Mendelian Genetics

  • 3,407 views
Uploaded on

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
3,407
On Slideshare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
35
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. Mendelian Genetics ! •! Gregor Mendel (1822-1884),! •!Augustinian monk,! »! Botanist,! •! Pisum sativa,! •!Garden pea,! »! 1st “Model Organism”.! Meiosis without recombination. Independent assortment of homologs to different Poles creates gamete diversity 2n (n = # of diff chromosome pairs)! A Aa a The two possible alignments a aA A of nonhomologous chromosomes at metaphase I ! B Bb b B Bb b Anaphase I! A A a a a a A A Metaphase II! B B b b B B b b Anaphase II! A A a a a a A A B B b b B B b b
  • 2. Dominant vs. Recessive Traits ! x! P! F1! The trait that appears in the F1 generation is the DOMINANT trait.! The trait that disappears in the F1 generation is termed RECESSIVE.! Nomenclature ! •! Dominant unit factors are designated with a capital letter, often (but not always) with the first letter of the description,! –! Y = yellow, ! –! V = violet, ! –! T = tall, ! •! Recessive unit factors are represented by small letters,! –! y = green, ! –! v = white, ! –! t = dwarf, !
  • 3. •! alternate forms,!
  • 4. Many plants are hermaphrodites. They carry both male ! and female sex organs, and can self-fertilize!
  • 5. Monohybrid Cross ! •! Mating between individuals that differ in only one trait, ! –! yellow pea x green pea,! –! violet flower x white flower! –! tall x dwarf! –! round seed x wrinkled seed! –! full pod x constricted pod! –! etc.!
  • 6. How did Mendel explain the 1:3 and 1:2:1 ratios? ! Homozygous Dominant B/B Heterozygous B/b Homozygous recessive b/b
  • 7. Can also use two probability rules to predict Mendel’s results." ! (probability: the # of times expect an event / # times event could happen.) •! The Product rule (a.k.a. the AND rule): The probability of two independent outcomes occurring simultaneously is the product of their individual probabilities.! •! The Sum rule (a.k.a. the OR rule): The probability of two mutually exclusive outcomes occurring is the sum of their individual probabilities.! The probability of two heterozygous parents having a homozygous recessive offspring is the product of the probabilities of inheriting a recessive allele from the female AND a recessive allele from the male.! (1/2)(1/2) = 1/4! The probability of two heterozygous parents having a heterozygous offspring is the product of the probabilities of inheriting a dominant allele from the female AND a recessive allele from the male, + the product of the probabilities of inheriting a recessive allele from the female AND a dominant allele from the male.! (1/2)(1/2) + (1/2)(1/2) = 1/2!
  • 8. Meiosis without recombination. Independent assortment of homologs to different Poles creates gamete diversity 2n (n = # of diff chromosome pairs)! A Aa a The two possible alignments a aA A of nonhomologous chromosomes at metaphase I ! B Bb b B Bb b Anaphase I! A A a a a a A A Metaphase II! B B b b B B b b Anaphase II! A A a a a a A A B B b b B B b b
  • 9. But we can use the product rule to save time and space!! The probability of getting round, yellow peas is the probability of getting round peas AND the probability of getting yellow peas.! (3/4)(3/4) = 9/16! The probability of getting round, green peas is the probability of getting round peas AND the probability of getting green peas.! (3/4)(1/4) = 3/16! The probability of getting wrinkled, yellow peas is the probability of getting wrinkled peas AND the probability of getting yellow peas.! (1/4)(3/4) = 3/16! The probability of getting wrinkled, green peas is the probability of getting wrinkled peas AND the probability of The testcross Suppose the genotype of Y-R- is YyRr. This can be determined by crossing to a truebreeding stock that is homozygous for recessive alleles of both genes. gametes yr YyRr YR yellow round Yyrr Yr yellow wrinkled YyRr yyRr yR green round yyrr yr green wrinkled The four possible gamete types are found to be present in a 1: 1: 1: 1 ratio. In a testcross the genotype of the progeny = phenotype
  • 10. Extensions to simple mendelian analysis! Penetrance and expressivity! ---Penetrance: The percentage of individuals with a given genotype! ! expressing the expected phenotype. This depends on the whole genotype! of the individual causing specific allelic interactions, and on the! environment.! ---Expressivity: Determines the extent to which a given genotype is! ! expressed phenotypically in an individual.! 1. Incomplete dominance! 2. Codominance! 3. Multiple alleles! 4. Recessive epistasis! 5. Dominant epistasis! 6. Recessive lethality! 7. Gene interactions in different pathways! 8. Complementary gene action! 9. Duplicate genes! 10. Pleiotropy! Penetrance! …the frequency at which ! individuals with a given ! genotype manifest a specific ! phenotype.! •! 4 of 6 dogs, or 66% of the population shows the phenotype, at some level,! •! penetrance is usually referred to as a percentage.! all the same genotype!
