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Today: some things Mendel did not tell us… plus Mapping and Epigenetics –Exam #3 W 7/30 in class (bonus #2 due)– Single genes controlling a single trait are unusual. Inheritance of most genes/traits is much more complex… Dom. Rec. Rec. Dom. Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple. Fig 4.4 Sickle-cell anemia is caused by a point mutation Fig 4.7 Sickle and normal red blood cells Fig 4.7 Sickle-Cell Anemia: A dominant or recessive allele? Mom = HS Dad H or S H HH Mom or S HS HS SS S=sickle-cell H=normal Fig 4.7 Dad = HS possible offspring 75% Normal 25% Sickle-cell Coincidence of malaria and sickle-cell anemia Fig 24.14 Sickle-Cell Anemia: A dominant or recessive allele? Mom = HS Dad H or S H HH Mom or S HS HS SS S=sickle-cell H=normal Fig 4.7 Dad = HS possible offspring Oxygen transport: 75% Normal 25% Sickle-cell Malaria resistance: 75% resistant 25% susceptible The relationship between genes and traits is often complex Complexities include: • Complex relationships between alleles Fig 3.18 Sex determination is normally inherited by whole chromosomes or by number of chromosomes. X/Y chromosomes in humans The X chromosome has many genes; the Y chromosome only has genes for maleness. Human sex chromosomes (includes Mic2 gene) Fig 4.14 Sex-linked traits are genes located on the X chromosome Color Blind Test Sex-linked traits: Genes on the X chromosome A= normal; a= colorblind normal similar to Fig 4.13 colorblind No one affected, female carriers Sex-linked traits: Genes on the X chromosome A= normal; a= colorblind normal normal 50% of males affected, 0 % females affected similar to Fig 4.13 Sex-linked traits: Genes on the X chromosome A= normal; a= colorblind normal similar to Fig 4.13 colorblind 50% males affected, 50% females affected Sex-linked traits: Genes on the X chromosome A= normal ; a= colorblind No one affected, female carriers similar to Fig 4.13 50% of males affected, 0 % female affected 50% males affected, 50% females affected Fig 3.18 males and females may have different numbers of chromosomes Tbl 7.1 dosage compensation Fig 7.4 The epithelial cells derived from this embryonic cell will produce a patch of white fur While those from this will produce a patch of black fur At an early stage of embryonic development Mammalian X-inactivation involves the interaction of 2 overlapping genes. Promotes compaction Prevents compaction Barr body compaction is heritable within an individual The Barr body is replicated and both copies remain compacted • A few genes on the inactivated X chromosome are expressed in the somatic cells of adult female mammals – Pseudoautosomal genes (Dosage compensation in this case is unnecessary because these genes are located both on the X and Y) – Up to a 25% of X genes in humans may escape full inactivation • The mechanism is not understood Lamarck was right? Sort of… Epigenetics: http://www.pbs.org/wgbh/nova/sciencenow/3411/02.html Image from: http://www.sparknotes.com/biology/evolution/lamarck/section2.rhtml Genomic Imprinting • Genomic imprinting is a phenomenon in which expression of a gene depends on whether it is inherited from the male or the female parent • Imprinted genes follow a non-Mendelian pattern of inheritance – Depending on how the genes are “marked”, the offspring expresses either the maternallyinherited or the paternally-inherited allele **Not both Genomic Imprinting: Methylation of genes during gamete production. A hypothetical example of imprinting a B* A=curly hair A* b a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females a B* A* b A hypothetical example of imprinting a B* A=curly hair A* b a=straight hair B=beady eyes b=normal *=methylation A* in males B* in females A*a bB* a B* A*a bB* A* b A hypothetical example of imprinting a B* A=curly hair A* b a=straight hair B=beady eyes A*a bB* a B* A*a bB* b=normal *=methylation A* in males B* in females A*a bB Aa bB* A* b A hypothetical example of imprinting a B* A=curly hair A* b a=straight hair B=beady eyes A*a bB* similar to Fig 7.10 a B* A* b A*a bB* b=normal *=methylation A*a bB Aa bB* A* in males B* in females A*b, A*B, ab, aB Ab, AB*, ab, aB* Thus genomic imprinting is permanent in the somatic cells of an animal – However, the marking of alleles can be altered from generation to generation Imprinting and DNA Methylation • Genomic imprinting must involve a marking process • At the molecular level, the imprinting is known to involve differentially methylated regions – They are methylated either in the oocyte or sperm • Not both • For most genes, methylation results in inhibition of gene expression –However, this is not always the case Changes in methylation during gamete development alter the imprint Haploid female gametes transmit an unmethylated gene Fig 7.11 Haploid male gametes transmit a methylated gene To date, imprinting has been identified in dozens of mammalian genes Tbl 7.2 Tbl 7.2 Imprinting plays a role in the inheritance of some human diseases: Prader-Willi syndrome (PWS) and Angelman syndrome (AS) –PWS is characterized by: reduced motor function, obesity, mental deficiencies –AS is characterized by: hyperactivity, unusual seizures, repetitive muscle movements, mental deficiencies Usually, PWS and AS involve a small deletion in chromosome 15 –If it is inherited from the mother, it leads to AS –If it is inherited from the father, it leads to PWS • AS results from the lack of expression of UBE3A (encodes a protein called EA-6P that transfers small ubiquitin molecules to certain proteins to target their degradation) – The gene is paternally imprinted (silenced) • PWS results (most likely) from the lack of expression of SNRNP (encodes a small nuclear ribonucleoprotein that controls gene splicing necessary for the synthesis of critical proteins in the brain) – The gene is maternally imprinted (silenced) The deletion is the same in males and females, but the expression is different depending on who you received the normal version from. Fig 7.12 The relationship between genes and traits is often complex Complexities include: • Multiple genes controlling one trait Two genes control coat color in mice Fig 4.21 The interaction of these two proteins explains their affect on a single trait (in fruit flies). Variation in Peas Fig 3.2 Fig 2.8 Inheritance of 2 independent genes Approximate position of seed color and shape genes in peas Y y Gene for seed color r Chrom. 