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Glorious Genetics with a Marvelous Monk Named Mendel Chapter 14.

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Presentation on theme: "Glorious Genetics with a Marvelous Monk Named Mendel Chapter 14."— Presentation transcript:

1 Glorious Genetics with a Marvelous Monk Named Mendel Chapter 14

2 Science as a Process  1. Gregor Mendel –monk turned scientist, worked with garden peas to study inheritance –father of modern genetics

3 Science as a Process  2. Why Peas??? –Come in many varieties (i.e. purple/white flowers, round/wrinkled) –Easy to control parentage  sex organs are in flowers and each flower has both male (stamens) and female (carpals) parts

4 Science as a Process  3. How did he do it? –Mendel removed Stamens before plants could self- fertilize. The plant now only has the female parts. –He then he put the pollen from another plant onto the now “female” flower and made offspring (seeds) –focused on either/or characters (there were only two varieties of each trait) –started with true-breeding plants = purple flowered plants that produced only purple flowered offspring

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6 Terrific Terminology  Character - heritable feature (i.e. flower color)  Trait - variant of a character (purple, white)  Hybridization - mating or crossing of two varieties  True-breeding- after many generations of self pollination, parent plant produces only the same variety of offspring (i.e. purple plant makes only purple offspring )

7 Terminology - cont  Monohybrid cross - looks at only one trait at a time  P generation - parental (original cross)  F1 generation - offspring from P generation  F2 generation - results if the F1 plants are allowed to self - fertilize

8 Mendel’s Shocking results  P Generation - Purple x White flower Produced  F1 Generation - ALL PURPLE (GASP!!!)  Let F1 self fertilize  F2 Generation Purple, 224 White (WOW!)

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10 So Who Cares???  Most scientists thought that traits blended together. So after the Parents bred, the F1 should have been pale purple, but Mendel’s results disproved this. Even more surprising was that the white flowers showed up again in the F2  Mendel continued his studies with 6 other characters (round/wrinkled, tall/short etc) and found the same results in all characters.

11 What does it all mean???  Alternate versions of genes account for variations in inherited characters  ex. There is a gene for flower color in peas. This gene exists in two forms, purple or white.  allele - alternate form of a gene

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13 How does it work??? 1.Alternative versions of genes account for variations in inherited characteristics  Different alleles are caused by slight changes in nucleotide sequences for a gene on the DNA  This change results in a slightly different protein (causing the difference in appearance)  The gene is in the same place on the chromosome, but the order of the nitrogen bases (A, T, G, C) is different.

14 How does it work??? 2.An organism inherits two alleles for each trait, one from each parent (one copy in sperm/pollen, the other copy from egg).  The offspring can inherit two of the same alleles, or two different alleles 3.If the two alleles differ, then the Dominant allele determines the organisms appearance and the Recessive allele is not seen in the organisms’ appearance –(Mendel’s F1 generation looked purple because purple color is dominant over white color in peas)

15 Why only one trait??? 4.Two alleles for each character separate during meiosis and only one is passed on to the next generation = Law of Segregation

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17 Terrific Terminology cont  Homozygous - organism with two of the same alleles for a character  Homozygous Dominant - AA Homozygous recessive - aa  Heterozygous = organism with two different alleles for a character = Aa  Genotype = genetic makeup (actual genes the organism has)  Phenotype = physical appearance  Punnett Square - box used to show all possible offspring resulting from the cross between two parents

18 Punnett Square review  Purple flowers are dominant to white flowers. Mendel crossed a true breeding (homozygous) purple flower to a true breeding (homozygous) white flower. What are the possible genotype and phenotype results of this cross?

19 The Answer  A = purplea = white  P gen. = AA x aa  Genotype results  4/4 Aa  Phenotype results  4/4 Purple AaAa AaAa A A a a

20 More fun!!! Next, Mendel let the F1 generation self fertilize. What are the possible results of this cross?  Cross Aa x Aa  Genotype results  1/4 AA 1/2 Aa 1/4 aa  Phenotype results  3/4 Purple, 1/4 White  Do the results above agree with Mendel’s results? AAAa Aaaa Aa A a

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22 How can you tell if the plant is Homozygous or Heterozygous???  Do a Test-cross  Cross your unknown purple plant with a white plant.  If the offspring are all purple, you know your original plant is homozygous (AA).  If 50% of the offspring are purple and 50% are white, your original plant is heterozygous (Aa)  Do the squares if you don’t believe me!!!

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24 Dihybrid Crosses - look at two traits at once  Used to determine if traits assort independently of one another  Try these (remember the 16 squares!!)  Purple flowers are dominant to white flowers.  Round seeds are dominant to wrinkled seeds.  Cross a homozygous purple round plant with a homozygous white wrinkled plant.

