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Chapter 14: Mendel and the Gene Idea

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  1. Chapter 14: Mendel and the Gene Idea 14.1 Mendel used the scientific approach to identify two laws of inheritance 14.2   The laws of probability govern Mendelian inheritance 14.3 Extensions to Mendelian Genetics:Inheritance patterns are often more complex than predicted by simple Mendelian genetics 14.4   Many human traits follow Mendelian patterns of inheritance

  2. Before we begin: Define these terms: • Gene: functional unit of heredity - a segment of DNA located in a specific site on a chromosome that gives the instrcutions to make one (or more) enzyme or other protein. • Dive into the chromosome here! ! • Phenotype:Observable characteristics of an organism, for example, hair type, eye color, height. • Genotype: The genetic identity of an individual, including alleles that do not show as outward characteristics. • Locus:The position on a chromosome where a gene, or some other sequence, is located. • Allele:One of two or more alternative forms of a gene; for example, allele "A" may produce a tall plant, while allele "a" gives a short plant. • Dominant allele: An allele that is fully expressed in the phenotype, regardless of the characteristics of its counterpart allele. • Recessive allele:An allele which is not expressed in the phenotype unless two copies are present in the individual, ie, in homozygous recessive condition. • Homozygous:the state of having two identical alleles of a particular gene (eg AA, aa). • Heterozygous: the state of having two different alleles of a particular gene (eg Aa). • Mendel’s Law of Segregation:Homologous chromosomes separate during the formation of gametes, and are distributed to different gametes, so that every gamete receives only one member of the pair. • Mendel’s Law of Independent Assortment:Homologous chromosomes are random distributed at the metaphase plate such that a mixture of maternal and paternal homologous assort to each gamete.

  3. Figure 14. 3 and 14.5 Mendel’s Peas 1. Parental Generation(P):Purple flowers x white flowers: Mendel bred (cross-pollinated) two types of pea plants that were true-breeding for flower color. 2.F1 Generation: When Mendel grew seed from this First Filial generation (F1) and looked at their flower color, he saw 100% purple flowers. Result:In the (F1)generation, the white trait was masked. 3. F2 GenerationMendel went one step farther, allowing the F1 to “self”.Result: The white trait re-appeared in the F2 generation in a ratio of ~3 purple plants to 1 white. Next Slide: Mendel’s Genius: He realized that these results were explainableif three things were true.

  4. Alleles controlling Flower color: Purple White Alleles controlling Pea shape: Wrinkled Round Figure 14.4: Mendel’s Genius: He realized that these results were explainable if three things were true. He hypothesized that: 1. Every trait (like flower color) is controlled by two "heritable factors” (genes), one from each parent -Each form of the gene is called anallele, and produces a slightly different protein 2.If the two alleles differ, one is dominant (observed in appearance or physiology) and one is recessive (only observed when an individual has two copies of the recessive allele). Dominant traits mask the appearance of recessive traits. 3.Alleles are randomly segregated into eggs and sperm when the homologues separate during Meiosis I, allowing all possible combinations of factors to occur in the gametes. Mendel's Law of Segregation- When gametes (eggs or sperm) are formed, the two factors (alleles) separate, and only one factor (allele) is present in each gamete. [ie: the homologues separate during meiosis] Homologous Chromosomes

  5. Table 14.1 The Results of Mendel’s F1 Crosses for Seven Characters in Pea Plants Result:For a monohybrid cross (single trait), Mendelian Genetics predicts that a 3:1 ratio (Dominant : recessive) will be seen in the F2 generation

  6. Phenotype (Observable characteristics) Genotype (Gene alleles) Purple PP (homozygous) 1 Pp (heterozygous) 3 Purple 2 Pp (heterozygous) Purple pp (homozygous) White 1 1 Ratio 3:1 Ratio 1:2:1 Figure 14.6 Phenotype versus genotype