1/7 R Chrom. 7/7 Gene for seed shape There must be a better way… Fig 2.9 Inheritance can be predicted by probability Section 2.2, pg 30-32 Sum rule • The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities • Consider the following example in mice • Gene affecting the ears • Gene affecting the tail – De = Normal allele – de = Droopy ears – Ct = Normal allele – ct = Crinkly tail • If two heterozygous (Dede Ctct) mice are crossed • Then the predicted ratio of offspring is – – – – 9 with normal ears and normal tails 3 with normal ears and crinkly tails 3 with droopy ears and normal tails 1 with droopy ears and crinkly tail • These four phenotypes are mutually exclusive – A mouse with droopy ears and a normal tail cannot have normal ears and a crinkly tail • Question – What is the probability that an offspring of the above cross will have normal ears and a normal tail or have droopy ears and a crinkly tail? • Applying the sum rule – Step 1: Calculate the individual probabilities P(normal ears and a normal tail) = 9 (9 + 3 + 3 + 1) = 9/16 P(droopy ears and crinkly tail) = 1 (9 + 3 + 3 + 1) = 1/16 – Step 2: Add the individual probabilities 9/16 + 1/16 = 10/16 • 10/16 can be converted to 0.625 – Therefore 62.5% of the offspring are predicted to have normal ears and a normal tail or droopy ears and a crinkly tail Product rule • The probability that two or more independent events will occur is equal to the product of their respective probabilities • Note – Independent events are those in which the occurrence of one does not affect the probability of another • Consider the disease congenital analgesia – Recessive trait in humans – Affected individuals can distinguish between sensations • However, extreme sensations are not perceived as painful – Two alleles • P = Normal allele • p = Congenital analgesia • Question – Two heterozygous individuals plan to start a family – What is the probability that the couple’s first three children will all have congenital analgesia? • Applying the product rule – Step 1: Calculate the individual probabilities • This can be obtained via a Punnett square P(congenital analgesia) = 1/4 – Step 2: Multiply the individual probabilities 1/4 X 1/4 X 1/4 = 1/64 • 1/64 can be converted to 0.016 – Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia Different genes are not always independent Crossingover Meiosis I (Ind. Assort.) Meiosis II 4 Haploid cells, each unique Fig 5.1 The haploid cells contain the same combination of alleles as the original chromosomes The arrangement of linked alleles has not been altered Fig 5.1 These haploid cells contain a combination of alleles NOT found in the original chromosomes These are termed parental or nonrecombinant cells This new combination of alleles is a result of genetic recombination These are termed recombinant cells Linked alleles tend to be inherited together Crossing over produces new allelic combinations Recombinants are produced by crossing over For linked genes, recombinant frequencies are less than 50 percent Homologous pair of chromosomes Does this pedigree show recombination or linkage? Does this pedigree show recombination or linkage? Longer regions have more crossovers and thus higher recombinant frequencies Some crosses do not give the expected results =25% 42% 41% 9% 8% These two genes are on the same chromosome By comparing recombination frequencies, a linkage map can be constructed = 17 m.u. Linkage map of Drosophila chromosome 2: This type of map, with mapping units more than 50, can only be put together by making comparisons of linked genes. Today: some things Mendel did not tell us… plus Mapping and Epigenetics –Exam #3 W 7/30 in class (bonus #2 due)– Lecture ended here, but I am leaving in the following material so you can get a preview of the mapping problem we will work on to start class on M 7/28. Fig 5.2 A much greater proportion of the two types found in the parental generation The probability of crossing over can be used to determine the spatial relationship of different genes Double recombinants arise from two crossovers Recombinant Double recombinants can show gene order What is the relationship between these 3 genes? What order and how far apart? similar to Fig 5.3, also see Fig 5.9, and pg 115-117 What is the relationship between these 3 genes? What order and how far apart? similar to Fig 5.3 Double crossover Which order produces the double crossover? Which order produces the double crossover? We have the order. What is the distance? Recombinants between st and ss: (50+52+5+3)/755 =14.6% Recombinants between ss and e: (43+41+5+3)/755 =12.2% Put it all together… 26.8 m.u. st ss 14.6 m.u. e 12.2 m.u. Drosophila linkage map Linkage map of Drosophila chromosome 2 Recombination is not completely random. physical distance Yeast chromosome 3 linkage map Alignment of physical and recombination maps Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple. {Meiosis: producing gametes} For life to exist, the information (genes) must be passed on. {Mitosis: producing more cells}