Mendelian Genetics 1 Gregor Mendel The Father of Genetics 2 Gene • A discrete unit of hereditary information consisting of a specific nucleotide sequence in DNA (or RNA in some viruses) 3 Alleles • •Alleles are alternate forms of a gene. • •Examples: tall and short for plant height or purple or white for flower color. • •Every trait has at least two alleles- one from each parent. • •The location of an allele on a chromosome is known as its locus (loci = plural form). 4 Genotype • •The letters that represent (symbolize) the trait being investigated. The genetic makeup of an organism. • •Examples: Bb, BB, bb 5 Phenotype • The actual representation of the genes. The Physical appearance or traits in an organism resulting from its genetic makeup (what you see). • Examples: tall, purple flower or white flower, blond hair, freckles, etc. 6 Dominant • The allele that is fully expressed in an organism (observed). • •Represented by capital letters. • •Tall = T 7 Recessive • •The allele that is masked by the dominant allele. • •Represented by lower case letters. • •Short = t 8 Homozygous • When both alleles (letters) are the same. • BB = Homozygous Dominant • bb = Homozygous recessive 9 Heterozygous • When the alleles (letters) are different. • One upper case letter and one that is lower case. • Bb = Heterozygous 10 Example Problem • Round = R • wrinkled = r • If a plant has round seeds, do we know what its genotype is? • It could be RR or Rr • If a plant has wrinkled seeds, do we know what its genotype is? • Yes, it is rr. 11 Punnett Square • A method for finding predicted outcomes and probabilities for offspring from any cross. • A chart for predicting the traits of offspring. 12 Some more terms: • P-generation is the parental generation. • The p-generation produce the F1 generation. • The F1 generation crossed with itself produces the F2 generation. 13 Example Problem 14 • In foxes, red coat color is determined by the dominant gene R; silver-black coat is determined by the recessive gene r. A homozygous (pure) red male is crossed with a silver-black female. (The P generation). 1.What is the genotype of the female? 15 What are the genotype percentages of their offspring? •First…make a Punnett square for showing your work 16 Example Problem R R r Rr Rr r Rr Rr 17 Assignment • Section 32-3 18 Assignment 33-4 & 33-5 19 B = Brown b = blue #1 B b b Bb bb b Bb bb 1 point 20 1 point #1 • 1/2 or 50% chance of blue-eyed. • 1/2 or 50% chance of Brown-eyed. 21 T = Tall #2 1 point t = short T t t Tt tt t Tt tt 22 1 point #2 What fraction of offspring would be tall? • 1/2 or 50% would be Tall. 23 W = White #3 1 point w = black W w W WW Ww w Ww ww 24 1 point #3a What fraction of the offspring will be white? • 3/4 or 75% will be white. 25 1 point #3b What fraction of the offspring will be black? • 1/4 or 25% will be black. 26 1 point #3c What fraction of each genotype will you get? • 1/4 or 25% will be WW • 1/2 or 50% will be Ww • 1/4 or 25% will be ww. 27 1 point #3d What fraction of each phenotype will you get? • 3/4 or 75% will be White. • 1/4 or 25% will be black. 28 #4 RR = Red WW = White RW = Roan 1 point R R W RW RW W RW RW 29 1 point #4 Give the fraction of each color of offspring? • 1/1 or 100% will be Roan - RW. 30 #5 RR = Red WW = White RW = Roan 1 point R W W RW WW W RW WW 31 1 point #5 Give the fraction of each color of offspring? • 1/2 or 50% will be Roan - RW. • 1/2 or 50% will be White - WW. 32 #1a 1 point G = Green g = red G g G GG Gg g Gg gg 33 1 point #1a Give the fraction of each genotype of offspring? • 1/4 or 25% will be GG. • 1/2 or 50% will be Gg. • 1/4 or 25% will be gg. 34 #1b 1 point G = Green g = red G g g Gg gg g Gg gg 35 1 point #1b Give the fraction of each genotype of offspring? • 1/2 or 50% will be Gg. • 1/2 or 50% will be gg. 36 1 point #2a What is the fraction of each phenotype? • 3/4 or 75% will be Green. • 1/4 or 25% will be red. 37 1 point #2b What is the fraction of each phenotype. • 1/2 or 50% will be Green • 1/2 or 50% will be red. 38 b b B Bb Bb b bb bb #3 1 point •The mother had to be heterozygous or Bb since the couple had a blue eyed child. 39 1 point #4 •There would be a 50% chance that the 2nd child from the couple would have a brown eyes. 40 #5 • Spotted = S • white = s 2 points The couple had two spotted and two white kittens. • Mother = ss since she is white. • Father’s Genotype would be Ss. • Father’s Phenotype would be Spotted 41 #6 5 points • Man has blue eyes - bb. • G-Ma has blue eyes - bb. • Woman has brown eyes - Bb. • 50% of children would be Bb. (Brown) • 50% of children would be bb. (Blue) 42 Please put a score on top of their paper. • Put the number correct out of 28 43 Punnett Squares • Why are punnett squares useful? • We can use a punnett square to predict the probable genotypes and phenotypes for offspring from a genetic cross. • Genotype = What is inside the genes, the make-up. • Phenotype = The outward expression of the genes. 44