  7. Doing a genetic cross (Monohybrid Cross = 1 gene): (Most of you have done this in lab already) Geneticists use letters to represent alleles. *Dominant trait = Capital letter ; recessive trait = a lowercase letter. *The same letter is used to indicate both alleles. *Remember, the definition of “Dominant” means…”SEEN IN THE PHENOTYPE” Examples = Flower color: P= purple, p= white = Seed color: Y= yellow, y = green = Seed shape: W = wrinkled, w = round In Humans... = Widow's peak: W = widow's peak, w = continuous hairline (which are you?) = Freckles: F = Freckles, f = no freckles (which are you?) = Earlobes: E = unattached, e = attached (which are you?) = Cystic fibrosis C = no CF, c = cystic fibrosis E-Z steps for doing genetics problems: 1. Indicate the genotype of the parents using letters 2. Determine what the possible gametes are 3. Determine the genotype and phenotype of the children after reproduction. To consider every type of offspring possible, use a Punnett Square in which all possible types of sperm are lined up vertically and all types of eggs are lined up horizontally: 4. Fill in the squares by "multiplying" the alleles from mom and dad:

  8. Human Genetics: Recessively Inherited Disorders 1. Autosomal recessive human disorders: Show up only in the homozygous recessive person (aa). With two carrier parents, there is a 1/4 chance of an affected child. There is 2/4 chance of a carrier child (50%) A.Cystic fibrosis: Mutation in CFTR (Chloride channel) B.Tay-Sachs: Mutation in Hexaminidase A gene.(Jewish - Ashkenazi) C. Sickle-cell disease: Mutation in hemoglobin gene. The most common inherited disease of African-Americans (1:400 affected). With these parents, what is the chance of - giving birth to an affected child? - giving birth to a child who is also a carrier? Heterozygote advantage; For some disease alleles, being a heterozygote (‘carrier’) offers protection against another disease. Examples: 1. Sickle cell disease: Ss carriers are resistant to malaria 2. Cystic fibrosis: Cc carriers are resistant to cholera unaffected carrier mother

  9. Heterozygote advantage One in 10 Africans are carriers (heterozygotes) for the sickle cell trait, even though the recessive condition, SCD, can be fatal. Why is the incidence so high? Sadly, over 1 million people / year (mostly children, pregnant women) die from Malaria (2,700 per day)…. People who are HbBHbB= normal hemoglobin, but may succumb to malaria if they live in a region with endemic malaria. HbSHbS= have sickle cell diseaseSCD and are anemic, in pain, and may suffer strokes or death. HbBHbS= heterozygotes suffer mild symptoms of SCD, but if bitten by a mosquito carrying the malaria parasite, the parasite life cycle is interrupted due to the difficult-to-digest, irregular HbS In parts of the world where malaria is common, the incidence of SCD is common because there is an advantage to being a carrier. P. falciparum malaria Frequency of Hbs allele 1–10% 10–20%

  10. Human Genetics: Dominantly Inherited Disorders Autosomal dominant human disorders: child will show the phenotype if he / she receives just 1 allele from either parent (OR if a spontaneous mutation appears in the zygote). A.Achondroplasia (dwarfism):Cause: Mutation in FGFR3 (fibroblast growth factor receptor 3 gene) severely limits bone growth. AA = Dominantlethal - fatal. Aa = dwarfism. aa = no dwarfism. 99.96% of all people in the world are (aa). (Fig 14.15) B. Polydactyly(extra fingers or toes): PP or Pp = extra digits, aa = 5 digits. 98% of all people in the world are (pp). C. Huntington's choreaCause: Mutation of Huntingtingene damages nerves and brain. Dominantlethal (HH = fatal). Heterozygotes (Hh) live to be ~40 or so, then their nervous system starts to degenerate. (Woody Guthrie). D. Marfan's syndrome - Cause: Defect in Fibrillin gene involved in connective tissue; affects skeletal system, eyes, and CV system. (Pleiotropic). *Abraham Lincoln, Flo Hyman With these parents, what is the chance of - giving birth to an affected child? - giving birth to a child who is also a carrier? E. Progeria -Cause: Mutation of Lamin-A Gene. Defective Lamin-A makes the nucleus unstable, leading to aging, CV disease, stroke (Note: since children with Progeria don’t live to reproduce, this is an example of a spontaneous (rather than inherited), Dominant disorder.

