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Chapter 9 Patterns of Inheritance

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1 Chapter 9 Patterns of Inheritance
Pre-AP Biology Ms. Haut

2 Modern Theory of Heredity
Based on Gregor Mendel’s fundamental principles of heredity Parents pass on discrete inheritable factors (genes) to their offspring These factors remain as separate factors from one generation to the next

3 Experimental genetics
Began with Gregor Mendel’s quantitative experiments with pea plants These plants are easily manipulated. These plants can self-fertilize. Figure 9.2 Figure 9.3

4 Mendel’s Discoveries Developed true-breeding lines—populations that always produce offspring with the same traits as the parents when parents are self-fertilized Mendel then crossed two different true-breeding varieties. Counted his results and kept statistical notes on experimental crosses Figure 9.4

5 Mendel’s Law of Segregation
Mendel performed many experiments. 1st Law of genetics The two members of an allele pair segregate (separate) from each other during the production of gametes. Figure 9.5

6 Based on Mendel, we’ve developed four hypotheses from the monohybrid cross:
There are alternative forms of genes, called alleles. For each characteristic, an organism inherits two alleles, one from each parent. If 2 alleles differ, one is fully expressed (dominant allele); the other is completely masked (recessive allele) Gametes carry only one allele for each inherited characteristic. 2 alleles for each trait segregate during gamete production

7 Useful Genetic Vocabulary
Homozygous—having 2 identical alleles for a given trait (PP or pp) Heterozygous—having 2 different alleles for a trait (Pp); ½ gametes carry one allele (P) and ½ gametes carry the other allele (p) Phenotype—an organism’s expressed traits (purple or white flowers) Genotype—an organism’s genetic makeup (PP, Pp, or pp)

8 Monohybrid Crosses

9 Ratio 3.15:1 3.14:1 3.01:1 2.96:1 2.95:1 2.82:1 2.84:1 x x x 3:1 x x x x

10 Punnett Square tt t Tt x tt T t Tt Tt Tt tt tt
Cross a heterozygous tall pea plant with a dwarf pea plant. T = tall, t = dwarf tt t Tt x tt T t Tt Tt Tt tt tt

11 The Testcross CAUTION:
The cross of an individual displaying the dominant phenotype to a homozygous recessive parent Used to determine if the individual is homozygous dominant or heterozygous CAUTION: Must perform many, many crosses to be statistically significant Figure 9.10

12 Genetic Alleles and Homologous Chromosomes
Have genes at specific loci. Have alleles of a gene at the same locus. Figure 9.7

13 Mendel’s Law of Independent Assortment
During gamete formation, the segregation of the alleles of one allelic pair is independent of the segregation of another allelic pair Law discovered by following segregation of 2 genes (dihybrid cross)

14 Dihybrid Cross Figure 9.8

15 Gamete formation AB Ab aB AaBbCc AaBb ab ABC aBC ABc aBc AbC abC Abc

16 Mendelian Inheritance Reflects Rules of Probability
Rule of Multiplication: The probability that independent events will occur simultaneously is the product of their individual probabilities. What is the probability that you will roll a 6 and a 4? 1/6 x 1/6 = 1/36 chance

17 Mendelian Inheritance Reflects Rules of Probability
Question: In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability that the offspring will be homozygous recessive? Answer: Probability that an egg from the F1 (Pp) will receive a p allele = ½ Probability that a sperm from the F1 will receive a p allele = ½ Overall probability that 2 recessive alleles will unite at fertilization: ½ x ½ = ¼

18 Mendelian Inheritance Reflects Rules of Probability
Works for Dihybrid Crosses: Question: For a dihybrid cross, YyRr x YyRr, what is the probability of an F2 plant having the genotype YYRR? Answer: Probability that an egg from a YyRr parent will receive the Y and R alleles = ½ x ½ = ¼ Probability that a sperm from a YyRr parent will receive the Y and R alleles = ½ x ½ = ¼ Overall probability of an F2 plant having the genotype YYRR: ¼ x ¼ = 1/16

19 Mendelian Inheritance Reflects Rules of Probability
Rules of Addition: The probability of an event that can occur in two or more independent ways is the sum of the separate probabilities of the different ways. What is the probability that you will roll a 6 or a 4? 1/6 + 1/6 = 2/6 or 1/3 chance

20 Mendelian Inheritance Reflects Rules of Probability
Question: In a Mendelian cross between pea plants that are heterozygous for flower color (Pp), what is the probability that the offspring will being a heterozygote? Answer: There are 2 ways in which a heterozygote may be produced: the dominant allele may be in the egg and the recessive allele in the sperm, or the dominant allele may be in the sperm and the recessive allele in the egg.

21 Mendelian Inheritance Reflects Rules of Probability
Probability that the dominant allele will be in the egg with the recessive in the sperm is ½ x ½ = ¼ Probability that the dominant allele will be in the sperm with the recessive in the egg is ½ x ½ = ¼ Therefore, the overall probability that a heterozygote offspring will be produced is ¼ + ¼ = ½

22 Pedigree Analysis Analysis of existing populations
Studies inheritance of genes in humans Useful when progeny data from several generations is limited Useful when studying species with a long generation time

23 Pedigree Analysis

24 Dominant Pedigree: For dominant traits:
II III For dominant traits: Every affected individual has at least one affected parent; Affected individuals who mate with unaffected individuals have a 50% chance of transmitting the trait to each child; Two affected individuals may have unaffected children.

