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Presentation on theme: "CHAPTER 13 LECTURE SLIDES"— Presentation transcript:

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2 Chromosomes, Mapping, and the Meiosis–Inheritance Connection
Chapter 13

3 What Carried Mendel’s “Factors”?
Carl Correns – 1900 First suggests central role for chromosomes One of papers announcing rediscovery of Mendel’s work Walter Sutton – 1902 Chromosomal theory of inheritance Based on observations that similar chromosomes paired with one another during meiosis

4 T.H. Morgan – 1910 Working with fruit fly, Drosophila melanogaster
Discovered a mutant male fly with white eyes instead of red Crossed the mutant male to a normal red-eyed female All F1 progeny red eyed = dominant trait

5 Morgan crossed F1 females x F1 males
F2 generation contained red and white- eyed flies But all white-eyed flies were male Testcross of a F1 female with a white-eyed male showed the viability of white-eyed females Morgan concluded that the eye color gene resides on the X chromosome FIRST DIRECT EVIDENCE FOR CHROMOSOMAL THEORY OF INHERITANCE! – showed that genes lie on chromosomes



8 Sex Chromosomes Sex determination in Drosophila is based on the number of X chromosomes 2 X chromosomes = female 1 X and 1 Y chromosome = male Sex determination in humans is based on the presence of a Y chromosome Having a Y chromosome (XY) = male

9 © BioPhoto Associates/Photo Researchers, Inc.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. X chromosome Y chromosome 2.8 µm © BioPhoto Associates/Photo Researchers, Inc. Humans have 46 total chromosomes 22 pairs are autosomes 1 pair of sex chromosomes Y chromosome highly condensed Recessive alleles on male’s X have no active counterpart on Y “Default” for humans is female Requires SRY gene on Y for “maleness”

10 Hemophilia Disease that affects a single protein in a cascade of proteins involved in the formation of blood clots Form of hemophilia is caused by an X-linked recessive allele Heterozygous females are asymptomatic carriers Allele for hemophilia was introduced into a number of different European royal families by Queen Victoria of England


12 Dosage compensation Ensures an equal expression of genes from the sex chromosomes even though females have 2 X chromosomes and males have only 1 In each female cell, 1 X chromosome is inactivated and is highly condensed into a Barr body Females heterozygous for genes on the X chromosome are genetic mosaics


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15 Chromosome theory exceptions
Mitochondria and chloroplasts contain genes Traits controlled by these genes do not follow the chromosomal theory of inheritance Genes from mitochondria and chloroplasts are often passed to the offspring by only one parent (mother) Maternal inheritance In plants, the chloroplasts are often inherited from the mother, although this is species dependent

16 Genetic Mapping Early geneticists realized that they could obtain information about the distance between genes on a chromosome Based on genetic recombination (crossing over) between genes If crossover occurs, parental alleles are recombined producing recombinant gametes



19 Review of Expectations – 2 genes on same chromosome
R - Round, r-oval W-White, w-black P1 RRWW x rrww F1 RrWw F2 - testcross RW rw RrWw rrww F2 Expected if Same 1 Chromosome Round_White oval_black

20 Review of Expectations- 2 genes on diff. chromosomes
R - Round, r-oval W-White, w-black P1 RRWW x rrww F1 RrWw F2 - testcross RW Rw rW rw RrWw Rrww rrWw rrww F2 Expected if 1 Indep. Asst. Round_White Round_black oval_White oval_black

21 Review of Expectations- 2 genes on same chromosome – but now with crossover events
R - Round, r-oval W-White, w-black P1 RRWW x rrww F1 RrWw F2 - testcross RW Rw rW rw RrWw Rrww rrWw rrww F2 Expected if <1 <<1 Crossover between genes Round_White Round_black oval_White oval_black RECOMBINANTS

22 Alfred Sturtevant Undergraduate in T.H. Morgan’s lab
Put Morgan’s observation that recombinant progeny reflected relevant location of genes in quantitative terms As physical distance on a chromosome increases, so does the probability of recombination (crossover) occurring between the gene loci

23 Probability of crossover depends on distance

24 Recombination Frequency
Genotype of gametes formed by double heterozygous parent (Cc Ss) C S c s C s c S CcSs ccss Ccss ccSs Appearance (phenotype) Colored, Smooth colorless, wrinkled Colored, wrinkled colorless, Smooth % total recombinants if Independent Assortment 25% 50% if closely linked (no crossover) -- 0% if there's some distance between them 45% 5% 10% if there's more distance between them 40% 20% if there's even more distance between them 35% 15% 30% Now they're so far apart they almost sort independently due to the high probability of crossover events C c S s C c S s C c S s C c S s C c S s C c S s C c S s

25 Multiple crossovers If homologues undergo two crossovers between loci, then the parental combination is restored Leads to an underestimate of the true genetic distance Relationship between true distance on a chromosome and the recombination frequency is not linear


27 Working examples- note the cute way to cross to test for linkage
1. In tomatoes tall growth habit is the result of a dominant gene, D, dwarf growth to its recessive allele, d. Smooth epidermis is due to a dominant gene, P, pubescent epidermis to its recessive allele, p. A homozygous tall smooth variety was crossed with a dwarf pubescent variety. The F1 were test crossed to dwarf pubescent. The results of the test cross were as follows: tall, smooth - 96 tall, pubescent - 4 dwarf, smooth - 3 dwarf, pubescent - 95 Are these genes linked (on the same chromosome)? If so, what is the percent of crossing over?

