Transcript Genetics 2
Final Concepts for Chapter 11 Mendelian Genetics Allele Dominant Recessive P-generation F1 generation F2 generation Law of independent assortment Law of segregation Chromosomes Pure breed Trait •Codominance •Complete dominance •Dihybrid cross •Genotype •Genotypic ratio •Heterozygous •Homozygous •Incomplete dominance •Monohybrid cross •Phenotype •Phenotypic ratio •Probability •Punnett square •Testcross •Expected/predicted results •Actual/observed results •Karyotype •Amniocentesis •Linked genes •Sex-linked disorders •Autosomal disorders Independent Assortment vs. Linked Genes Mendel did not know about chromosomes when he proposed the Law of Independent Assortment. The pea traits he studied happened to be located on different chromosomes – so they did assort independently. Independent Assortment vs. Linked Genes Question: How many traits do you have? Question: How many chromosomes (per cell) do you have? Question: Is it possible to have only one trait per chromosome? – No, lots of genes are carried or linked together on the same chromosome. Independent Assortment vs. Linked Genes Do the punnett square for the following cross – assume independent assortment. Cross two heterozygous tall, heterozygous red flowered plants T=tall R=red flower t= short r = white flower Independent Assortment vs. Linked Genes What is the phenotypic ratio? TR TR Tr tR tr TtRr x TtRr Tr tR tr TTRR TTRr TtRR TtRr TTRr TTrr TtRr Ttrr TtRR TtRr ttRR ttRr TtRr Ttrr ttRr ttrr Independent Assortment vs. Linked Genes 9:3:3:1 ratio 9 3 3 1 = = = = tall and red tall and white short and red short and white PROBABILITY: From this cross, 48 offspring were produced. 1. How many offspring would you expect to be tall and red? 2. How many would expect to be tall and white? 3. How many would you expect to be short and white? Independent Assortment vs. Linked Genes Now, do the following cross BUT the genes for tallness and red flowers are linked. Cross two heterozygous tall, heterozygous red flowered plants T=tall t= short R=red flower r = white flower Independent Assortment vs. Linked Genes Hint TtRr X TtRr T t R r Is it possible to produce a Tr gamete? Independent Assortment vs. Linked Genes TtRr X TtRr TR TR tr TTRR TtRr tr TtRr ttrr What is the phenotypic ratio? 3:1 3 = Tall and Red 1 = Short and white Independent Assortment vs. Linked Genes So… out of the 48 offspring, if the genes are linked, how many would be – 1. tall and red? – 2. tall and white? – 3. short and red? – 4. short and white? Answer: tall and red = 36 short and white = 12 EXPECTED RESULTS! tall/white = 0 short/red = 0 Independent Assortment vs. Linked Genes Is it possible for our Actual Results to show any flowers that are tall/white or short/red? Yes – how? Crossing over Crossing over occurs in meiosis Pieces of the chromosomes actual switch places. Complete vs Incomplete Dominance Codominance – the alleles are equally dominant Roan Cow Human Blood Type Sex-linked Traits Traits carried on the X chromosome Fill in the genotypes on the pedigree. Autosomal disorders Disorders carried on non-sex chromosomes (first 22 pairs) Some are autosomal dominant – Huntington’s disease Most are autosomal recessive – Sickle-cell anemia – Cystic fibrosis Question: How do you know if the pure bred dog you just paid big bucks for is actually pure? GG? Gg? Test Cross Cross using a homozygous recessive individual with a dominant individual to determine if the dominant individual is heterozygous or homozygous dominant (pure) Why use a homozygous recessive individual? Test Cross Do the punnett squares for each case: GG x gg Gg x gg Test Cross All offspring produced should show the dominant characteristics if the dominant parent is pure (GG) for the trait. Class Quiz – all or nothing! Key: T = tall and t = short 1. Cross a heterozygous tall plant with a homozygous dominant tall plant. Show punnett square ONLY. 2. Cross a heterozygous tall plant with a short plant. Show punnett square and PHENOTYPE ratio. A man and woman marry. They have five children, 2 girls and 3 boys. The mother is a carrier of hemophilia, an Xlinked disorder. She passes the gene on to two of the boys who died in childhood and one of the daughters is also a carrier. Both daughters marry men without hemophilia and have 3 children (2 boys and a girl). The carrier daughter has one son with hemophilia. One of the non-carrier daughter’s sons marries a woman who is a carrier and they have twin daughters. What is the percent chance that each daughter will also be a carrier? Hemophilia is a sex-linked trait. Draw the pedigree. Do a punnett square cross for each of the couples to determine…. What is the genotype ratio? What is the phenotype ratio? Answers to front page of quiz 1 point each Part A 1.M 2. H 3. K 4. A 5. B 6. C 7. E 8. F 9. D 10. I Answers to front page of quiz 1 point each Part B 1. d 2. b 3. a 4. h 5. m 6. e 7. c 8. k 9. i 10. f A G g Answers to the back 3pts G g GG Gg Gg gg Gg X Gg Phenotype Ratio: – 3:1 – 3= green eyes – 1 = yellow eyes B Answers to the back 3pts Aa X Aa A a What is the phenotypic ratio? A a AA Aa Aa aa 3:1 3 = Big Ears 1 = Small Ears Or for the G’s 3:1 3= green eyes 1 = yellow eyes A Answers to the back 2 points Bb X bb b b B b Bb bb Bb bb B Answers to the back 2 points Tt X tt T t t t Tt tt Tt tt A 1 pt. for the correct genotypes to set up the cross ½ pt. for each correct gamete combo =6 pts. Total bbgg x BbGg bg BG Bg bG bg bg bg bg BbGg BbGg BbGg BbGg Bbgg Bbgg Bbgg Bbgg bbGg bbGg bbGg bbGg bbgg bbgg bbgg bbgg B 1 pt. for the correct genotypes to set up the cross ½ pt. for each correct gamete combo =6 pts. Total Bbgg x BbGg BG Bg Bg bg bg Bg bG bg BBGg BBgg BbGg Bbgg BBGg BBgg BbGg Bbgg BbGg Bbgg bbGg bbgg BbGg Bbgg bbGg bbgg A 4 points ( 1 pt. each correct ) What is the phenotypic ratio? 1:1:1:1 or 4:4:4:4 Black hair / green eyes = 4 Black hair / yellow eyes = 4 brown hair / green eyes = 4 brown hair / yellow eyes = 4 B points ( 1 pt. each correct ) What is the phenotypic ratio? 3:3:1:1 or 6:6:2:2 Black hair / green eyes = 6 Black hair / yellow eyes = 6 Brown hair / green eyes = 2 Brown hair / yellow eyes = 2