  • 11. Expressivity! …the degree or range ! within which a phenotype! of a specific genotype! is expressed! range of phenotypes! all the same genotype! Phenotypic ratios reflect genotypic ratios!
  • 12. Dominance relationships are determined by the environment and the level of analysis. At the molecular level sickle cell and normal hemoglobin are codominant. ! Pleiotropic mutations have effects on several different characters. In this case the dominance relationships depend on the phenotype examined.!
  • 13. The genotypes puppies! coat Epistasis and behind color in labrador retrievers In epistasis the genotype at one locus masks the ability to determine the genotype at a second locus. Epistasis relationships are often found in biochemical pathways.! Recessive epistasis! Disruption of an early step in the pathway blocks the ability to determine the genotype with respect to genes involved in later steps! E! B! X! Y! Z! = 9: 3: 4! colorless! Brown! Black! substrate! pigment! pigment! ee epistatic to bb
  • 14. The functions of E and B •! E encodes melanocortin receptor 1 -- promotes production of eumelanin leaving dogs black or chocolate brown; null allele (e) means mc1r is not functional and dogs only produce phaeomelanin (yellow) •! B regulates distribution of melanin in hair shaft. B promotes dense packing of melanin (black) and b promotes less dense packing (chocolate brown). Dominant epistasis produces ratios of 12:3:1 or 13:3! 12:3:1---A dominant allele in one gene, B, masks the contribution of a second gene, A, (blocks color ! formation) regardless of the A allele. Thus in bb individuals the A- and aa genotypes can express . themselves, resulting in a total of three different phenotypes. Dominant epistasis: one possible mechanism! = 12 (white): 3(yellow): 1(green)! B! X! A! white! green! yellow! B acts to inhibit the processing of a white precursor! into a green product.! In the absence of the B gene product the A product processes! the green precursor into a yellow product.!
  • 15. Dominant epistasis continued! 13:3---A dominant allele for B blocks a process (red color formation) that also requires at least one Dominant allele of a second gene A. Thus only individuals that are A-bb show a red color. Dominant epistasis: a second possible mechanism! B! = 13: 3! X! A! white1! ! white2! ! red! B acts to inhibit the processing of the white1!precursor! ! into the white2!product.! ! In the absence of the B gene the A product processes! the white2!precursor into a colored product.! ! Manx cat genetics X Average litter: 2 tailless kittens : 1 tailed kitten
  • 16. Manx cat genetics X Average litter: 2 tailless kittens : 2 tailed kittens Manx cat genetics X Average litter: 2 tailless kittens : 2 tailed kittens
  • 17. •! Is the tailless allele recessive or dominant? –! DOMINANT (dominant at the phenotypic level, incomplete dominance at the molecular level) •! Is the tailless cat homozygous or heterozygous? –! HETEROZYGOUS How do we explain the ratio of phenotypes in the F1 generation? X Average litter: 2 tailless kittens to 1 tailed kitten
  • 18. Recessive lethality (associated with a dominant allele)! Manx is homozygous lethal Tailless Male tailless Manx X tailless Manx M m Mm M MM Tailless Tailless Female Mm mm m Tailless Tailed 2 tailless : 1 tailed We can use a Punnett Square to show that a self cross of dihybrids producing gametes in a 1: 1: 1: 1 will generate a 9: 3: 3: 1 phenotypic ratio. ! female gamete! RY Ry rY ry! RR YY ! RR yY ! rR YY ! rR yY ! RY! round, yellow round, yellow round, yellow round, yellow ! ! ! ! RR Yy ! RR yy ! rR Yy ! rR yy! Ry! round, yellow! round, green! round, yellow! round, green! male! gamete! Rr YY ! Rr yY ! rr YY ! rr yY! rY! round, yellow round, yellow ! ! wrinkled, wrinkled, yellow ! yellow ! Rr Yy ! Rr yy! rr Yy! rr yy! ry! round, yellow! round, green! wrinkled, wrinkled, yellow ! green !