  11. Fig 14.7: Is there a way to distinguish between a purple heterozygote (Pp) and a purple homozygous dominant(PP)? Yes = A testcross: Mate the unknown genotype (Purple, but PP or Pp?) with a recessive genotype: white (pp) in this case. If offspring are 100% purple: the unknown genotype was homozygous dominant(PP), to begin with; and 100% of the offspring will be heterozygotes. If offspring are 50% purple And 50% white: the unknown genotype was a heterozygote (Pp), To begin with. So now you know!

  12. Fig 14.8: Studying Two Traits at once: (Dihybrid Cross) AA or Aa = Purple; aa = white BB or Bb = Tall; bb = short Result: Classic 9:3:3:1 ratio:9 Purple & Tall, 3 Purple & short, 3 white & Tall, 1 white & short One really important thing that Mendel noticed from this type of cross was that the two traits (like flower color, height) are inherited independently - not together as a unit. This has become known as Mendel's Law of Independent Assortment - Genes for various traits assort into gametes independently (due to homologues lining up randomly at the metaphase plate).

  13. (See Fig 14.9) A simpler way to do a Dihybrid Cross or beyond... (or, how to avoid making anything larger than a simple Punnett square) Mendelian genetics reflect the laws of probability For a dihybrid cross - the chance that 2 independent events will occur together is the product of their chances of occurring separately. The chance of yellow (YY or Yy) seeds = 3/4 (the dominant trait) The chance of round (RR or Rr) seeds  = 3/4 (the dominant trait) The chance of green (yy) seeds = 1/4 (the recessive trait) The chance of wrinkled (rr) seeds = 1/4 (the recessive trait) So... (Remember multiplication with fractions..its easy! Just multiply across the top and the bottom! ) The chance of yellow and round = 3/4 x 3/4 = 9/16 The chance of yellow and wrinkled = 3/4 x 1/4 = 3/16 The chance of green and round = 1/4 x 3/4 = 3/16 The chance of green and wrinkled = 1/4 x 1/4 = 1/16 9:3:3:1 - Sound familiar? And no Punnett square needed (whew). Therefore, you can avoid doing a Punnett square if you can reduce the problem to a series of probability statements. Next, you guessed it, a tri-hybrid cross

  14. P Generation White CWCW Red CRCR  CR Gametes CW Pink CRCW F1 Generation 1⁄2 1⁄2 Gametes CW CR Sperm 1⁄2 1⁄2 CR CW Eggs F2 Generation 1⁄2 CR CR CR CR CW 1⁄2 Cw CR CW CW CW Figure 14.10 Incomplete dominance in snapdragons Incomplete Dominance: 2 incompletely dominant alleles(no ‘recessives’) CR and CW The F1 heterozygote (CRCW) is intermediate in phenotype (pink!) compared to the two parentals. Why does this happen? Just like with Purple:white, this comes about when one of the alleles lacks the ability to produce a pigment, but in this case, the loss IS detectable in the phenotype. Note that traits do not ‘blend’: all 4 colors result in the F2.

  15. Table 14.2 Determination of ABO Blood Group by Multiple Alleles (Co-dominance) ABO Blood Types3 alleles; (Each dominant allele codes for the linkage of a glycophorin sugar molecule to the surface of RBCs.) 2 are dominant (IA and IB ), 1 is recessive (i) This results in 4 possible phenotypesWhich do YOU have?

