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Chpt. 14 Mendelian Genetics. How are traits passed from parent to offspring?

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Presentation on theme: "Chpt. 14 Mendelian Genetics. How are traits passed from parent to offspring?"— Presentation transcript:

1 Chpt. 14 Mendelian Genetics

2 How are traits passed from parent to offspring?

3 Chpt. 14 Mendelian Genetics WE … know it is genes on chromosom es…

4 Chpt. 14 Mendelian Genetics MENDEL had no idea what a chromosome even was!!!!!

5 Chpt. 14 Mendelian Genetics 1843 - entered an Augustinian monaster y.

6 Chpt. 14 Mendelian Genetics 1846 - assigned to a High School as a teacher… didn’t pass the teacher test. BACK TO THE MONASTERY

7 Chpt. 14 Mendelian Genetics 1851 - 53 Tried to “find himself”, sent to Univ. of Vienna

8 Chpt. 14 Mendelian Genetics 1851 - 1853 two profs. influence d him and he began research. Statistics Professor and a Botany Professor

9 Chpt. 14 Mendelian Genetics 1851 - 53 His research was “controversi al”he FLUNKED OUT! Mendel became an emotional wreck!!

10 Chpt. 14 Mendelian Genetics 1854 - BACK to the monastery, HOWEVER, this time…

11 Chpt. 14 Mendelian Genetics 1854 - the Monsignor allowed him to research inheritance …

12 Mendel’sMethods: Worked with Worked with Pea Plants observed several characters

13 Either - Or traits

14

15 Mendel’sMethods: Cross-pollinated two contrasting true breeding pea varieties pg. 567 anatomy

16 P F 1 4:0 F 2 3:1 True breeding

17 P F 1 4:0 F 2 3:1 hybridization True breeding Mating two true-breeding varieties

18 White did not disappear in F1! Purple is just DOMINANT over white

19 We now know : Alternative versions of a gene

20 Results - Summary In all crosses, the F 1 generation shows only one of the traits regardless of which was male or female (4:0 ratio). The other trait reappeared in the F 2 at ~25% (3:1 ratio).

21 Mendel’s Laws = (alleles) Alternate versions of genes (alleles) account for variations. Organisms inherit two alleles (one from each parent)- for each character. When alleles differ, the dominant is fully expressed.

22 Mendel’s Laws = Each pair of alleles separate during gamete formation by chance- 50% probability that each version will be the one involved in the fertilization. Aa A a Germ cell after meiosis II A a

23 PP Law of Segregation pp

24 --> VO CA BU LA RY -->

25 Testcross= Cross of a phenotypicall y dominant with a homozygote recessive

26 Testcross= Cross of a suspected heterozygot e with a homozygote recessive u ex: T_ x tt If TT: all offspring will be dominant If Tt: offspring will be 1 Dominant : 1 recessive

27 6 Mendelian Crosses are Possible (these are your expected): Cross Genotype Phenotype TT X tt all Tt all Dominant Tt X Tt 1TT:2Tt:1tt 3 Dom: 1 Res TT X TT all TT all Dominant tt X tt all tt all Recessive TT X Tt 1TT:1Tt all Dominant Tt X tt 1Tt:1tt 1 Dom: 1 Res Sorry,but College Board who runs AP, says you must memorize this… so does Hobby, so does your Freshman collegiate Bio instructor!!

28 Lets try it with one trait: COLOR

29 What if you are tracking more than ONE trait???

30 Do the traits “travel” together when gametes are formed?

31 In other words, does seed color & seed shape always stay together?

32 Does the “R” allele always stay with the “Y” allele?

33 Law of Independent Assortment

34 Mendel’s Each pair/types of alleles segregates into gametes independently. Color and seed shape are not a “package deal”!`

35 Dihybrid Cross u Cross with two genetic traits. u Need 4 letters to code for the cross. u ex: TtRr u Each Gamete - Must get 1 letter for each trait. u ex. TR, Tr, tR, tr = possible gametes

36 How big of a Punnett Square? Number of Gametes possible Critical to calculating the results of higher level crosses. Look for the number of heterozygous traits.

