Chapter 8: Patterns Of Inheritance. Johann Gregor Mendel set the framework for the study of genetics.

Transcription

1 Chapter 8: Patterns Of Inheritance Johann Gregor Mendel set the framework for the study of genetics.

2 Mendel s Laws I. The Study of Genetics Has Ancient Roots A. Hippocrates (c.460-c.370 BC), father of western medicine 1. particles, pangenes, travel from all parts of the body to the eggs or sperm a. pangenes are passed on to the next generation b. changes that occur in someone during their life will be passed on this way II. Genetics began in an abbey garden A. in the early part of the 19 th century, blending hypothesis was widely accepted 1. maternal & paternal hereditary materials are mixed in the offspring a. like red & white paint making pink paint 1) cannot separate red & white paint from pink a) over time a population should become genetically uniform B. Gregor Mendel documented evidence for the particulate hypothesis of heredity 1. parents pass on to their offspring discrete heritable units, genes, that retain their separate identities in the offspring 2. this hypothesis views genes like a deck of cards they can be shuffled and passed down in undiluted form C. In 1843, at the age of 21, Gregor Mendel entered an Augustinian monastery

3 D. In 1851, he left the monastery to study physics & chemistry for 2 years at the University of Vienna 1. Christian Doppler, a physicist, encouraged his students to learn science through experimentation & to use mathematics to help explain natural phenomena 2. Franz Unger, a botanist, encouraged Mendel s interest in the causes of variation in plants E. Gregor Mendel lived & worked in an abbey in Brunn, Austria (now Brno, in the Czech Republic) F. the field of genetics began in the 1860 s when Gregor Mendel deduced the fundamental principles of genetics by breeding garden peas (began in 1857) 1. peas have a short generation time

4 2. easy to control matings among plants 3. each mating leads to a large number of offspring 4. peas exhibit characters w/ different traits a. flower color (character): purple & white (traits) b. seed color (character): yellow & green (traits) c. seed shape (character): round & wrinkled (traits) Mendel identified seven pea plant characteristics.

5 Phenotypes are physical expressions of traits that are transmitted by alleles. Capital letters represent dominant alleles and lowercase letters represent recessive alleles. The phenotypic ratios are the ratios of visible characteristics. The genotypic ratios are the ratios of gene combinations in the offspring, and these are not always distinguishable in the phenotypes.

6 G. Mendel would start an experiment with the crossing of 2 different true-breeding plants P generation 1. plants that after multiple generations of self-fertilization had produced the same trait a. this resulted in the F 1 generation (hybrids) 2. F 1 s were allowed to self-pollinate or were cross-pollinated a. this resulted in the F 2 generation w/ a 3:1 ratio 3. From this work, Mendel revealed important aspects of heredity Mendel s Experimental Process a. he documented a particulate mechanism for inheritance b. law of segregation c. law of independent assortment P Generation: (true-breeding parents) F 1 Generation: All F 1 s are purple Fertilization among F 1 s F 2 Generation: 3 purple: 1 white

7 III. Mendel s law of segregation describes the inheritance of a single character A. Mendel s results lead him to develop 4 hypotheses 1. there are alternative versions of genes that account for variations in inherited characters a. P & p (in the case of flower color) 2. for each character, an offspring inherits 2 alleles 1 allele from each parent a. offspring can inherit 2 dominant alleles, PP, homozygous dominant b. offspring can inherit 2 recessive alleles, pp, homozygous recessive c. offspring can inherit 2 different alleles, Pp, heterozygous *Genotype: PP = Phenotype: purple flower *Genotype: Pp = Phenotype: purple flower *Genotype: pp = Phenotype: white flower P Generation: (true-breeding parents) Explanation of Mendel s Experimental Process Genetic makeup (alleles) F 1 Generation: all are purple and heterozygous: Pp F 2 Generation: 75% purple: 25% white PP Gametes: All P All p Gametes: 1/2 P Alleles ½ p segregate Fertilization among F 1 s during meiosis Eggs from F 1 : P p Sperm from F 1 : P p PP Pp Pp pp Pp pp

