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MENDELIAN GENETICS Laws of Heredity. A.Origins of Genetics Passing characteristics from parent to offspring is called heredity Accurate study of heredity.

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Presentation on theme: "MENDELIAN GENETICS Laws of Heredity. A.Origins of Genetics Passing characteristics from parent to offspring is called heredity Accurate study of heredity."— Presentation transcript:


2 A.Origins of Genetics Passing characteristics from parent to offspring is called heredity Accurate study of heredity began with Austrian monk Gregor Mendel at his monastery gardens

3 Mendel used different varieties of garden pea plant – Could predict patterns of heredity which form modern-day genetics principles – Garden peas have eight observable characteristics with two distinct traits that Mendel counted and analyzed with each cross or breeding

4 Pea Characteristics & Traits

5 In nature, pea plants self pollinate since both reproductive organs (male stamen [pollen] & female pistil) are internal – Mendel physically removed stamens & dusted pistils with pollen from chosen plants to observe results by cross- pollination

6 Parent plants are P generation All offspring are F generation (from Latin filialis for son/daughter) – F 1 generation = first offspring (children) – F 2 generation = second offspring (grandchildren) F1F1 F2F2 F1F1 F2F2 F2F2 F1F1 PP F2F2

7 Cross-pollinated two pure-bred plants with one very different trait (purple vs. white flowers) in P generation – Examined each F 1 plants trait & counted them Allowed F1 generation to self-pollinate to produce F2 generation – Examined each F 2 plants trait & counted them P F1F1F1F1 F1F1F1F1 F2F2F2F2 F2F2F2F2 Monohybrid Cross

8 Mendel collected tons of data – his results are reproducible – Monohybrid cross of white flowers and purple flowers in P generation produced 100% purple flowers in F 1 generation – Self-pollination of F 1 produced 705 purple flowers & 224 white flowers in F 2 generation F1F1F1F1 P F2F2F2F2

9 Mendels ratios still hold true today – Crossing pure bred traits in monohybrid cross will ALWAYS express only one trait in F 1 – Self-pollinating F 1 will ALWAYS result in 3:1 ratio 705:224 = 3:1

10 B.Heredity Theories & Laws Mendel knew all ideas about blending characteristics was bogus – Developed four hypotheses: 1. individual has two copies of gene, one from each parent From DadFrom Mom

11 2. there exists alternative versions of genes called alleles represented by letters From DadFrom Mom AbCDeFAbCDeF ABCdeFABCdeF

12 3. if two different alleles occur together, one may be expressed while other is not – dominant and recessive – UPPERCASE alleles are dominant alleles » Trait gets expressed ALWAYS – lowercase alleles are recessive alleles » Trait only gets expressed if dominant is not present B trait will be expressed e trait will be expressed From DadFrom Mom AbCDeFAbCDeF ABCdeFABCdeF

13 –W–When both alleles are identical, individual is considered homozygous for that trait Homozygous dominant = both dominant (UPPERCASE) Homozygous recessive = both recessive (lowercase) –W–When alleles are different, individual is considered heterozygous for that trait AbCDeFAbCDeF ABCdeFABCdeF

14 TT = ? Homozygous dominant Tt = ? Heterozygous tt = ? Homozygous recessive XX = ? Homozygous dominant rr = ? Homozygous recessive QUIZ YOURSELF

15 4. when gametes (sperm/eggs or spores) are formed, alleles for each trait separate independently during meiosis – Occurs during anaphase A a HH p P b b

16 C.General Rules for Genes Each gene is given allele letter – Letter is always first letter of dominant trait Ex: yellow peas are dominant over green peas Y = yellow, y = green Ex: purple flowers are dominant over white flowers P = purple, p = white YY yy Yy PP Pp pp

17 Ex: In roses, pink petals are dominant over white petals, and tall stems are dominant over short stems. P = pink, p = white T = tall, t = short Cross a male heterozygous pink tall with a female homozygous white short Male = PpTt Female = pptt

18 Some traits are dominant – only one dominant allele needed in genome to show phenotype Some traits are recessive – both recessive alleles needed to express phenotype Dominant TraitRecessive Trait Polydactyl (P)Non-polydactyl (p) No Hitchhikers Thumb (T)Hitchhikers thumb (t) Tongue rolling (R)No tongue rolling (r) Free-hanging ear lobe (F)Attached ear lobe (f) Widows Peak (W)No widows peak (w) Brown eyes (B)Blue or green eyes (b) Left thumb on top (L)Right thumb on top (l) Plus four others!

