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CHAPTER 14 MENDEL AND THE GENE IDEA Gregor Mendel 1822-1884 “My scientific labors have brought me a great deal of satisfaction, and I am convinced that before long the entire world will praise the result of these labors." A Cloning Poem Mary had a little lamb, its fleece was slightly gray, It didn't have a father, just some borrowed DNA. It sort of had a mother, though the ovum was on loan, It was not so much a lambkin, as a little lamby clone. And soon it had a fellow clone, and soon it had some more, They followed her to school one day, all cramming through the door. It made the children laugh and sing, the teachers found it droll, There were too many lamby clones, for Mary to control. No other could control the sheep, since their programs didn't vary, So the scientists resolved it all, by simply cloning Mary. But now they feel quite sheepish, those scientists unwary, One problem solved, but what to do, with Mary, Mary, Mary! History of genetics: Leeuwenhoek (1632-1723) Homunculus – contained inside a sperm (spermist)!!!!!!! History of genetics: Charles Darwin Body cells shed 'gemmules' which collect in the reproductive organs prior to fertilization History of genetics: “Blending” hypothesis –Genetic material contributed by each parent mixes in a manner analogous to the way blue and yellow paints blend to make green. Mendel brought an experimental and quantitative approach to genetics (1800’s) • Pea plants have several advantages for genetics. Many phenotypes (physical traits) Can control fertilization Short lifecycle Mendel’s Pea Plant Experiments Phenotype/trait: feature of an organism that can be seen. Ex. flower color (purple/white) Vocab Chromosome 1 from Mom Chromosome 1 derived from Mom and Dad Chromosomes A gene locus A pair of alleles 3 gene pairs A and a IMPORTANT: There are at least 2 genes for every trait (called alleles) Genotype: Genetic make-up of individual for a given phenotype. Ex. AA or aa Alleles: Alternative forms of the same gene (A and a). Gene is - Flower Color. Homozygous—having 2 of the same allele A A Homozygous Dominant (AA) a a Homozygous Recessive (aa) Heterozygous—having 2 different alleles A a Heterozygous (Aa) Dominant Allele: Allele that is fully expressed in a heterozygote (A) Aa Recessive Allele: Allele that is masked in a heterozygote (a) Aa True-breeding—always producing offspring with the same traits as the parents when the parent plants are self-fertilized Which plant/s are true breeding? AA plant, Aa plant, aa plant True Breeding AA X AA will always give AA Aa X Aa - Hmmm aa X aa - will always give aa True Breeding How to cross pea plants Punnett Square: used to predict probabilities of offspring genotype and phenotype Parent Cross: Parents are true breeding: this means they are AA (pink) or aa (white) Mendel isolated true breeding plants by growing them are selffertilizing to check for the offspring that resembled parents All offspring in the F1 will be: HETEROZYGOUS Parent cross produces F1 A A a Aa Aa a Aa Aa Genotype ratio: 0AA : 1Aa : 0aa Phenotype ratio: 1 pink: 0 white Probability: 1 pink : 0 white Frequency : 100% pink : 0% white Trait = flower color Pink = dominant White = recessive Female parent Male Parent A A Both parents are heterozygous for the trait a A a a a Aa A F1 = A a A a A A Monohybrid Cross A a A a Aa aa A aa A a AA Aa a Monohybrid cross Offspring (F2): Phenotype ratio: 3 pink: 1 white Genotype ratio: 1 AA: 2 Aa: 1 aa a Aa aa a Aa aa Probability:3/4 pink: 1/4 white Test or Back Cross: Given: Mom is not Deaf (only 2 alleles) 50% pups are deaf What is the genotype of Mom ? Test Cross - Unknown parent is crossed with 1) Assume unknown is DD D D d Dd Dd d Dd Dd HOMOZYGOUS RECESSIVE Genotype ratio: 0DD : 1Dd : 0dd Phenotype ratio: 1 norm : 0deaf