8 3. if the 2 inherited alleles are different, then 1 allele determines the organism s appearance & is called the dominant allele & the other allele has no noticeable effect on the organism s appearance & is called the recessive allele 4. a sperm/egg carries only 1 allele for each inherited character because allele pairs separate (segregate) during gamete formation 1. anaphase I & II of meiosis (law of segregation) B. Punnett square provides probabilities of outcomes from a mating In one of his experiments on inheritance patterns, Mendel crossed plants that were true-breeding for violet flower color with plants true-breeding for white flower color (the P generation). The resulting hybrids in the F 1 generation all had violet flowers. In the F 2 generation, approximately three quarters of the plants had violet flowers, and one quarter had white flowers. P Generation: (true-breeding parents) F 2 Generation: 75% purple: 25% white Explanation of Mendel s Experimental Process Genetic makeup (alleles) F 1 Generation: all are purple and heterozygous: Pp PP Gametes: All P All p Gametes: 1/2 P Alleles ½ p segregate Fertilization among F 1 s during meiosis Eggs from F 1 : P p Sperm from F 1 : P p PP Pp Pp pp Pp pp

9 IV. Homologous chromosomes bear the alleles for each character A. same length, centromere placement & code for the same characters B. humans have 23 pairs of homologous chromosomes C. loci: locations of genes along a chromosome

10 V. Law of independent assortment is revealed by tracking 2 characters at once A. Mendel wondered what would happen if he crossed a true-breeding yellow-round (YYRR) seed producing plant w/ a true-breeding green-wrinkled (yyrr) seed producing plant 1. he knew he would get only yellow-round F 1 s 2. but what would he get in the YyRr F 2 s a. would the YR genes stay together? 1) dependent assortment b. would the YR genes ever sort into different gametes? 1) independent assortment

11 B. Labrador retriever coat color & progressive retinal atrophy (PRA) 1. cross between 2 labs that are heterozygous for both characters a. BbNn (female) x BbNn (male) 1) 1 character = 4 squares; 2 characters = 16 squares FOIL: BbNn x BbNn to derive gametes produced by each lab BN Bn bn bn BN Bn bn bn Mating of two heterozygous labs (black coat & normal vision) BbNn x BbNn BBNN BBNn BbNN BbNn BBNn BBnn BbNn Bbnn Phenotypic ratio of the offspring 9 Black coat, normal vision 3 Black coat, blind (PRA) 3 Chocolate coat, normal vision 1 Chocolate coat, blind (PRA) BbNN BbNn bbnn bbnn "Labrador retriever bulaj1". Licensed under CC BY-SA 1.0 via Wikimedia Commons - ile:labrador_retriever_bulaj1.jpg#/me dia/ "Labrador Retriever chocolate Hershey sit (cropped)" by Rob Hanson - Hershey - Chocolate Lab. Licensed under CC BY 2.0 via Wikimedia Commons - hocolate_hershey_sit_(cropped).jpg#/media/ BbNn Bbnn bbnn bbnn

12 VI. Geneticists can use a testcross to determine unknown genotypes A. mating an individual w/ an unknown genotype to a homozygous recessive individual B. cannot tell the genotype of a black lab 1. mating it w/ a chocolate lab should reveal black lab s genotype What is the genotype of the black dog? Testcross Genotypes B? or bb Two possibilities for the black dog: BB or Bb Gametes b Bb B or B b b Bb bb Offspring All black 1 black:1 chocolate "Labrador retriever bulaj1". Licensed under CC BY-SA 1.0 via Wikimedia Commons - ile:labrador_retriever_bulaj1.jpg#/me dia/ "Labrador Retriever chocolate Hershey sit (cropped)" by Rob Hanson - Hershey - Chocolate Lab. Licensed under CC BY 2.0 via Wikimedia Commons - hocolate_hershey_sit_(cropped).jpg#/media/