28 Worked solution P1 == DDPP x ddpp Parental gametes DP and dp
F1 == DdPp TestCross DP Dp dP dp DdPp Ddpp ddPp ddpp Tall, smooth Tall, pubes dwarf, smooth dwarf, pubes 96 4 3 95 Same Chromosome because mostly parental types % Crossover = (4+3) / ( ) "=" 7 recombinants 198 total offspring % crossover

29 2. In rabbits color is due to a dominant gene, D, albinism to its recessive allele, d. Black is the result of a dominant gene, B, brown to its recessive allele, b. Brown (homozygous) rabbits are crossed with albinos carrying black in the homozygous state. The F1 are crossed to double recessive. From many such crosses the total results are: black - 68 brown - 132 albino - 200 Are these genes linked?

30 Worked solution P1 == DDbb x ddBB Parental gametes Db and dB
F1 == DdBb TestCross Db DB db dB Ddbb DdBb ddbb ddBb Colored, brn Colored, blk Albino, brown Albino, black 132 68 ? 200 albinos= 68+132 Same Chromosome because mostly parental types % Crossover = (68+68?) / ( (68+132) ) "=" 136 recombinants 400 total offspring 34 % crossover

31 In corn purple plant color is due to a dominant gene, P, green plant color to recessive allele, p. Normal leaves are due to a dominant gene, N, narrow leaves to its recessive allele, n. Homozygous green plants with homozygous normal leaves were crossed with homozygous purple plants having narrow leaves. The F1 were crossed with green plants with narrow leaves. The results of this test cross were as follows: purple plants with normal leaves - 197 purple plants with narrow leaves - 201 green plants with normal leaves - 199 green plants with narrow leaves - 203 Are these genes linked? If so what is the percent cross over?

32 Worked solution-note that parental gametes are mixed
P1 == ppNN x PPnn Parental gametes pN and Pn F1 == pPNn TestCross pN pn PN Pn ppNn ppnn PpNn Ppnn green, norm Grn, narrow Purple, norm Purp, narrow 199 203 197 201 Different Chromosomes because equal probabilities

33 In corn tallness is due to a dominant gene, D, dwarfness to its recessive allele, d. Normal leaves is dominant, N, crinkly leaves is recessive, n. Homozygous tall, crinkly-leaved corn was crossed to dwarf, homozygous normal-leaved corn. The F1 which was tall normal leaved, was crossed to double-recessive dwarf crinkly-leaved corn. The results were as follows: Tall crinkly-leaved - 83 Tall, normal-leaved - 19 dwarf crinkly-leaved - 17 dwarf normal-leaved - 81 Are these two pairs of genes linked? If so, what is the percent crossing over?

34 Worked solution-note that parental gametes are mixed
P1 == DDnn x ddNN Parental gametes Dn and dN F1 == DdnN TestCross Dn DN dn dN Ddnn DdNn ddnn ddNn Tall, crinkly Tall, norm dwarf, crinkly dwarf, norm 81 19 17 83 Same chromosome because mostly parental types (19+17)/(200) = 36/200=18%

35 Constructing maps The distance between genes is proportional to the frequency of recombination events recombination recombinant progeny frequency total progeny 1% recombination = 1 map unit (m.u.) 1 map unit = 1 centimorgan (cM) =

36 Three-point testcross
Uses 3 loci instead of 2 to construct maps Gene in the middle allows us to see recombination events on either side In any three-point cross, the class of offspring with two crossovers is the least frequent class In practice, geneticists use three-point crosses to determine the order of genes, then use data from the closest two-point crosses to determine distances


38 Human genome maps Data derived from historical pedigrees
Difficult analysis Number of markers was not dense enough for mapping up to 1980s Disease-causing alleles rare Situation changed with the development of anonymous markers Detected using molecular techniques No detectable phenotype

39 SNPs Single-nucleotide polymorphisms
Affect a single base of a gene locus Used to increase resolution of mapping Used in forensic analysis Help eliminate or confirm crime suspects or for paternity testing


41 Sickle cell anemia First human disease shown to be the result of a mutation in a protein Caused by a defect in the oxygen carrier molecule, hemoglobin Leads to impaired oxygen delivery to tissues

42 Heterozygotes appear normal
Homozygotes for sickle cell allele exhibit intermittent illness and reduced life span Heterozygotes appear normal Do have hemoglobin with reduced ability Sickle cell allele is particularly prevalent in people of African descent Proportion of heterozygotes higher than expected Confers resistance to blood-borne parasite that causes malaria

43 Nondisjunction Failure of homologues or sister chromatids to separate properly during meiosis Aneuploidy – gain or loss of a chromosome Monosomy – loss Trisomy – gain In all but a few cases, do not survive

44 Smallest autosomes can present as 3 copies and allow individual to survive
13, 15, 18 – severe defects, die within a few months 21 and 22 – can survive to adulthood Down Syndrome – trisomy 21 May be a full, third 21st chromosome May be a translocation of a part of chromosome 21 Mother’s age influences risk


46 Nondisjunction of sex chromosomes
Do not generally experience severe developmental abnormalities Individuals have somewhat abnormal features, but often reach maturity and in some cases may be fertile XXX – triple-X females XXY – males (Klinefelter syndrome) XO – females (Turner syndrome) OY – nonviable zygotes XYY – males (Jacob syndrome)


48 Genomic imprinting Phenotype exhibited by a particular allele depends on which parent contributed the allele to the offspring Specific partial deletion of chromosome 15 results in Prader-Willi syndrome if the chromosome is from the father Angelman syndrome if it’s from the mother

49 Detection Pedigree analysis used to determine the probability of genetic disorders in the offspring Amniocentesis collects fetal cells from the amniotic fluid for examination Chorionic villi sampling collects cells from the placenta for examination




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