  16. AaBbCc AaBbCc aabbcc Aabbcc AABBCc AABBCC AaBbcc AaBbCc AABbCc 20⁄64 15⁄64 Fraction of progeny 6⁄64 1⁄64 Figure 14.12 Polygenic inheritance: skin color 2 or more genes act additively to produce a phenotype Ex: human height, skin color Skin pigmentation

  17. Pleiotropy:(Greek: pleion, more) How 1 gene can affect multiple aspects of the phenotype.Ex: Sickle Cell Disease, Cystic Fibrosis, Marfan’s Disease Many organ systems affected, since each mutation changes a protein involved in many bodily functions

  18. 9 Black :(BBEE, BbEE, BbEe) 4 Yellow : (BBee, Bbee, bbee) 3 Chocolate (bbEE, bbEe) Fig 14.11 Epistasis: Presence of alleles at one locus affect the expression of alleles at a second locus. (Epi = above) Other examples: Albinism in humans[In class] Can use probability to solve these; Variations on a 9:3:3:1 result (9:3:4; 9:7, 15:1)Make sure you can work an epistasis genetics problem!Bonus: What genotype was the yellow labBB, Bb, or bb? The nose knows!

  19. Fig 15-10: Sex-linked disorders in humans: X Chromosome: • Females: Inherit 2 X chromosomes, one from each parent - 2 alleles for all genes (Fig 15.9) • Males: Inheritone X chromosome (from their mother), so only 1 allele for every gene, Technically, males are hemi-zygous - neither homozygous nor heterozygous at the X! • If that one allele is mutated, the male will be affected (show the trait). • Daughters are almost always unaffected (carrier females) but have a 50% chance of giving birth to an affected son and a 50% chance of giving birth to a carrier daughter. • Affected females are more rare because they must inherit two recessive alleles. • List the 5 possible genotypes and phenotypes: • Color blindness: sex-linked recessive Defect in opsin gene; XA = normal vision, Xa = color blind • Duchenne muscular dystrophy: Affects 1:3500 US. Males. Children lack Dystrophin (X chromosome). Progressive muscle weakness and loss of coordination leads to death in early adulthood. • Hemophilia:Affected children lack the blood clotting factor VIII. Queen Victoria was a carrier; thus her sons had a 50% chance of being affected and her daughters had a 50% chance of being carriers.

  20. Fig 15.11: Dosage compensation in females: X-Chromosome Inactivation (XCI) Females normally inactivateone of their 2 Xs in early embryonic development (XCI). The inactivated X-chromosome condenses into aBarr Body This inactivation is a random event, thusfemales (this means all K101 females) are Mosaics 50% of our cells have one active X, and 50% have the other active X. Female calico kitty: mosaic! Dosage Compensation = inactivation of 1 X chromosome per cell that equalizes the expression of X-linked genes in males and females 1 Nucleus 2 2 Barr body (randomly inactivated chromosome) Followed by mitosis 3 50% of cells 50% of cells

  21. Just a mention about Multifactorial Diseases: (and see Fig 14.13) Cancer, Autism and Schizophrenia are multi-factorial - they are not 'caused' by one dominant or recessive gene, but may be the result of several inherited genetic susceptibilities PLUS environmental factors. Example: Cancer: Can be an inherited susceptibility - loss of tumor suppressors (BRCA1, BRCA2, p53), oncogene activation, OR a spontaneous mutation PLUS environmental factors . Example: Autism: at least 4 genes linked to this neurodevelopmental condition, particularly on Chromosomes 2, 3, 7, 15, and X. Some have to do with language development, some with brain development, some with producing brain neurotranmitters. Example: Schizophrenia: several genes linked to the development of this disorder, including Chromosome 22plus a combination of family genetics and environmental factors.