13 VII. Genetic traits in humans can be tracked through family pedigrees A. the homozygous recessive condition codes for attached ear lobes First generation (grandparents) Ff Ff ff Ff Second generation (parents, aunts & uncles) Third generation (two sisters) Female Male Attached Free FF or Ff ff ff Ff Ff ff ff FF or Ff

14 VIII. Many inherited traits in humans are controlled by a single gene A. a trait that is encoded in a dominant gene does not mean that it is better or most common 1. a dominant gene can mask a recessive gene a. the heterozygote will display the dominant phenotype 1) Pp codes for purple flower color in pea plants B. a trait that is most common in a population is called a wild type trait 1. not having freckles 2. normal pigmentation C. a trait that is less common in a population is called the mutant trait 1. having freckles Dominant Traits Recessive Traits 2. albinism Freckles No freckles Widow s peak No widow s peak

15 D. autosomal disorders in humans 1. carried on chromosomes other than the sex chromosomes 2. some are recessive disorders a. show up only in a homozygous recessive individual 1) usually born to parents that are both heterozygous a) i.e., parents are carriers b. Cystic fibrosis is most common in Caucasians 1) allele is carried by ~1/31 Americans 2) excessive secretion of thick mucus a) can interfere w/ breathing, digestion & liver function b) person is more at risk for bacterial infections

16 3. some are dominant disorders a. show up in homozygous dominant & heterozygous individuals b. at least one parent had to have had the disorder c. polydactyly: having extra fingers & toes d. webbed fingers & toes e. achondroplasia: a form of dwarfism 1) head & torso develop normally 2) arms & legs do not develop normally 3) occurs in heterozygotes (Aa) 4) AA individuals causes death of the embryo 5) more than 99.99% of the population are homozygous recessive, aa f. dominant alleles that cause death are much less common than lethal recessives g. Huntington s disease 1) does not appear until middle age 2) degenerative disorder of the nervous system 3) genetic test is available

17 Human Autosomal Disorders Recessive Disorder Major Symptoms Incidence *Albinism *Lack of pigment in the skin, hair & eyes 1 in 22,000 *Cystic Fibrosis *Excess mucus in the lungs, digestive tract, liver; 1 in 2,500 liver; increased risk of infections (Caucasians) *Galactosemia *Accumulation of galactose in tissues; 1 in 100,000 mental retardation; eye & liver damage *Phenyketonuria *Cannot convert the amino acid phenylalanine to the amino 1 in 10,000 (PKU) acid tyrosine; lack of normal skin pigmentation; mental retardation *Sickle-cell disease *RBC s sickle under oxygen tension; damage to many tissues 1 in 400 (African-Americans) *Tay-Sachs disease *Lipid accumulation in brain cells; mental deficiency; 1 in 3,500 blindness, early death (Jews: central Europe) Dominant Disorder Major Symptoms Incidence *Achondroplasia *Dwarfism 1 in 25,000 *Alzheimer s disease Mental deterioration; usually strikes in middle age (familial) *Huntington s disease *Mental deterioration & uncontrollable movements; 1 in 25,000 strikes in middle age

18 Variations In Mendel s Laws I. Incomplete dominance results in intermediate phenotypes A. Mendel s heterozygous F 1 s revealed the dominant phenotype 1. an example of complete dominance B. F 1 s that have 2 different genes that are incompletely dominant to one another reveal a phenotype that is intermediate to either true-breeder 1. pink snapdragons 2. mild hypercholesterolemia disease HH Homozygous for ability to make LDL receptors Hh Heterozygous for ability to make LDL receptors hh Homozygous for inability to make LDL receptors LDL LDL LDL LDL LDL Max. # of LDL receptors Normal Cell Mild disease Fewer LDL receptors No LDL receptors Severe disease

19 II. Many genes have more than 2 alleles in the population A. ABO blood group system has 3 alleles in the population 1. I A, I B & i a. I A codes for a specific carbohydrate to coat the surface of rbc b. I B codes for a different carbohydrate than I A c. i does not code for a carbohydrate to coat rbc surface 2. a person w/ blood type A can be either homozygous or heterozygous a. I A I A or I A i 3. a person w/ blood type B can be either homozygous or heterozygous a. I B I B or I B i 4. a person w/ blood type AB can only be heterozygous a. I A I B 5. a person w/ blood type O will have an ii genotype 6. alleles I A & I B are codominant 7. alleles I A & I B are completely dominant to i 8. universal recipient & donor