Probability: 1 norm : 0 deaf Frequency : 100% norm : 0% deaf Test or Back Cross: Given: Mom is not Deaf (only 2 alleles) 50% pups are deaf 2) Assume unknown is Dd D d d Dd dd d Dd dd Unknown parent is Crossed with HOMOZYGOUS RECESSIVE Now the unknown Parent (mom) is Dd Genotype ratio: 0DD : 1Dd : 1 dd Phenotype ratio: 1 norm : 1 deaf Probability: 1/2 norm : 1/2 deaf Frequency : 50% norm : 50% deaf Mendel’s Law of Segregation Alleles (A and a) separate in meiosis (gamete formation) since homologous chromosomes are separated AA X aa No segregation All AA (purple) Or aa (white) AA X aa Meiosis segregation gametes The paired condition is restored by the random fusion of gametes at fertilization. F1 A A a All Aa (purple) Aa Aa Aa Aa a AABB purpleflowered tall AB X ab F1 OUTCOME: All F1 plants purple-flowered, tall (AaBb heterozygotes) AaBb aabb whiteflowered dwarf Dihybrid Cross AaBb X AaBb (Hybrids for both traits) AaBb AaBb 1/4 AB 1/4 Ab 1/4 aB 1/4 ab 1/4 AB 1/16 AABB 1/16 AABb 1/16 AaBB 1/4 Ab 1/16 AABb 1/16 AAbb 1/16 AaBb 1/16 Aabb 1/16 AaBB 1/16 AaBb 1/16 aaBB 1/16 aaBb 1/16 AaBb 1/16 Aabb 1/16 aaBb 1/16 aabb 1/4 aB 1/4 ab 1/16 AaBb purple-flowered, tall purple-flowered, dwarf white-flowered, tall white-flowered, dwarf Phenotype ratio:9:3:3:1 Mendel’s Law of independent assortment :when there are 2 or more allelle pairs, each pair of alleles segregates into gametes independently Mendelian inheritance reflects rule of probability Rule of multiplication - chance that two or more independent events will occur together – Compute probability of each independent event. – Multiply the individual probabilities – The probability that two coins tossed at the same time will land heads up is 1/2 x 1/2 = 1/4 – The probability that a heterogyzous pea plant (Aa) will produce a white-flowered offspring (aa) is 1/2 x 1/2 = 1/4 • The rule of multiplication also applies to dihybrid crosses. – For a heterozygous parent (YyRr) the probability of producing a YR gamete is 1/2 x 1/2 = 1/4 – The probability that an F2 plant will have a YYRR genotype from a heterozygous parent is 1/4 YR x 1/4 YR = 1/16 – The probability that an F2 plant will have a YyRr genotype from heterozygous parents is (1/4 YR x 1/4 yr) + (1/4 yR x 1/4 Yr) + (1/4 yr x 1/4 YR) + (1/4 Yr x 1/4 yR) = 4/16 = 1/4 • The rule of addition - the probability of an event that can occur two or more different ways is the sum of the separate probabilities of those ways. • The probability of a heterozygote in a Monohybrid cross is 1/4 + 1/4 = 1/2 • Probability of finding two recessive phenotypes for at least two of three traits resulting from a trihybrid cross between PpYyRr and Ppyyrr. – There are five possible genotypes that fulfill this condition: ppyyRr, ppYyrr, Ppyyrr, PPyyrr, and ppyyrr. •The probability of producing a ppyyRr offspring: –The probability of producing pp = 1/2 x 1/2 = 1/4. –The probability of producing yy = 1/2 x 1 = 1/2. –The probability of producing Rr = 1/2 x 1 = 1/2. –Therefore, the probability of all three being present (ppyyRr) in one offspring is 1/4 x 1/2 x 1/2 = 1/16. •For ppYyrr: 1/4 x 1/2 x 1/2 = 1/16. •For Ppyyrr: 1/2 x 1/2 x 1/2 = 2/16 •for PPyyrr: 1/4 x 1/2 x 1/2 = 1/16 •for ppyyrr: 1/4 x 1/2 x 1/2 = 1/16 •Therefore, the chance of at least two recessive traits is 6/16. Exceptions to Mendelian Rules: 1) Incomplete dominance - heterozygotes show a distinct intermediate phenotype. Genotype ratio: 1RR : 2Rr : 1 rr Phenotype ratio: 1 Red: 2 Pink: 1 white R r R RR Rr r Rr rr Exceptions to Mendelian Rules: 2) Codominance - two alleles affect the phenotype in separate, distinguishable ways.. Genotype ratio: 1RR : 2Rr : 1 rr Phenotype ratio: 1 Red: 2 Pink: 1 white R r R RR Rr r Rr rr How does a genotype become a phenotype? • Gene = DNA segment • DNA codes for a mRNA => protein • Protein is directly or indirectly