  22. Two homologous chromosomes undergo synapsis in meiosis Fig 15.5: “Linked Genes” Mendel showed that genes are inherited independently, but in 1910,Thomas Hunt Morgan (Nobel Prize 1933) and Alfred Sturtevant showed that Independent assortment does not apply if 2 genes are very close together “Linked” on a chromosome.Crossing Over during meiosis results in recombination between genes. The closer two genes are on a chromosome, the more likely that they are linked and will travel together. The farther apart they are, the more likely they will be separated. Can use this information to create maps of gene order on chromosomes (See Fig 15.7. 15.8 for more info) Crossing-over between a pair of homologous chromatids Meiosis I Meiosis II Parental Recombinant Recombinant Parental Four haploid cells produced

  23. Making a physical map of a chromosome based on recombination frequency (crossing over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a estimate of how closely linked genes were on the Drosophila chromosomes. Recombination frequency: Frec (or % crossing over)) 391 recombinants X 100= 17% 2300 offspring Thus, we can say that these genes are 17 “Centimorgans” or “map units” apart (cM or m.u.) Try it: Problem 1 DdEe x ddee 65 Ddee, 25 ddEe, 256 DdEe, 266 ddee Problem 2: AaEe x aaee 424 Aaee, 430 aaEe, 70 AaEe, 76 aaee Make a map of these 3 genes, assuming E is in the middle 0 50 100 Grey, normal wings BbVv BV bv Bv bV Grey, normal Black, vestigial Grey, vestigial Black, normal bv Black, vestigial wings bbvv bbvv Bbvv bbVv BbVv 575 575 575 575 Expected results: independent assortment Parentals Recombinants 965 944 185 206 Actual results

  24. Making a physical map of a chromosome based on recombination frequency (crossing over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a estimate of how closely linked genes were on the Drosophila chromosomes. • Recombination frequency: Frec (or % crossing over)) • 391 recombinants X 100= 17% • 2300 offspring • Thus, we can say that these genes are 17 “Centimorgans” or “map units” apart (cM or m.u.) • Try it: Problem 1 • DdEe x ddee • 65 Ddee, 25 ddEe, 256 DdEe, 266 ddee • 1. Circle the parentals • 2. Add the recombinants =65 + 25 =90 • 3. Add all the offspring = 612 total • 4. 90 / 612 X 100 = 14.7 D-E are 14.7 m.u. apart • Problem 2: • AaEe x aaee • 424 Aaee, 430 aaEe, 70 AaEe, 76 aaee • 1. Circle the parentals • 2. Add the recombinants =424 + 430 =854 • 3. Add all the offspring = 1000 total • 4. 854 / 1000 X 100 = 85.3 • E-A are 85.3 m.u. apart • Make a map of these 3 genes, assuming E is in the middle • D E A • 0 50 100 Grey, normal wings BbVv BV bv Bv bV Grey, normal Black, vestigial Grey, vestigial Black, normal bv Black, vestigial wings bbvv bbvv Bbvv bbVv BbVv 575 575 575 575 Expected results: independent assortment Parentals Recombinants 965 944 185 206 Actual results

  25. Hardy-Weinberg ‘Equlilbrium’The Hardy-Weinberg equation : Basic assumptions1. Mutation is not occurring2. Natural selection is not occurring3. The population is infinitely large4. All members of the population breed5. All mating is totally random6. Everyone produces the same number of offspring7. There is no migration in or out of the population(This is never fully true for any population, but If you know the allele frequency,Can determine the genotype frequenciesAllele Frequency: Dom + rec = 1 (p + q = 1)Genotype Frequency: p2 + 2pq + q2 = 1

  26. Hardy Weinberg:Basic Relations1. Two alleles at a gene - A and a2. Frequency of the A allele = p3. Frequency of the a allele = q4. Allele frequencies: p + q = 15. Genotype frequencies: p2 (AA) + 2pq (Aa) + q2 (aa) = 1Problem 1. Allele B is present in a large, random mating population at a frequency of 54 %1. What is the frequency of allele b? ______________2. What proportion of the population is expected to be -homozygous for B (BB)? __________ ______ -homozygous for b (bb)? __________ -heterozygous? (Bb) ______________Problem 2. In a large, random-mating population, the frequency of homozygous recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals. What is the frequency of 1. the recessive disease aa? ______________2. the recessive allele a? ______________3. the dominant allele A? ______________4. the heterozygous individual Aa? ______________5. the homozygous dominant individual AA? ______________

