Mendel s Laws Results of Monohybid Cosses Inheitable factos o genes ae esponsible fo all heitable chaacteistics Phenotype is based on Genotype Each tait is based on two genes, one fom the mothe and the othe fom the fathe Tue-beeding individuals ae homozygous ( both alleles) ae the same Law of Dominance In a coss of paents that ae pue fo contasting taits, only one fom of the tait will appea in the next geneation. All the offsping will be heteozygous and expess only the dominant tait. RR x yields all R (ound seeds)
Law of Segegation Duing the fomation of gametes (eggs o spem), the two alleles esponsible fo a tait sepaate fom each othe. Alleles fo a tait ae then "ecombined" at fetilization, poducing the genotype fo the taits of the offsping. Law of Independent Assotment Alleles fo diffeent taits ae distibuted to sex cells (& offsping) independently of one anothe. This law can be illustated using dihybid cosses.
Dihybid Coss A beeding expeiment that tacks the inheitance of two taits. Mendel s Law of Independent Assotment a. Each pai of alleles segegates independently duing gamete fomation n b. Fomula: 2 (n = # of heteozygotes) Question: How many gametes will be poduced fo the following allele aangements? n Remembe: 2 (n = # of heteozygotes) 1. RYy 2. AaBbCCDd 3. MmNnOoPPQQRssTtQq Answe: 2 n 1. RYy: 2 = 2 = 4 gametes RY Ry Y y n 3 2. AaBbCCDd: 2 = 2 = 8 gametes ABCD ABCd AbCD AbCd abcd abcd abcd abcd n 6 3. MmNnOoPPQQRssTtQq: 2 = 2 = 64 gametes
Dihybid Coss Taits: Seed shape & Seed colo Alleles: R ound winkled Y yellow y geen RY Ry Y y RYy x RYy RY Ry Y y All possible gamete combinations RY Ry Y y RY R Y y
Dihybid Coss RY Ry Y y RY RRY RRYy RYY RYy Ry RRY RRy RYy Ryy Y RY RYy YY Yy y RYy Ryy Yy yy copyight cmassengale Round/Yellow: 9 Round/geen: 3 winkled/yellow: 3 winkled/geen: 1 9:3:3:1 phenotypic atio
Test Coss A mating between an individual of unknown genotype and a homozygous ecessive individual. Example: bbc x bbcc BB = bown eyes Bb = bown eyes bb = blue eyes CC = culy hai Cc = culy hai b bc b Possible Results: bc b C b bbcc bbcc bc b c o b bbcc bbcc
Summay of Mendel s Laws LAW DOMINANCE PARENT CROSS TT x tt tall x shot OFFSPRING 100% Tt tall SEGREGATION Tt x Tt tall x tall 75% tall 25% shot INDEPENDENT ASSORTMENT RGg x RGg ound & geen x ound & geen 9/16 ound seeds & geen pods 3/16 ound seeds & yellow pods 3/16 winkled seeds & geen pods 1/16 winkled seeds & yellow pods copyight cmassengale 55
Incomplete Dominance (no tait completely shows, hence incomplete) F 1 hybids have an appeaance somewhat in between the phenotypes of the two paental vaieties. Example: snapdagons (flowe) ed (RR) x white () RR = ed flowe = white flowe R All R = pink (heteozygous R R R R poduces the F 1 geneation R R R copyight cmassengale
Incomplete Dominance R R R Put atios in the following ode: RR:R: o Red:Pink:White Genotype Ratio: : : Genotype Ratio: : : Phenotype Ratio: : : Phenotype Ratio: : :
Codominance (Both Taits Show Simultaneously, Taits Shae Dominance) Two alleles ae expessed (multiple alleles) in heteozygous individuals. Example: blood type expesses codominance o complete dominance depending on the blood types 1. type A = I A I A o I A i 2. type B = I B I B o I B i 3. type AB = I A B I 4. type O = ii I B I B Example: homozygous male Type B (I B I B ) x heteozygous female Type A (I A i) I A i 1/2 = I A I B I A I B I B i 1/2 = I B i I A I B I B i copyight cmassengale 61
Moe Codominance Poblems Example : male Type O (ii) x female type AB (I A I B ) I A I B i i I A i I A i I B i I B i 1/2 = I A i 1/2 = I B i Question: If a boy has a blood type O and his siste has blood type AB, what ae the genotypes and phenotypes of thei paents? boy - type O (ii) X gil - type AB (I A I B )
Codominance Answe : I A i I B I A I B Paents: genotypes = I A i and I B i phenotypes = A and B i ii Sex-linked Taits Taits (genes) located on the sex chomosomes Sex chomosomes ae X and Y XX genotype fo females XY genotype fo males Many sex-linked taits caied on X chomosome The Y chomosome is much smalle than the X chomosome. Thus, the X chomosomes caies moe genes than the Y chomosome
Sex-linked Taits Example: Eye colo in fuit flies fuit fly eye colo XX chomosome - female Xy chomosome male Sex-linked Tait Poblem Example: Eye colo in fuit flies (ed-eyed male) x (white-eyed female) X R Y x X X Remembe: the Y chomosome in males does not cay taits. RR = ed eyed R = ed eyed = white eyed XY = male XX = female X R Y X X
Sex-linked Tait Solution: X X X R X R X X R X 50% ed eyed female 50% white eyed male Y X Y X Y Female Caies = Female = Male Half Puple = Caie Full Puple = Hemophiliac
Beed the P geneation 1 Genetic Pactice Poblems tall (TT) x dwaf (tt) pea plants Beed the F geneation 1 tall (Tt) vs. tall (Tt) pea plants
P 1 Geneation Solution: tall (TT) vs. dwaf (tt) pea plants t t T Tt Tt poduces the F 1 geneation T Tt Tt All Tt = tall (heteozygous tall) F 1 Geneation Solution: tall (Tt) x tall (Tt) pea plants T t poduces the F 2 geneation T T Tt 1/4 (25%) = TT 1/2 (50%) = Tt 1/4 (25%) = tt t Tt t 1:2:1 genotype 3:1 phenotype