37 Equation to determine number of gametes possible: The formula 2 n can be used, where “n” equals the number of heterozygous traits possible in a cross. ex: TtRr, n=2 2 2 or 4 different kinds of gametes are possible. 2 2 or 4 different kinds of gametes are possible. TR, tR, Tr, tr TR, tR, Tr, tr

38 Dihybrid Cross TtRr X TtRr u each parent can produce 4 types of gametes. TR, Tr, tR, tr u cross is a 4 X 4 with 16 possible offspring. FOIL

39 Results of TtRr X TtRr u 9 Tall, Red flowered u 3 Tall, white flowered u 3 short, Red flowered u 1 short, white flowered Or: 9:3:3:1

40 Comment u Ratio of Tall to short is 3:1 (b/c Tt X Tt) u Ratio of Red to white is 3:1 (b/c Rr X Rr) u The cross is really a product of the ratio of each trait multiplied together. (3:1) x (3:1) FOIL

41 PROBABILITY RULES OF INHERITANCE &

42 Rule of Multiplication- Rule of Multiplication- used when determining chances that two independentevents w/ occur together simultaneously. used when determining chances that two independent events w/ occur together simultaneously.

43 Rule of Multiplication- Rule of Multiplication- two independentevents w/ occur together simultaneously. two independent events w/ occur together simultaneously. two independentevents ex. height & flower color two independent events ex. height & flower color

44 Rule of Multiplication - Rule of Multiplication - Determine the probability of 1.Determine the probability of each independent event (ex. allele in each sperm and allele in each egg) multiply the probabilities of those independent events. 2. multiply the probabilities of those independent events.

45 Rule of Multiplication - Rule of Multiplication -ex. P generation = Pp x Pp “what is the probability of a white offspring??” “what is the probability of a white offspring??” Can only get this one way… (pp)

46 Rule of Multiplication- 1. P generation = Pp x Pp chance of Ovum p = 1/2 2. chance of Ovum p = 1/2 1/4 pp P p p P chance of Sperm p = X 1/2 chance of Sperm p = X 1/2

47 Rule of Multiplication- Rule of Multiplication- used when determining chances that two independentevents w/ occur together simultaneously. used when determining chances that two independent events w/ occur together simultaneously.

48 u The probability of getting a tall offspring is ¾ (Tt x Tt). u The probability of getting a red offspring is ¾ (Rr x Rr). u The probability of getting a TALL red offspring is ¾ x ¾ = 9/16 u Example: TtRr X TtRr

49 Rule of Multiplication - Rule of Multiplication - YyRr is in a germ cell: “What is the probability that a gamete will be YR?” 1. YyRr is in a germ cell: “What is the probability that a gamete will be YR?” 1/2 a chance for Y 2. 1/2 a chance for Y 1/4 YR x 1/2 a chance for R

50 Rule of Multiplication- Rule of Multiplication- used when determining chances that two independentevents w/ occur together simultaneously. used when determining chances that two independent events w/ occur together simultaneously.

51 But, what about, for example, heterozygote parents? Yy x Yy But, what about, for example, heterozygote parents? Yy x Yy What is the probability that these parents would produce a heterozygote offspring?

52 Rule of Addition - Rule of Addition - can occur in 2 or more different ways Used when determining the probability of an event… an event that can occur in 2 or more different ways.

53 Rule of Addition - Rule of Addition - each 1. Compute the probability for each way in which each event can happen. Add 2. Add the probabilities for each of those ways.

54 Rule of Addition - Rule of Addition - P 1 = Yy x Yy ex. “what is the probability of a heterozygous offspring?” Ovum Y = 1/2 Sperm y = 1/2 Ovum y = 1/2 Sperm Y = 1/2

55 Rule of Addition- Rule of Addition- 1/4 + 1/4 = 1/2 2/4 = 1/2 There is a 1/2 chance for a Yy offspring Ovum Y = 1/2 x Sperm y = 1/2 Ovum y = 1/2 x Sperm Y = 1/2

56 Rule of Addition - Rule of Addition - can occur in 2 or more different ways Used when determining the probability of an event… an event that can occur in 2 or more different ways.