responsible for phenotype Blue pigment gene makes enzyme that makes Blue Pigment White areas - gene for Blue pigment enzyme is inactive = no Blue pigment • Dominant gene is not overpowering the recessive gene • Cystic fibrosis - chloride transport protein is missing in cells that have the recessive gene • Dominant gene is not always more common than recessive • Polydactyly, Huntington’s Disease, Achondroplasia • Blood group Inheritance: Plasma - clear liquid portion of blood (has Antibodies) RBC (Red Blood Cell) carries Antigens on its surface • Blood group Inheritance: Antigen A Antigen B Antigen A and B • Blood group Inheritance: • Blood group Inheritance: Blood group AB = Universal Recipient; O – Universal donor During blood transfusion only RBCs are transferred from donor to recepient Agglutination with O group because the NO plasma has A agglutination and B antibody with AB group – so no other because the groups can plasma has no donate to ‘O’ antibody • Blood group Inheritance: Gene Antigen Blood Group IA IA or IA i A A Allele Relation IA > i IB IB or IB i B B IB > i (Dom - rec) (Dom - rec) IA IB A and B AB ii None O IA = IB (Codominance) Fig. 14.10 Exception to Mendelian Rules 3) Pleiotropic genes- affecting more than one phenotypic character. – Sickle-cell disease - one gene mutation, many symptoms . •Anemia •Jaundice •Blindness •Infections •Delayed growth Exception to Mendelian Rules 4) Epistasis - one gene alters another genes phenotypic expression Cc or Cc = coat color cc = no coat color BB, Bb = Black bb = brown What phenotype will ccBB bb? Not 9:3:3:1 in a Dihybrid Cross but 9 black: 3 brown: 4 white Fig. 14.11 Exception to Mendelian Rules 6)) Quantitative characters - Trait depends on HOW MANY genes are expressed Skin color 3 allelles AABBCC- dark AaBBCC NORMAL DISTRIBUTION AABbCC AABBCc AaBbCc aaBBCC AAbbCC AABBcc What kind of Dominance? aaabbcc - light Fig. 14.12 7) Phenotype depends on environment and genes (nature vs nurture). Pedigree Analysis • Map out family traits and predict genotypes • Parents are WwFf, what are the chances of a wwff child? • 1/4 ww x 1/4 ff = 1/16 Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings Predict the probability that a child with WwFf parents will have a widow’s peak and attached earlobes. • The chance of having a widow’s peak is 3/4 (1/2 [Ww] + 1/4 [WW]). • The chance of having attached earlobes is 1/4 [ff]. • This combination has a probability of 3/4 + 1/4 = 3/16. • Cystic Fibrosis is a recessive disorder. If a carrier marries a person with Cystic Fibrosis , what is the probability that their child will be of Normal phenotype? Ff X ff 50% chance of Normal phenotype Ff ? Ff Ff ? Ff ? ff ? • Amniocentesis- beginning at the 14th to 16th week of pregnancy (assess the presence of a gnetic disease). – Fetal cells extracted from amniotic fluid are cultured and karyotyped • Chorionic villus sampling (CVS) can faster karyotyping, 8th - 10th week of pregnancy. – This technique extracts a sample of fetal tissue from the chrionic villi of the placenta. Fig. 14.17b • Other techniques, ultrasound and fetoscopy, allow fetal health to be assessed visually in utero. • Both fetoscopy and amniocentesis cause complications in about 1% of cases. – These include maternal bleeding or fetal death. • Some genetic tests can be detected at birth by simple tests that are now routinely performed in hospitals. • One test can detect the presence of a recessively inherited disorder, phenyketonuria (PKU). – This disorder occurs in one in 10,000 to 15,000 births. – Individuals with this disorder accumulate the amino acid phenylalanine and its derivative phenypyruvate in the blood to toxic levels. – This leads to mental retardation. – If the disorder is detected, a special diet low in phenyalalanine usually promotes normal development.