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How does chromosome behavior account Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? for Mendel’s Principles.

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Presentation on theme: "How does chromosome behavior account Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? for Mendel’s Principles."— Presentation transcript:

1 How does chromosome behavior account Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? for Mendel’s Principles ? (The chromosomal theory of inheritance)

2 The chromosomal theory of inheritance:

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5 The Chromosome Theory of Inheritance: genes (allele pairs) are on chromosomes and homologous chromosomes segregate during meiosis (principle of segregation) and reunite during fertilization. If allele pairs are on different chromosomes they will sort independently (principle of independent assortment) due to independent orientation of the tetrads at metaphase I. If the allele pairs are on the same chromosome, they will sort dependently and travel together into the gametes.

6 Draw out meiosis and indicate how this process accounts for the laws of segregation and independent assortment. Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel?

7 Make sure you can quickly draw this and explain how the laws of segregation and independent assortment are described by meiosis and specifically where in meiosis. Meiosis accounts for the laws of segregation and independent assortment. The two alleles of each gene separate. Alleles of genes on non- homologous chomosomes assort independently during gamete formation.

8 What have we been ignoring thus far during dependent assortment? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? Crossing-Over

9 You perform an experiment where you cross a double heterozygous plant with purple flower and long pollen with another of the same genotype. P = purple flower p = red flower L = long pollen l = round pollen PpLl x PpLl 1. Determine the expected phenotypic ratio if the allele pairs were on different chromosomes – independent assortment. This is identical to Mendel’s classic F1 dihybrid cross resulting in a 9 purple,long : 3 purple, round : 3 red, long : 1 red, round 2. Determine the expected phenotypic ratio if the allele pairs were linked (P is linked to L). Possible gametes: PL or pl for both Making a Punnett square or just putting together the possible fertilization events gives you the following offspring: PPLL, PpLl, PpLl, ppll or a 3 purple,long : 1 red, round phenotypic ratio

10 Neither prediction fits this observation… How can you account for this observation? Come up with a model for how this can occur. Take a look at the actual offspring:

11 If the alleles did sort independently, you would expect to see the 9:3:3:1 phenotypic ratio. Therefore they are likely sorting dependently, but how do we account for the purple-round and red-long offspring?... CROSSING OVER P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes Synapsis during Prophase I

12 If the alleles did sort independently, you would expect to see the 9:3:3:1 phenotypic ratio. Therefore they are likely sorting dependently, but how do we account for the purple-round and red-long offspring?... CROSSING OVER P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes Predict the resulting gametes…

13 If the alleles did sort independently, you would expect to see the 9:3:3:1 phenotypic ratio. Therefore they are likely sorting dependently, but how do we account for the purple-round and red-long offspring?... CROSSING OVER P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes You get the same gametes as if it were independent assortment because of crossing over, but what is going to be different?

14 If the alleles did sort independently, you would expect to see the 9:3:3:1 phenotypic ratio. Therefore they are likely sorting dependently, but how do we account for the purple-round and red-long offspring?... CROSSING OVER P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes They are not made in a 1:1:1:1 ratio since crossing over doesn’t always occur between the allele pairs… The majority will be parental type…

15 Which gametes need to fuse in order to get the recombinant offspring? P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes Whatever the PL gamete fertilizes, the outcome is the same… the purple, long phenotype.

16 What do you think determines the number of recombinant offspring? P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes The number of gametes with the recombinant chromosome of course…

17 What do you think determines how often crossing over occurs between the P/p and L/l genes? P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes The number of crossing over events between the allele pairs is directly related to how far apart the allele pairs are on the chromosome… The further apart they are, the more likely it is that crossing over will occur BETWEEN them.

18 P L p l P Lp l P l p L Recombinant Gametes Parental Type Gametes Therefore, the more recombinant offspring you get, the_________________ apart the allele pairs are. further

19 Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? If I have three or more pairs of alleles, I want to know where they are on a chromosome relative to each other. How can we do that?

20 Drosophila melanogaster Thomas Hunt (T. H.) Morgan (The modern day pea plant) The Fly Room

21 Fruit fly experiments demonstrating the role of crossing-over in inheritance Fig. 9.19C Predict the offspring phenotypic ratio for dependent assortment where G is linked to L. and recombination frequency G = gray body (dominant) g = black body L = long wings (dominant) l = vestigial,short wings Characteristics: Traits: Body Color Wing shape GgLl (wild type female) x ggll (male) Gametes: GL, gl gl Offspring: GgLl or ggll (both are parental type) 1 Gray, long wings : 1 black, vestigial/short wings

22 Fig. 9.19C Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency 1 Gray, long wings : 1 black, short wings IT DOESN’T MATCH…WHY? CROSSING OVER… Draw the possible gametes in each parent (show chromosomes). Indicate the parental-type and recombinant chromosomes.

