Practical Biology Revision Notes PDF

Practical revision Terminology:- Phenotype: - Any measurable characteristic or distinctive trait possessed by an organism. Genotype: - All of genes possessed by an individual constitute its genotype. Pure line: - A group of individuals with similar genetic background (breeding). It...

Practical revision Terminology:- Phenotype: - Any measurable characteristic or distinctive trait possessed by an organism. Genotype: - All of genes possessed by an individual constitute its genotype. Pure line: - A group of individuals with similar genetic background (breeding). It's produced by self-fertilization or mating closely related individuals for many generations (inbreeding) usually produces a population which is homozygous at nearly all loci. Homozygous: - Produced by union of gametes carrying identical alleles. A homozygote produces only one kind of gamete. Heterozygous: - Produced by union of gametes carrying different alleles. Recessive factor: - Whenever one of a pair of alleles can come to phenotypic expression only in a homozygous genotype. Dominant factor: - The allele which can phenotypically express itself in the heterozygote as well as in the homozygote. 1  Mendel's first law "Law of Segregation" Members of a pair of factors segregate from each other during the formation of sex cells in an individual. 1. Complete Dominance A condition in which the allele can phenotypically express itself in the heterozygote as well as in the homozygote. Upper letters are commonly used to designate the dominant and lower case for recessive alleles Example: red dominant to white One allele completely mask the effect of another one Parent: red X white RR rr F1: Rr Red F2: RR Rr Rr rr Red Red Red White Number of gamete =2 Number of phenotype=2 Number of genotypes= 3 Phenotypic ratio = ¾ red : ¼ white 2/4 Genotypic ratio = AA : Aa : ¼ aa Testcross As both homozygous dominant and heterozygous individuals has the same phenotype, a testcross may be used to distinguish between them. In the testcross, the unknown individual is mated to homozygous recessive ones (testcrosses cross = Rr X rr >>>>>>>>> 1/2 Rr red : 1/2 rr white) 2 2. Incomplete Dominance When parents with contrasting traits are crossed together, their offspring are intermediate in phenotype. One allele not completely mask the other one and heterozygous phenotype is intermediate character of both homozygous. Example: plumage colour in chicken Parent Black X white BB bb F1 Bb Grey F2 BB Bb Bb bb Black Gery White Number of gamete =2 Number of phenotype=3 Number of genotypes= 3 Phenotypic ratio = 14 : 2 : 1/4 Genotypic ratio = 1 : 2/4 : 1/4 Test cross = no 3. Codominance Both alleles express their phenotype in heterozygous Example cattle coat color Parent Red X White CRC R C WC W F1 CRC W Roan (mixture of red and white) F2 CRC R CRC W CRC W CWC W Red Roan (mixture of red and white) White Number of gamete =2 Number of phenotype=3 Number of genotypes= 3 Phenotypic ratio = 1 : 24 : 1 Genotypic ratio = 1 : 24 : 14 Test cross = no 3 4. Lethal Alleles Genes, which cause the death of homozygous individual Recessive lethal: Homozygous is lethal ( die). Has 2 types: A. Recessive lethal with dominant phenotype: Example : yellow coat color in mice Parents : Yy X Yy Yellow yellow F1: YY Yy Yy yy Die yellow yellow agouti 2 1 Number of gamete =2 Number of phenotype=2 Number of genotypes=2 phenotypic ratio = 2/3: 1/ 3 Genotypic ratio = 2/3 : 1/3 Test cross = no B. Recessive lethal with recessive phenotype: Example: sicklet cell anaemia Parents: Ss X Ss Carrier Carrier F1: SS Ss Ss ss Normal carrier die 1/3 2/3 4  5. Multiple Alleles Gene has more than allele only 2 express diploid number in individual but multiple alleles expressed in population. Number of genotypes= n (n+1) / 2 Number of heterozygous genotype= n (n-1) / 2 Number of homozygous genotype=n n= number of alleles Example Multiple allelic series ((A >B > C > D> E > F)) calculate the following ? n= 6 Number of genotypes= n (n+1) / 2 = 6 (6+1) / 2 = 21 Number of heterozygous genotype= n (n-1) / 2 = 6 (6-1) / 2 =15 Number of homozygous genotype= number of alleles = 6 Numbers and Ratios lethal Co-dominant In Complete Complete case dominant dominant 2A,a Number of gametes 2 3 3 2 Number of phenotypes 2 3 Number of AA , Aa , aa genotypes 3 4 Number of AA, Aa, Aa, aa individuals 1/3 : 2/3 or 1/4 : 2/4 : 1/4 Genotypic 2/3:1/3 AA : Aa : aa ratio 1/3 : 2/3 or 1/4 : 2/4 : 1/4 3/4 : 1/4 Phenotypic 2/3:1/3 ratio No 1/2 : 1/2 Test cross 5  Chemicals and equipments For nutrition of cells Fetal calf serum Protect from infection Penicillin and streptomycine Activate cell division Phytohaemagglutinin M Denaturation temperature 94-98 ˚C Annealing temperature 50-65 ˚C Extension temperature 72 ˚C. destruct protein Proteinase K solution break the cell wall of cell Lysis buffer washing the disc in spin column Wash buffer dissolve the DNA and cross from spin column Elution Buffer participate DNA Absolute alcohol To incubate the samples at specific Heat block temperature to facilitate lysis process For centrifuge the samples at high speed to Eppendorf centrifuge extract DNA For mixing the sample with lysis solutions Vortex For measuring different solution used in Pipettes with different range reaction It provides ions to ensure electrical TAE Buffer conductivity in gel electrophoresis DNA of known molecular weight used to DNA Marker (ladder) comparing DNA bands in samples with its molecular weight It is dissolved in buffer and heated, then cools Agarose powder formed a gelatinous solid molecules Fluorescent chemical that enter between Ethidium bromide base pairs in a double stranded DNA molecule used to detect DNA following gel electrophoresis. 6  PCR reaction volume Volume 50 Volume 20 Volume 40 Master mix 25 𝞵 10 𝞵 20 𝞵 DNA 4𝞵 2𝞵 4𝞵 Forward primer 1.5 𝞵 1𝞵 1.5 𝞵 Reverse primer 1.5 𝞵 1𝞵 1.5 𝞵 Distilled water 18 𝞵 6𝞵 13 𝞵 PCR reaction tempratures DNA amplification by Taq Polymaerase enzyme Step Temperature Time Cycle 1.Initial denaturation 95 c 10 min 1 2. Cycling steps Denaturation 95 c 30 sec 35 Annealing 55 c 30 sec cycle Extension 72 c 30 sec 3. Final 72 c 10 min 1 extension 4. Final hold 4c Indefinite time The equation calculate the melting temperature Tm = {40(G+C) + 20(A+T)} C 7  Sex determination XX- XO XX-XY ZZ-ZW Haplodiploid system system system System Males (XO). Male is the Male is Male (n), produced The letter O heterogametic homogametic from unfertilized haploid = absence of sex (XY) (ZZ). eggs and contain a single a sex set of chromosome. 16 chromosomes. Female: Female is the Female is Female (2n) are produced (XX), homogametic heterogametic from fertilized eggs and sex (XX). (ZW) therefore are diploid. Bees, Wasps Insect Humans, Birds Ants Drosophila Snakes Many other Butterflies organisms Amphibians Some fishes Sex Determination in Drosophila Presence of the Y chromosome does not determine maleness in Drosophila Sex is determined by a balance between genes on the autosomes and genes on the X chromosome which determined by the number of X chromosomes divided by the number of haploid sets of autosomal chromosomes (X: A ratio). 1. X: A ratio of 1.0 produces a female. 2. X: A ratio of 0.5 produces a male. 8  Gene expression problem The diagram below represents the main features of a gene. 1. What would be the length of the primary DNA 100 X 9 = 900 base 2. What would be the length of the primary RNA transcript (before any processing) for the following gene? DNA – promotor = 900 – 100 = 800 base 3. What would be the length of the mRNA in the cytoplasm (assume a poly A tail of 200 nucleotides). Remove promoter Remove introns Add poly A tail (5 x 100) + 200 = 700 base 4. How many amino acids would be in the encoded protein? (Sum of Exons ÷ 3) – 1 = Sum of Exons = 3 x 100 = 300 (300 ÷ 3 ) -1 = 99 amino acid 5. This gene belong to prokaryotes or eukaryotes and why? Eukaryote Because it contain introns in its structure 9