25 How do you do it???  A = purplea = white  R = roundr = wrinkled  AARR x aarr  First find the gametes  Gametes  ARar

26 Dihybrid crosses (cont)  Next – make the square – This one’s easy AaRr AR ar

27 Dihybrid crosses (cont) Genotype results 1/1 AaRr Phenotype results Phenotype results 1/1 Purple, Round

28 Dihybrid crosses (cont)  Now let the plants self-fertilize  AaRr x AaRr  What will the gametes be?  AR  Ar  aR  ar

29 Yikes – 16 boxes!!! AR Ar aR ar AR Ar aR ar

30 AARRAARrAaRRAaRr AARrAArrAaRrAarr AaRRAaRraaRRaaRr AaRrAarraaRraarr AR Ar aRar AR Ar aR ar

31 Now Count Now Count  Genotype results  1/16 AARR  2/16 AaRR  2/16 AARr  4/16 AaRr  1/16 AArr  2/16 Aarr  1/16 aaRR  2/16 aaRr  1/16 aarr

32 More counting  Phenotype results  9/16 Purple, Round  3/16 Purple, wrinkled  3/16 white, Round  1/16 white, wrinkled

33 Law of Independent Assortment  If you look at each of the characters individually you get the correct ratios  Purple 3: White: 1  Round 3: Wrinkled: 1  Law of Independent Assortment – Alleles sort independently of one another during meiosis ( traits are not linked together)

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35 Laws of Probability (shortcuts we never learned in Biology)  Rule of Multiplication – used to determine the chance that two or more independent events will occur together in some specific combination  To figure this out – find the probability of one event occurring and multiply it by the probability of the other event(s) occurring

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37 Try it  You are crossing two plants that are heterozygous (Aa) for purple color. What is the chance of getting a white (aa) offspring?  Chance of plant 1 giving a = ½  Chance of plant 2 giving a = ½  Chance of white offspring aa = ¼  Does it work??? Do the square to confirm

38 Now try with a dihybrid  You are crossing two plants that are heterozygous Purple, Round (AaRr). What is the chance of getting a plant that is aarr?  Chance of plant 1 being ar = ¼  Chance of plant 2 being ar = ¼  Chance of aarr = 1/4 x 1/4 = 1/16  Check the results in the dihybrid problem

39 Rule of Addition  The probability of an event that can occur in two or more different ways is the sum of the separate probabilities of those ways.  Cross two plants that are heterozygous (Aa). What is the chance that the offspring will be heterozygous?  Chance of Aa = 1/4 (if plant 1 gives A, plant 2 gives a)  Chance of Aa = 1/4 (if plant 2 gives A, plant 1 gives a)  Chance of heterozygous = 1/4 + 1/4 = 1/2

40 Now for a tough one  Parents AaRrTt x Aarrtt  Looking for the chance of at least two recessives  aarrTt =  Aarrtt =  aarrtt =  aaRrtt =  AArrtt =

41 Solution AaRrTt x Aarrtt  aarrTt = 1/4 x 1/2 x 1/2 = 1/16  Aarrtt = 1/2 x 1/2 x 1/2 = 1/8 = 2/16  aarrtt = 1/4 x 1/2 x 1/2 = 1/16  aaRrtt = 1/4 x 1/2 x 1/2 = 1/16  AArrtt = 1/4 x 1/2 x 1/2 = 1/16  Total chance of at least 2 recessives = 6/16 or 3/8

42 Now use the rules of probability to determine the following  PKU is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers (Aa x Aa), what is the probability of each of the following:  All three of her children will be of normal phenotype  3/4 x 3/4 x 3/4 = 27/64

43 Cont  One or more of the three children will have the disease  64/64 – 27/64 = 37/64  All three children will have the disease  1/4 x 1/4 x 1/4 = 1/64  At least one child will be phenotypically normal  64/64 – 1/64 = 63/64

44 Exceptions to Mendel’s findings  Mendel looked only at traits following simple inheritance (complete dominance). Mendel’s laws lay the foundation for modern genetics and the basic principles are true, but there are exceptions to his rules.

45 Incomplete Dominance  Heterozygous organisms show a blending of two other traits.  Ex. – red flower x white flower = pink flower  Try a few problems  Red color is incompletely dominant to white. Heterozygous flowers are pink.  What would be the results of a cross between a red and white flower?