25 Dominant Disorders: A Fifty-Fifty Chance
The affected parent has a single defective gene (D), which dominates its normal counterpart (n). Each child has a 50 percent risk of inheriting the faulty gene and the disorder.

26 Recessive Pedigree: For recessive traits:
II III For recessive traits: An individual who is affected may have parents who are not affected All the children of two affected individuals are affected; In pedigrees involving rare traits, the unaffected parents of an affected individual may be related to each other.

27 Recessive Disorders: One Chance in Four
Both parents carry a single defective gene (d) but are protected by the presence of a normal gene (N) Two defective copies of the gene are required to produce a disorder. Each child has a 50 percent chance of being a carrier like both parents and a 25 percent risk of inheriting the disorder.

28 Recessive Human Disorders
Sickle-cell anemia; autosomal recessive Caused by single amino acid substitution in hemoglobin Abnormal hemoglobin packs together to form rods creating crescent-shaped cells Reduces amount of oxygen hemoglobin can carry

29 Genetic Testing & Counseling
Genetic counselors can help determine probability of prospective parents passing on deleterious genes Pedigree analysis Genetic screening Fetal testing Karyotype Chemical testing

30 Pedigree Analysis Neither Jan nor Bill knew they each carried the faulty CF gene until they had Sue, as there were no other family members who had the condition. Jan is currently 8 weeks pregnant. What is the probability the baby will have CF? Be a carrier?

31 Genetic Screening DNA is examined using direct gene testing to see if the mutation in each allele of the gene involved can be identified Since a mutation is detected, testing can be offered to Jan and Bill in this or future pregnancies. It will be possible to examine the baby’s DNA for the mutation

32 Amniocentesis and chorionic villus sampling (CVS)
Allow doctors to remove fetal cells that can be tested for genetic abnormalities (karyotype/chemical testing) Some risk of complications—so reserved for those with higher possibility of genetic disorder Figure 9.1

33 Fetal imaging Ultrasound—uses sound waves to produce a picture of the fetus

34 Variations to Mendel’s First Law of Genetics
Incomplete dominance—pattern of inheritance in which one allele is not completely dominant over the other Heterozygote has a phenotype that is intermediate between the phenotypes of the homozygous dominant parent and homozygous recessive parent

35 Incomplete Dominance in Snapdragon Color
F2 Genotypic ratio: 1 CRCR: 2 CRCW: 1 CWCW Phenotypic ratio: 1 red: 2 pink: 1 white Figure 9.16

36 Variations to Mendel’s First Law of Genetics
Codominance—pattern of inheritance in which both alleles contribute to the phenotype of the heterozygote Roan Cattle

37 In chickens, black feather color (BB) is codominant to white feather color (WW).  Both feather colors show up in a checkered pattern in the heterozygous individual (BW). Cross a checkered hen with a checkered rooster. What are the genotypic and phenotypic ratios?

38 Example of Codominance
Ex: Feather colors in chickens Black (BB) x White (WW) = Black and White checkered Chicken B W BB BW B BW WW W

39 Multiple Alleles Some genes may have more than just 2 alternate forms of a gene. Example: ABO blood groups A and B refer to 2 genetically determined polysaccharides (A and B antigens) which are found on the surface of red blood cells (different from MN blood groups) A and B are codominant; O is recessive to A and B

40 Multiple Alleles for the ABO Blood Groups
3 alleles: IA, IB, i Figure 9.18

41 Blood Types The immune system produces blood proteins
That may cause clotting when blood cells of a different type enter the body. Figure 9.19

42 Example of ABO Blood Groups
Ex: Feather colors in chickens Black (BB) x White (WW) = Black and White checkered Chicken IA IB IA IA IA IB IA IA i IB i i

43 Pleiotropy The ability of a single gene to have multiple phenotypic effects (pleiotropic gene affects more than one phenotype)

44 Epistasis Interaction between 2 nonallelic genes in which one modifies the phenotypic expression of the other. If epistasis occurs between 2 nonallelic genes, the phenotypic ratio resulting from a dihybrid cross will deviate from the 9:3:3:1 Mendelian ratio

45 CC, Cc = Melanin deposition
cc = Albinism BB, Bb = Black coat color bb = Brown coat color A cross between heterozygous black mice for the 2 genes results in a 9:3:4 phenotypic ratio 9 Black (B_C_) 3 Brown (bbC_) 4 Albino (__cc)

46

47 Polygenic Traits Skin pigmentation in humans
--3 genes with the dark-skin allele (A, B, C) contribute one “unit” of darkness to the phenotype. These alleles are incompletely dominant over the other alleles (a, b, c) --An AABBCC person would be very dark; an aabbcc person would be very light --An AaBbCc person would have skin of an intermediate shade Figure 9.21

48 Polygenic Trait

49

50 Chromosome Theory of Inheritance
Based on Mendel’s observations and genetic studies and cytological evidence Genes are located at specific positions on chromosomes. The behavior of chromosomes during meiosis and fertilization accounts for inheritance patterns.