23 Fig. 9.19C Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency CROSSING OVER…

24 Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency What do the number of recombinant phenotypes relative to the total offspring tell you? 1. They tell you the number of times crossing over occurred between the G/g and L/l allele pairs in the FEMALE ONLY…why? The male was chosen to be double homozygous recessive on purpose so that crossing over does not matter because the same gametes form either way. Thus, from this experiment one can determine the frequency of crossing over in a SINGLE INDIVIDUAL.

25 Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency That’s great, but who cares…what is this useful for? 1. Remember, the further apart the allele pairs are, the more recombinant offspring you should get. 2. Therefore, one can do this with another pair of alleles like G/g and P/p. If the result is fewer recombinants than with G and L, P is closer to G… 3. You can determine where allele pairs are relative to each other on a chromosome = chromosome mapping (linkage mapping)!!

26 Fig. 9.19C Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency The recombinant frequency can be determined by taking the number of recombinants and dividing by the total offspring and multiplying by 100.

27 Fig. 9.19C Fruit fly experiments demonstrating the role of crossing-over in inheritance and recombination frequency

28 Fruit fly (Drosophila) nomenclature 1. Genes are typically named after the mutant (non-wild type) allele. In this case black (b) and vestigial (vg). 2. The wild type is then indicated with a “+”. Therefore grey would be b + and long wings would be vg +.

29 Fruit fly (Drosophila) nomenclature SUMMARY: Grey (wild type; wt)b+b+ Black (mutant; mut)b Long Wings (wt)vg + Vestigial Wings (mut)vg

30 Fruit fly (Drosophila) nomenclature This is the same figure as shown previously just using Drosophila gene nomenclature.

31 9:3:3:1 phenotypic ratio Independent Assortment

32 9:3:3:1 phenotypic ratio Independent Assortment Linked Genes (no recombination)

33 9:3:3:1 phenotypic ratio Independent Assortment Linked Genes (WITH recombination)

34 9:3:3:1 phenotypic ratio Independent Assortment Linked Genes (WITH recombination)

35 9:3:3:1 phenotypic ratio Independent Assortment Linked Genes (WITH recombination)

36 How can we use recombination frequency to determine the position of genes on a chromosome (gene mapping)? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? You have determined the frequency between G/g and L/l (17%). There is a third gene on the same chromosome (E/e). How can you determine the frequency between G and E/e? By doing this testcross: GgEe x ggee

37 How can we use recombination frequency to determine the position of genes on a chromosome (gene mapping)? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? Let’s say you determine the recombinant frequency to be 9%. What does that tell you? It tells you that recombination occurred less frequently between G and E and therefore they are closer together than G and L. Let’s try to map this on a chromosome… (convince yourself that it is not possible yet) 9% G E L 17% or E G L 9%

38 How can we use recombination frequency to determine the position of genes on a chromosome (gene mapping)? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? What else do we need to do to figure out the order on the chromosome of G, E and L? G E L 17% E G L 9% We need to figure out the recombinant freq between L and E…is it ~8% or ~26%: LlEe x llee

39 How can we use recombination frequency to determine the position of genes on a chromosome (gene mapping)? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? You determine the recombinant offspring frequency to be 9.5%. Now can you map the genes?

40 How can we use recombination frequency to determine the position of genes on a chromosome (gene mapping)? Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? This is the only possible combination that fits the data (or l,e,g – e must be between them)… e

41 Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel?

42 Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? A geneticist wants to map the position of three dominant/recessive allele pairs (A/a, B/b, and F/f) relative to each other in D. melanogaster (fruit flies). For simplicity let’s say all the dominant alleles are on the same chromosome. Where should she begin? 1. Do AaBb x aabb and look for recombinant phenotypes to determine the frequency of crossing-over Her cross generates 1000 offspring of which 200 are recombinant. What is the recombination frequency? 200/1000 * 100 = 20% What should she do next?

43 Chapter 15 - Chromosomal Basis of Inheritance AIM: How does chromosome behavior relate to Mendel? A geneticist wants to map the position of three dominant/recessive allele pairs (A/a, B/b, and F/f) relative to each other in D. melanogaster (fruit flies). For simplicity let’s say all the dominant alleles are on the same chromosome. Where should she begin? 2. Do AaFf x aaff and BbFf x bbff Find recombinant phenotypes to determine the frequency of crossing-over The AaFf cross results in a 7% recombinant frequency Map the genes on the chromosome: The BbFf cross results in a 28% recombinant frequency The AaBb cross resulted in a 20% recombinant frequency B F A

44 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Sex-Linked Gene Inheritance

45 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Sex determination in humans obviously also uses the XY system where the Y chromosome determine if the individual is male. Do all sexually reproducing animals work the same…?

46 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Turns out the answer is no… Your book has a very brief explanation of these systems (Figure 15.9).

47 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Eye color is determined by a dominant/recessive single gene locus on the X chromosome. R = red-eye allele r = white-eye allele Q. Determine the possible offspring from a cross between a homozygous dominant red eye female and a white eye male.