19 Polydactyl (PP or Pp)

20 No Hitchhikers Thumb (T) TT or Tt tt

21 Tongue Rolling (R) RR or Rrrr

22 Free-Hanging Earlobes (F) FF or Ffff

23 Widows Peak (W) WW or Wwww

24 Brown Eyes (B) BB or Bb bb

25 Left Thumb on Top (L) LL or Llll

26 Mid-Digit Hair (H) HH or Hh hh

27 Cleft Chin (C) CC or Cccc

28 Dimples (D) DD or Dd dd

29 Freckles (F) FF or Ff ff

30 D.Laws of Heredity During meiosis (forming haploid gametes from diploid cells), chromatids separate during anaphase II – Law of segregation: two alleles for character separate when gametes are formed Male Parent (Tt) Female Parent (Tt) Alleles segregate (separate) into gametes T t t T

31 Alleles pair up in all combos Alleles segregate (separate) into gametes Alleles pair up in all combos

32 Mendel studied whether different characteristics were inherited together or separately – Conducted dihybrid crosses where two traits are studied – Concluded that traits NOT inherited together & developed law Law of independent assortment: alleles of different genes separate independently during gamete formation in meiosis

33 Law of Independent Assortment Male Parent (TtBb) Female Parent (TtBb) Traits separate independently TBTbtBtb TB Tb tB tb TTBB TTBbTtBb 16 total!

34 E.Punnett Square Easiest way to represent Laws of Segregation and Independent Assortment is through Punnett Square –C–Cross a homozygous dominant yellow pea with a green (homozygous recessive) –G–Genotypes: YY & yy (1 trait, 4 alleles = 4 combos) Y Y y y Yy Yy Step 1: separate alleles from genotypes & place on top & down side Step 2: determine possible combinations by crossing alleles

35 Have to analyze findings from crossings – Genotypic ratio: – Phenotypic ratio: Y Y y y Yy 4 Yy 4 yellow peas

36 Curi Family Eye Color My dad has green eyes My mom has brown eyes – Knowing that I have brown eyes, what is my GENOTYPE? Brown is dominant (B) Green is recessive (b) – Dad must be bb b B B Bb b

37 According to Punnett Square, all my parents children should have BROWN eyes – In reality, my brother has green eyes. What does this mean? Moms genotype must be Bb b BBb bb This means that there is a 50% (2/4 or ½) chance that each child my parents had could have green eyes. I lucked out. b bbb

38 Monohybrid Cross Examples 1. Aliens with two eyes are dominant over aliens with one eye. Cross a heterozygous two- eyed male with a homozygous one-eyed female. – Genotypic ratio: – Phenotypic ratio: T t t t TtTt tttt 2:2(2 Tt: 2 tt) 2:2(2 two eyes: 2 one eye)

39 2. Orange carrots are dominant over purple carrots. Cross a male purple carrot with a heterozygous orange carrot. – Genotypic ratio: – Phenotypic ratio: o o O o Oo Oo oo oo 2:2(2 Oo: 2 oo) 2:2(2 orange: 2 purple)

40 Dihybrid (two traits) cross can be trickier – Cross heterozygous purple flowers, heterozygous yellow pea with another of the same. PpYyPpYy – Genotypes: PpYy & PpYy (2 traits, 8 alleles = 16 combos!) Step 1: find possible gametes (two traits each!) for each parent by doing FOIL method (first, outside, inside, last) & place on top & down side PY Py pY py PY PypY py Step 2: determine possible combinations by crossing alleles, making sure same alleles are together PPYY PPYy PpYY PpYy PPYy PPyy PpYy Ppyy PpYY PpYy ppYY ppYy PpYy Ppyy ppYy ppyy