  27. Hardy Weinberg:Basic Relations1. Two alleles at a gene - A and a2. Frequency of the A allele = p3. Frequency of the a allele = q4. Allele frequencies: p + q = 15. Genotype frequencies: p2 (AA) + 2pq (Aa) + q2 (aa) = 1Problem 1. Allele B is present in a large random mating population at a frequency of 54% p1. What is the frequency of allele b? q _0.46________2. What proportion of the population is expected to be -homozygous for B (BB)? p2___0.291_____-homozygous for b (bb)? q2___0.216_____-heterozygous? (Bb) 2pq_ 2(.54)(.46) = 0.495__Problem 2. In a large, random-mating population, the frequency of homozygous recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals. 1. What is the frequency of the recessive disease aa? q290/1000 = 0.092. What is the frequency of the recessive allele a? q sq.rt (0.09) = 0.33. What is the frequency of the dominant allele A? p1 - 0.3 = 0.7 4. What is the frequency of the heterozygous individual Aa? 2pq2(.3)(.7) = 0.425. What is the frequency of the homozygous dominant individual AA? p2 = 0.49

  28. (Not on exam): EXCITING and SURPRIZING new results in Nature, March 17, 2005about the Sequence of the X-chromosome and X-chromosome inactivation. Nature434, 400 - 404 (17 March 2005); doi:10.1038/nature03479 X-inactivation profile reveals extensive variability in X-linked gene expression in females LAURA CARREL AND HUNTINGTON F. WILLARD In human females, up to 15% of all genes on the ‘inactive’ X escape XCI, and are expressed from both the active and the inactive X. An additional 10% of X-linked genes show variable patterns of inactivation and are expressed to different extents in different women. Nature434, 400 - 404 (17 March 2005); doi:10.1038/nature03440 The DNA sequence of the Human X chromosome The International Human Genome Sequence Consortium We have determined 99.3% of the euchromatic sequence of the human X-chromosome. We found 1,098 genes in this sequence, of which 99 encode proteins expressed in the testis. Nature434, 279 - 280 (17 March 2005); doi:10.1038/434279a Genome biology: She moves in mysterious ways CHRIS GUNTER The human X chromosome is a study in contradictions. The detailed sequence of the X, and a survey of inactivated genes in females, help to illuminate this unique 'evolutionary space'…

  29. Objectives: Define the terms phenotype, genotype, locus, allele, dominant allele, recessive allele, homozygous, and heterozygous. Describe Mendel’s Laws (principles) of segregation and independent assortment. Solve genetics problems involving monohybrid, dihybrid, (trihybrid) and test crosses. Use probability (3/4 x 3/4…) to predict the outcomes of genetic crosses. Be able to discuss the difference between autosomal dominant and autosomal recessive disorders, being able to give specific examples and work genetics problems. Solve genetics problems involving incomplete dominance, codominance / multiple alleles, and epistasis; Pleiotropy: discuss how a single gene can affect many features of the organism. Discuss the the inheritance of X-linked genes in mammals. Discuss what it means to be hemizygous, and what is means to be a mosaic. What is a Barr Body, and what is dosage compensation? Define linkage, and relate it to specific events in meiosis. Show how data from a test cross involving alleles of two loci (‘two point test cross’) can be used to distinguish between independent assortment and linkage. Know how to calculate allele frequency using the Hardy-Weinberg equation. Explain how prenatal testing, carrier screening, and newborn screening can be used to detect genetic conditions before and after birth.