48 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Notice that now because we have two variables to follow…allele pair and sex chromosome…we no longer just write the allele pair. Therefore the female in this case is homozygous dominant RR, each on an X chromosome = X R X R ), while the white eye male is just r on one X (X r ) as the Y chromosome does not have the same loci as X (X and Y are not homologous) = X r Y. Continue as normal- possible gametes, fertilize, etc…

49 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. There are two possible offspring: Red eye females Red eye males

50 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Q. Determine the possible offspring from cross between a red eye female heterozygous for the gene locus and a red eye male.

51 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance.

52 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. 50% male, half of which have red eyes and half with white. 50% red eye females, half heterozygous and half homozygous dominant

53 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Q. Determine the possible offspring from cross between a red eye female heterozygous for the gene locus and a white eye male.

54 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance.

55 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance.

56 Hemophilia AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. 1. A group of hereditary disorders where the blood does not clot (or coagulate) efficiently. 2. Typically caused by mutations in genes that code for clotting factors (proteins involved in blood clotting). 3. Incidence is 1 in 10,000 males approximately 4. Severity can vary 5. No cure, but how would you treat it? - Give regular infusions (injections) of the protein the person is deficient in (prophylaxis = treating before the problem)

57 Hemophilia in humans is due to a recessive X-chromosome mutation. What will be the results of mating between a normal (non-carrier) female and a hemophiliac male? AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. X H X H x X h Y XHXH XHXH XhXh Y XHXhXHXh XHXhXHXh XHXH YXHXH Y 50% chance of female carrier, 0% chance of male hemophiliac

58 Hemophilia in the Royal Families of Europe AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. A half-filled shape in a pedigree implies carrier

59 A red-eyed female Drosophila of unknown genotype was crossed with a white-eyed male fly. Half of the male and half of the female offspring were red-eyed, and half of the male and half of the female offspring were white-eyed. What was the genotype of the female fly? AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. 25% X R Y 25% X R X ? 25% X r Y 25% X r X r X r Y x ?? XrXr Y XrXrXrXr XrYXrYXRYXRY XrXr XRXR XRXrXRXr XRXrXRXr

60 Why do recessive sex-linked disorders affect mostly males? Red-green color blind test AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Because males only have one copy of the X chromosome and therefore only need to inherit one disease allele. They cannot be carriers. Ex) Color Blindness - Can be caused by mutations in many different genes typically affecting proteins in the cone cells (color sensing cells) of the eye. - Deuteranomaly – most common type (affects 5% of males) – sex linked on X chromosome

61 Why do recessive sex-linked disorders affect mostly males? Red-green color blind test AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Because males only have one copy of the X chromosome and therefore only need to inherit one disease allele. They cannot be carriers. Ex) Color Blindness - Why might the frequency of this allele be this high (5%)? - remember that sickle cell allele is 4% It is hypothesize that there might be an advantage to being color blind as these individuals appear to be better at spotting camouflaged objects…

62 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Color of the rainbow as viewed by a person with color-blindness (Deuteranomaly) Color of the rainbow as viewed by a normal person

63 AIM: What are sex-linked genes and what makes them different from autosomal genes? Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance.

64 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Morgan’s experiment involving sex-linkage reveled that genes were on chromosomes. Q. A cross was done between a wt red- eyed female and a true breed mutant white-eyed male. What offspring do you expect in terms of phenotype?

65 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. A. All red-eyed offspring as usual. Q. He then crossed the F1 generation. What do you expect in terms of phenotype? Morgan’s experiment involving sex-linkage reveled that genes were on chromosomes.

66 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. A. The standard 3 red : 1 white is observed, but look more closely…what do you notice? FemalesAll red-eyes Males50% red, 50% white How did Morgan explain this? Sex-linked genes (the gene for eye-color is on the X chromosome) Morgan’s experiment involving sex-linkage reveled that genes were on chromosomes.

67 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance Mendel’s work published never gaining popularity. Morgan’s experiment involving sex-linkage reveled that genes were on chromosomes. In order to understand the importance of this experiment, it must be put in the proper context: Mitosis figured out Meiosis figured out 1900 – Mendel’s work rediscovered 1902 – Chromosomal Theory of Inheritance gains popularity 1910 – Morgan and co-workers performed the experiment shown to the right giving us the first solid piece of evidence that genes are on chromosomes.

68 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. Morgan’s experiment involving sex-linkage reveled that genes were on chromosomes. How does this experiment indicate that genes are on chromosomes? 1.Morgan and others already figured out how sex determination worked in terms of chromosomes (XY – male, XX – female) in fruit flies. 2. The simplest way to account for the pattern in the F2 generation, which is clearly related to the sex of the fly, is to predict that the gene for eye color is on the X chromosome.

69 Chapter 15 - Chromosomal Basis of Inheritance AIM: Sex-linked gene inheritance. We now knew with a good level of certainty that genes, whatever they are, travel on chromosomes... Thomas Hunt Morgan received the Nobel Prize in 1933 for his work done at Columbia University in his famous fly room.


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