51 Figure 9.23

52 Genes on the same chromosome tend to be inherited together
Experiment Explanation: linked genes PpLI  PpLI Long pollen Observed Prediction Phenotypes offspring (9:3:3:1) Purple long Purple round Red long Red round Parental diploid cell PpLI Most gametes offspring Eggs 3 purple long : 1 red round Not accounted for: purple round and red long Meiosis Fertilization Sperm 284 21 55 215 71 24 P I P L Purple flower Certain genes are linked They tend to be inherited together because they reside close together on the same chromosome Figure 9.19

53 Crossing over produces new combinations of alleles
Crossing over can separate linked alleles Producing gametes with recombinant chromosomes Figure 9.25

54 Thomas Hunt Morgan Performed some of the early studies of crossing over using the fruit fly Drosophila melanogaster Experiments with Drosophila revealed linkage traits. Why Drosophila? Easily cultured Prolific breeders Short generation times Only 4 pairs of chromosomes, visible under microscope

55 Morgan’s experiments Demonstrated the role of crossing over in inheritance Figure 9.24

56 Morgan’s experiments Two linked genes
Can give rise to four different gamete genotypes. Can sometimes cross over. Figure 9.26

57 Geneticists use crossover data to map genes
Morgan and his students Used crossover data to map genes in Drosophila Figure 9.21 A

58 Linkage Map Alfred Sturtevant hypothesized that the frequency of recombinants reflected the distances between genes on a chromosome. The farther apart two genes are, the higher the chance of crossover between them and therefore a higher recombination frequency. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

59 Recombination frequencies
Can be used to map the relative positions of genes on chromosomes. Mutant phenotypes Short aristae Black body (g) Cinnabar eyes (c) Vestigial wings (l) Brown Long aristae (appendages on head) Gray (G) Red (C) Normal (L) Wild-type phenotypes Chromosome g c l 9% 9.5% 17% Recombination frequencies Figure 9.21 B Figure 9.21 C

60 Fig. 15.5b Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

61 The recombination frequency between cn and b is 9%.
Sturtevant used the testcross design to map the relative position of three fruit fly genes, body color (b), wing size (vg), and eye color (cn). The recombination frequency between cn and b is 9%. The recombination frequency between cn and vg is 9.5%. The recombination frequency between b and vg is 17%. The only possible arrangement of these three genes places the eye color gene between the other two. Fig. 15.6 Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

62 Sturtevant expressed the distance between genes, the recombination frequency, as map units.
One map unit (sometimes called a centimorgan) is equivalent to a 1% recombination frequency.

63 What is the sequence of these three genes on the chromosome?
A series of matings shows that the recombination frequency between the black-body gene (b) and the gene for short wings (s) is 36%. The recombination frequency between purple eyes (p) and short wings is 41%. The recombination frequency between black-body gene and purple eyes is 6%.

64 Answer B % S P % S B 6% P 6% + 36% = 42% P 6% B B % S P % S

65 This results from multiple crossing over events.
You may notice that the three recombination frequencies in our mapping example are not quite additive: 9% (b-cn) + 9.5% (cn-vg) > 17% (b-vg). This results from multiple crossing over events. A second crossing over “cancels out” the first and reduces the observed number of recombinant offspring. Genes father apart (for example, b-vg) are more likely to experience multiple crossing over events. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

66 If the recombination frequency is less than 50%, the genes are linked
Some genes on a chromosome are so far apart that a crossover between them is virtually certain. In this case, the frequency of recombination reaches is its maximum value of 50% and the genes act as if found on separate chromosomes and are inherited independently. If the recombination frequency is 50% or greater, the genes are not linked If the recombination frequency is less than 50%, the genes are linked Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

67 SEX CHROMOSOMES AND SEX-LINKED GENES
Chromosomes determine sex in many species In mammals, a male has one X chromosome and one Y chromosome And a female has two X chromosomes The Y chromosome Has genes for the development of testes (SRY) The absence of a Y chromosome Allows ovaries to develop Figure 9.28

68 Other systems of sex determination exist in other animals and plants
22 + XX X Figure 9.22 B 76 + ZW ZZ Figure 9.22 C 32 16 Figure 9.22 D

69 Sex-linked Genes XRXr XrY Any gene located on a sex chromosome
Not much crossing over between X and Y chromosomes so DNA passed on in tact For recessive trait on the X chromosome to be expressed: In females, must have 2 copies of the allele In males, one copy is enough XRXr XrY

70 Sex-linked genes Were discovered during studies on fruit flies.
Figure 9.23 A

71 The inheritance pattern of sex-linked genes
Is reflected in females and males Figure 9.30

72 X-Linked Disorders: Males are at Risk
A male receiving a single X-linked allele from his mother Will have the disorder A female Has to receive the allele from both parents to be affected

73 Hemophilia and the Romanov Family
Nicholas II, the Last Russian Tsar


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