues of cloning in humansA. Possible to only clone a specific organ?B. Does the whole person need to be cloned for organ donation? -if we take the heart, what do we do with the rest of the bodyC. Read p. 32 in the review guide for arguments for and against therapeutic cloning of humans.
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