57 Rules of additio n and multiplicatio n:

58 Variations on Mendel 1. Incomplete Dominance 2. Codominance 3. Multiple Alleles 4. Epistasis 5. Polygenic Inheritance

59 Law of Incompl ete Dominanc e

60

61 Multiple alleles Any of a set of three or more alleles, only two of which can be present in a diploid organism.

62 Multiple alleles clumped

63

64 Comment Rh blood factor is a separate factor from the ABO blood group. Rh+ = dominant Rh- = recessive A+ blood = dihybrid trait

65 Epistasis

66 Epistasis When 1 gene locus alters the expression of a second locus. ex: 1 st gene: C = color, c = albino 2 nd gene: B = Brown, b = black

67 Epistasis

68 “Labs”

69 In Gerbils too! CcBb X CcBb Black X Black F1 = 9 black (C_B_) 3 brown (C_bb) 4 albino (cc__)

70 Result Ratios often altered from the expected. One trait may act as a recessive because it is “hidden” by the second trait.

71 Problem Wife is type A Husband is type AB Child is type O Question - Is this possible? Comment - Wife’s boss is type O

72 Bombay Effect Epistatic Gene on ABO group. Alters the expected ABO outcome. H = dominant, normal ABO h = recessive, no A,B, reads as type O blood.

73 Genotypes Wife: type A (I A I A, Hh) Husband: type AB (I A I B, Hh) Child: type O (I A I A, hh) Therefore, the child is the offspring of the wife and her husband (and not the boss).

74 Bombay - Detection When ABO blood type inheritance patterns are altered from expected.

75 Polygenic Several genes affect one character ex. skin color, height

76 Result Mendelian ratios fail. Traits tend to "run" in families. Offspring often intermediate between the parental types. Trait shows a “bell-curve” or continuous variation.

77 Pleiotropy one gene, multiple phenotypic effects

78 Pleiotropy Cystic Fibrosis one gene, multiple phenotypic effects

79 Some DOMINANT genes, are not often expressed in a population

80

81

82 Does the “R” allele always stay with the “Y” allele?

83

84 = normal male = normal female female

85 = affected male = affected female female

86 Reproductive partners

87 siblings

88 Can determine: 1) Autosomal recessive disorder 2) Autosomal dominant disorder

89 Can determine: 3) Sex-Linked disorder

90

91 Genetic Screening Risk assessment for an individual inheriting a trait.Risk assessment for an individual inheriting a trait. Uses probability to calculate the risk.Uses probability to calculate the risk.

92 General Formal R = F x M x D R = risk F = probability that the female carries the gene. M = probability that the male carries the gene. D = Disease risk under best conditions.

93 Example Wife has an albino parent.Wife has an albino parent. Husband has no albinism in his pedigree.Husband has no albinism in his pedigree. Risk for an albino child?Risk for an albino child?

94 Risk Calculation Wife = probability is 1.0 that she has the allele.Wife = probability is 1.0 that she has the allele. Husband = with no family record, probability is near 0.Husband = with no family record, probability is near 0. Disease = this is a recessive trait, so risk is Aa X Aa =.25Disease = this is a recessive trait, so risk is Aa X Aa =.25 R = 1 X 0 X.25R = 1 X 0 X.25 R = 0R = 0

95 Carrier Recognition Fetal TestingFetal Testing –Amniocentesis –Chorionic villi sampling Newborn ScreeningNewborn Screening

96

97 Summary Know the Mendelian crosses and their patterns. Be able to work simple genetic problems (practice). Watch genetic vocabulary. Be able to read pedigree charts. Be able to recognize and work with some of the “common” human trait examples.


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