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21 III. A single gene may affect many phenotypic characters A. pleiotropy 1. sickle-cell disease a. cells sickle under oxygen tension 1) body rapidly destroys these abnormal cells a) causes anemia & weakening of the body 2) clog capillaries a) can damage the heart, brain & kidneys b. kills ~ 100,000/yr

22 c. most common in those of African descent 1) 1/400 African Americans 2) 1/10 are carriers a) heterozygote advantage in areas where malaria is a threat *Plasmodium enters rbc causes sickling body destroys most of the sickle cells/parasites d. caused by codominant alleles

23 IV. A single character may be influenced by many genes A. polygenic inheritance 1. let s pretend skin color is controlled by a total of 6 genes a. AABBCC codes for dark skin b. aabbcc codes for light skin c. AaBbCc codes for medium skin d. Aa, Bb & Cc are incompletely dominant to one another

24 V. The environment affects many characters A. nutrition height B. sun exposure skin color C. rbc count - altitude

25 I. Crossing over produces new combinations of alleles A. Thomas Hunt Morgan, fruit flies 1. Drosophila melanogaster 2. linked genes would produce 1:1 parental phenotype ratio 3. unlinked genes would produce a 1:1:1:1 ratio The Experiment Gray body, long wings (wild type) X Black body, vestigial wings The Genetics of the experiment G L g l g l g l GgLl ggll Female Gray Long Offsprin Black vestigial g Gray vestigial Male Black Long G L g l G l g L g l Eggs Sperm G L g l Offsprin g G l g L g l g l g l g l Parental phenotypes Recombinant phenotypes Parental Recombinant Recombination = 391 recombinants = 0.17 (17%) frequency 2,300 total offspring

26 Sex Chromosomes & Sex-Linked Genes I. Chromosomes determine sex in many species A. X-Y system in humans 1. male determines offspring gender because he produces 2 different kinds of gametes a. sperm: 22 autosomes + X chromosome; 22 autosomes + Y chromosome b. egg: 22 autosome + X chromosome c. Y chromosome has SRY gene which triggers testis development 1) codes for proteins that regulate other genes on the Y Parents (diploid ) Male 44 + XY Female 44 + XX Gametes (haploid) 22 + X 22 + Y 22 + X Offsprin g (diploid) 44 + XX Female 44 + XY Male

27 II. Incubation temperature for some reptiles determines gender A. green sea turtles B. crocodiles Male 22+ X Female 22 + XX Fir0002/Flagstaffotos Male Female Male Female 76 + ZZ 76 + ZW A Honey Bee. Author: Erik Hooymans.

28 III. Sex-linked genes exhibit a unique pattern of inheritance A. female fruit fly w/ red eyes: X w+ X w+ or X w+ X w B. female fruit fly w/ white eyes: X w X w C. male fruit fly w/ red eyes: X w+ Y D. male fruit fly w/ white eyes: X w Y E. most human sex-linked genes are on the X chromosome & not the Y 1. X has many more genes (1,100) than the Y (78) a. will usually consider X-linked genes 1) dads pass X-linked alleles on to his daughters 2) dads pass Y-linked alleles on to his sons 3) moms pass X-linked alleles on to both IV. Human sex-linked disorders affect mostly males

29 A. assuming N codes for normal color vision & n codes for color-blindness: 1. homozygous normal female x color-blind male a. X N X N x X n Y b. can a daughter be color-blind? No c. can a son be color-blind? No d. what is the p of having offspring w/ normal vision? 100% e. which gender will be a carrier? Female f. can males be carriers? no X N X N X n Y Eggs Sper m X N X N X n X N X n X N X n A. Y X N Y X N Y