46  RR x WW  Genotype = 4/4 RW  Phenotype = 4/4 pink  Try these two  Pink x Pink  Red x Pink RWRW RWRW RR W W

47 RRRW RWWW RRRRRWRW R W R W R R R W 1/4 RR, 1/2 RW, 1/4 WW 25% red, 50% pink, 25% white 1/2 RR, 1/2 RW 50% red, 50% pink

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49 Codominance  One gene is codominant to another. Heterozygous organisms show both traits.  Ex – Normal blood x sickle blood = both normal and sickle cells

50  Try some problems – Normal red blood cells are codominant to sickle cells. Heterozygous individuals (carriers) have both normal and sickle cells (also resistant to malaria – Cool!!)  Normal x Sickle  Sickle x Carrier  Carrier x Carrier

51 NSNS NSNS NSNSSSSS NNNSNSSS N N S S 4/4 NS = 100% carriers S S N S 2/4 = NS, 2/4 SS 50% carrier, 50% sickle NS N S 1/4 NN, 2/4, NS, 1/4 SS 25% normal, 50% carrier, 25% sickle

52 Multiple Alleles  More than two alleles exist for a trait  Ex – blood types – A, B, AB, O PhenotypeGenotypeAntibodies A I A I A or I A i Anti – B B I B IB or I B i Anti – A AB IAIBIAIBIAIBIAIBNone Oii Anti-A,Anti - B

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54 Try some probs  Homozygous A x O  AB x O  Hetero A x Hetero B

55 IIAiIIAi IIAiIIAi IIAiIIAi IIAiIIAi IAiIAiIAiIAi IBiIBiIBiIBi IAiIAiIAiIAi IBiIBiIBiIBi IAIBIAIBIAIBIAIB IBiIBiIBiIBi IAiIAiIAiIAiii IAIA IAIA i i IAIA IBIB i i IAIA i IBIB i 4/4 IAi 100% Type A B 2/4 I A i, 2/4 I B i 50% type A, 50% type B 1/4 I A I B, 1/4 I B i, 1/4 I A i, 1/4 ii 25% AB, 25% B, 25% A, 25% O

56 Pleiotropy  One gene can affect an organism in many ways  Ex – sickle cell disease causes many different symptoms in the patient

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58 Epistasis  A gene at one locus changes the expression of another gene  Ex – one gene causes pigment, the other causes the actual color

59 Sample problems  One dominant gene in mice causes pigment to appear (D). If the mouse is homozygous recessive (dd), no pigment will appear.  Another gene (B) causes black color, the recessive form (bb) causes brown color.  Cross DDbb x Ddbb – How many black mice will result?  Chance of pigment 1/1 x chance of black 0 = no black mice

60 Try another  DdBb x DdBb – How many brown mice will result  Chance of pigment 3/4 x chance of brown 1/4 = 3/16

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62 Polygenic  More than one gene controls a trait  Ex – hair color, skin color in humans  Result is a range of trait in offspring - i.e not just black or brown hair, but many shades in between

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64 Nature vs Nurture  Some phenotypes are affected by environmental factors  Ex – amount of red blood cells depends not only on genes, but on altitude, nutrition, physical activity

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66 Pedigree  Used to trace traits in organisms that don’t produce many offspring or have long generation span (like humans)  Square = male  Circle = female  Shaded = trait being traced – can be dominant or recessive

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69 Recessively inherited Disorders  Caused only when organism inherits two recessive alleles (aa)

70 Cystic Fibrosis  1/2500 whites of European descent.  Normal people can transport chloride into cells, channels are defective in disease.  Causes excessive mucous build up in many organs  Untreated = death by 15, treated can live longer

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72 . Tay-Sachs  Dysfunction in enzyme that doesn’t break down brain lipids.  Causes seizures, blindness, decreased motor performance  Usually death within a few years.  High incidence in Jewish people

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74 Sickle-cell disease  Most common in Africans.  Caused by substitution of one amino acid in hemoglobin.  Red Blood cells abnormally shaped, can’t carry oxygen affectively.  Treated with blood transfusions, but no cure.  Heterozygous individuals have both normal/sickle blood cells but show a resistance to Malaria (common in Africa).  Heterozygotes have an advantage in these areas which may be why the gene has remained in the population.

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76 Dominantly inherited disorders  Occur when at least one dominant gene is present in an organism.

77 Huntington’s disease  Degenerative disease of nervous system.  Gene has remained in population because illness doesn’t occur until yrs of age.  Many people with disease have children and pass disease to them before they even know they have it.

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79 Achondroplasia  Dwarfism - Aa = dwarf (.01%),  aa = normal size (99.99%)  AA = lethal (death before birth)  Many other dominant disease are lethal and kill organisms before birth

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81 Genetic testing  Prenatal - amniocentesis - chromosome analysis of amniotic fluid (weeks) –carrier recognition - determining whether parents are carriers  CVS - chorionic villus sampling - fetal tissue from placenta is used to do chromosome analysis (24 hrs)  Ultrasound - looks for developmental problems in organs

82 Newborn screening  many tests are done before babies leave hospital to indicate genetic disorders  PKU - tests for phenylketonuria - if detected, parents can alter diet of children to prevent symptoms (no phenylalanine)

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