  • 19. White and purple flowers of sweet pea
  • 20. Molecular mechanism of complementation Complementation testing 1 How many genes are required to generate a particular phenotype such as flower color ?! 1. Carry out a genetic screen for mutations that lack flower color. 2. Carry out complementation testing to determine which mutations effect the same gene. A + B+ A + B+ A + B+ A + B+ A + B+ Mutagenized gametes! A+ B- A+ B- A + B+ A + B+ A- B+ A + B+ mutagenized wildtype! Parental generation! gametes! X! Diploid A+ B- A + B+ A- B+ A + B+ plants A + B+ A + B+
  • 21. Complementation testing 2 F1! Diploid A- B+ A + B+ A+ B- plants A + B+ A + B+ A + B+ A! ! B+! A-! B+! A-! B! ! + ! ! ! ! + A! ! B+! + ! A! ! B+! A! ! B+! A! !B! ! + ! + ! + - Self cross genotypes! A! ! B+! A! ! B! ! A-! B! ! + ! + + ! + A! ! B+! + ! + ! + ! + - A! ! B+! A! ! B-! A! !B! ! 1: 2: 1! 1: 2: 1! F2! White X white = True breeding stock! A- B + A+ B- The complementation X! A - B+ A + B- test! A+ B- = all blue progeny = complementation! A - B+ (mutations are in different genes = complementation groups)! Complementation testing 3! !! B! ! X! A! ! B!! =!a!l!!b!u! !p!! g! ! y!=!c! ! ! l!e! ! ! !! !i!! ! ! A- + + - l l! e ro! en! om p ! me nt a t on A!! - B! ! + A! ! + B!! - w!!! !! ! ! ! i ld t y p e d! u! !e! he!! !o! ! ! o!! ! o! bl! ! ter! zyg! te A! B! A! ! + B! ! X! + Y! b!! ! ! lue p!! ! ! ! !! i gm e n t A!! B! ! X! X!! A B! - + Y! c! lor!es! o!! !l! ! s! c! l! r!es! o!o!l! ! s! b!! ! ! lu e p!o! u!t! r! d!c! p!o!u!t! r! d!c! p!! ! ! ! !! i g m e nt Y! b!u! ! l! e Y! A! ! X! p!! ! ! ! !! A! ! B!! X! + B! ! + A! B! igm en t + - A! X!! B ! !! A1- !! B+ ! !! A2- !! B+ X! =! a!l!w! !t! ! pr! ! e! ! ! =! no! ! ! ! ! l! ! ! ! !! t!! ! ! l! hi!e !og! ny ! ncomp!ementa!ion A! !! B! ! + A! !! 2- B! ! + 1- A! ! ! X! B A1 -! !! B+!! X Y ! c! l! r!es! o!o!l! ! s! p!o!u!t! r! d!c! ! ! !! ! ! ! !! no pigment ! !! A 2- !! B+ X ! X! Y ! A! B B !
  • 22. How many different genes are required for a particular trait? Genetic screens and complementation testing. Matings between different white-flowered lines Phenotype produce purple progeny Genotype Mutations at different genes Affected genes encode Molecular mechanism different proteins that are required for the same biochemical pathway Complementation can occur during cell fusion.
  • 23. Look at human pedigree genetics in book! Dominant trait ! Male! Female! Diseased! Healthy!
  • 24. Molecular mapping: codominant markers
  • 25. Dominant phenotype associated with A A highly polymorphic recessive trait Two unaffected individuals give rise To an affected offspring = recessive Het.! trait nontaster must be recessive Het.!
  • 26. A dominant trait with incomplete penetrance, or a recessive trait associated with a common population polymorphism? How could you test?!