30 2. heterozygous female x normal male a. X N X n x X N Y b. can a daughter be color-blind? No c. can a son be color-blind? Yes 1) & he will be hemizygous for this X linked gene d. what is the p of a daughter being a carrier? 50% 3. female (whose father was color-blind) x color-blind male a. X N X n x X n Y b. can a daughter be color-blind? Yes c. can a son be color-blind? Yes d. what is the p of a daughter being color-blind? 50% e. what is the p of a son being color-blind? 50% X N X N X n Y X N X n X N Y X N X n X n Y Eggs Sper m X N X N X n X N X n X N X n Y Sper m X N Y Sper m X n Y X N Y X N X N X N X N Y X N X N X n X N Y Eggs Eggs X N Y X N X n X n Y X n X n X n Y X n X n A. B. C.

31 Genetic Crosses/Punnett Squares/Pedigree I. In cattle, hornless (H) is completely dominant over the horned condition (h). A heterozygous hornless bull is bred to a horned cow. What is the probability of this leading to an offspring that will be able to produce horns? A. Determine how many different characters are in the problem B. Draw up the appropriate Punnett square. C. Place gamete genes of an individual on same side of Punnett square. D. Write in the appropriate offspring genotypes. E. Determine probability of outcome. *heterozygous hornless bull: H h *horned cow: h h *50 % probability of horned offspring Hh hh Hh hh

32 II. Give the probabilities of offspring blood types that are possible from one parent who is heterozygous for B type blood and the other parent who is type AB blood. A. type A blood: homozygous: I A I A ; heterozygous: I A i B. type B blood: homozygous: I B I B ; heterozygous: I B i C. type AB blood: I A I B D. type O blood: ii E. Identify the # of characters in the problem. F. Draw up the appropriate Punnett square. G. Place gamete genes of an individual on same side of Punnett square. H. Write in the appropriate offspring genotypes. I. Determine probabilities. J. heterozygous B: I B i K. type AB: I A I B L. 25% p offspring will have AB blood type I A I B I A i M. 25% p offspring will have A blood type N. 50% p offspring will have B blood type I B I B I B i

33 III. A woman and her husband have been notified that their child-to-be is going to be a boy. Red-green colorblindness is caused by a recessive X-linked allele (n). The woman has normal color vision (her father was color-blind) and her husband is color-blind. Is it possible for them to have a son with normal vision? What is the p of the outcomes for this son? Is it possible for them to have a daughter w/ normal vision? A. Determine how many different characters are in the problem. B. Draw up the appropriate Punnett square. C. Place gamete genes of an individual on same side of Punnett square. D. Write in the appropriate offspring genotypes. E. Determine probability of outcomes. F. woman: X N X n G. husband: X n Y H. It is possible for them to have a normal son. X N X n X n X n a. 50% p I. What about a normal daughter? a. 50% p X N Y X n Y

34 IV. In cattle, coat color is determined by codominant alleles: red (C R C R ), white (C W C W ) or roan (C R C W ). Roan cattle have both red and white hairs. Some cattle have horns (hh) or lack horns (H-). A cross between a roanhornless (heterozygous hornless) cow w/ a red-horned bull will lead to what expected probabilities for each of the different phenotypes? A. Determine how many different characters are in the problem. B. Draw up the appropriate Punnett square. C. Place gamete genes of an individual on same side of Punnett square, using FOIL. D. Write in the appropriate offspring genotypes. E. Determine probability of outcomes. F. roan-hornless cow: C R C W Hh G. red-horned bull: C R C R hh H. red-hornless: 4/16 C R H C R h C W H C W h I. red-horned: 4/16 J. roan-hornless: 4/16 C R h C R C R Hh C R C R hh C R C W Hh C R C W hh K. roan-horned: 4/16 C R h C R C R Hh C R C R hh C R C W Hh C R C W hh C R h C R C R Hh C R C R hh C R C W Hh C R C W hh C R h C R C R Hh C R C R hh C R C W Hh C R C W hh

35 V. Is the allele affecting individuals in this pedigree dominant or recessive and X-linked or autosomal?