41 Analyze results – Genotypic ratio: – Phenotypic ratio: PY Py pY py PY PypY py PPYY PPYy PpYY PpYy PPYy PPyy PpYy Ppyy PpYY PpYy ppYY ppYy PpYy Ppyy ppYy ppyy 1:2:2:4:1:2:1:2:1 (1 PPYY: 2 PPYy: 2 PpYY: 4 PpYy: 1 PPyy: 2 Ppyy: 1 ppYY: 2 ppYy: 1 ppyy) 9:3:3:1 (9 purple/yellow: 3 purple /green: 3 white/yellow: 1 white: green)

42 Simple dihybrid rules … Always will be maximum of 4 phenotypes – Trait A vs. Trait B, Trait C vs. Trait D = 4 phenotypes AC, AD, BC, BD Heterozygous AaBb vs. Heterozygous AaBb will always have same phenotypic ratio – 9 AB, 3aaB, 3Abb, 1aabb = 9:3:3:1 Heterozygous AaBb vs. Homozygous aabb will always have same phenotypic ratio – 4 AB, 4 aaB, 4 Abb, 4 aabb = 4:4:4:4

43 Dihybrid Cross Examples 1. Red ants are dominant over black ants, and long antennae are dominant over short antennae. Cross a black short antennae male with a heterozygous red long female. – Genotypes: – Gametes: rl, rl, rl, rl & RL, Rl, rL, rl rrll & RrLl

44 Genotypic ratio: Phenotypic ratio: rl rl rl rl rlRL RlrL RrLl RrLl RrLl RrLl Rrll Rrll Rrll Rrll rrLl rrLl rrLl rrLl rrll rrll rrll rrll 4:4:4:4 (4 RrLl: 4 Rrll: 4 rrLl: 4 rrll) 4:4:4:4 (4 red/long: 4 red/short: 4 black/long: 4 black/short

45 2. Green frogs are dominant over brown frogs, and spots are dominant over no spots. Cross a heterozygous green spotted female with the same type of male. – Genotypes: – Gametes: GS, Gs, gS, gs & GS, Gs, gS, gs GgSs & GgSs

46 Genotypic ratio: Phenotypic ratio: GS Gs gS gs GS Gs gS gs GGSS GGSs GgSS GgSs GGSs GGss GgSs Ggss GgSS GgSs ggSS ggSs GgSs Ggss ggSs ggss 1:2:2:4:1:2:1:2:1 9:3:3:1

47 Curi Family Eye Color & Ear Shape My father has green eyes and free-hanging ear lobes (homo or hetero?) while my mother has brown eyes (heterozygous) and free-hanging ear lobes (homo or hetero?). What are the possible outcomes for the children? – Know that my mother is heterozygous for brown eyes since my brother has green eyes – What about free hanging ear lobes? I have free hanging, but my brother and sister are attached! What does that mean about my parents? – Both parents MUST be heterozygous for free- hanging ears!

48 Genotypes: Gametes: bF bF bf bf BFbfBfbF BbFF BbFf BbFF BbFf BbFf Bbff BbFf Bbff bbff bbFF bbFf bbFF bbFf bbFf bbFf bbff ? ? ? ? ? ? BRO SIS me bF, bf, bF, bf & BF, Bf, bF, bf bbFf & BbFf

49 Getting ratios of genotypes & phenotypes is actually calculating probability – Probability: likelihood that particular event (genotype or phenotype) will occur – Calculated by dividing number of predicted outcomes by number of total outcomes – Ex: 3 peas are yellow, 1 is green Words: Ratios: Decimals: Percentages: Fractions: F.Probability ¾ yellow, ¼ green (add to 4/4) 3 out of 4 are yellow 3:1 yellow 0.75 yellow, 0.25 green (add up to 1.0) 75% yellow, 25% green (add to 100%)

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