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Chromosomal Basis of Inheritance

Chromosomal Basis of Inheritance. Featuring fruit fly: Drosophila Melanogaster. Basic Terms/Information about Drosophila. My diploid number is 2N = 8. Wild-type (+) refers to the most common phenotype in fruit flies. It’s usually dominant (but not necessarily).

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Chromosomal Basis of Inheritance

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  1. Chromosomal Basis of Inheritance Featuring fruit fly: Drosophila Melanogaster

  2. Basic Terms/Information about Drosophila My diploid number is 2N = 8 Wild-type (+) refers to the most common phenotype in fruit flies. It’s usually dominant (but not necessarily) Mutants are non wild-type traits

  3. Sex-Linkage or (X-linked) In fruit flies, (R) is the dominant wild-type gene for red eyes, and (r) is the recessive, mutant gene for white eyes. The gene is found on the “X” chromosome only. This is considered X-linked. The Y chromosome is shorter than the X Does the gene for eye color exist on the “Y” chromosome? Why or why not? These are the X and Y chromosomes of a male fly. How is the Y chromosome different from the X? No, because the gene for eye color is found on the longer segment of the X chromosome. This part is missing on the Y R r r XX XY White eye male Red eye female What would be the phenotype of this female fly? What would be the phenotype of this male fly?

  4. Sex-Linkage or (X-linked) • When genes are sex-linked, we include the X and Y as part of their genotype. For example, the allele for red eye is not “R” but is written as XR. How would you write the allele for white eye? Xr

  5. Writing X-linked Genotypes What is the possible genotype(s) for the eye color of this fly if it is a female? What is the possible genotype for the eye color of this fly if it is male? XRXR or XRXr XrXr XRY XrY Answer the above questions again for this fly.

  6. Sample Problems Example 1: What is the F1 genotypic and phenotypic ratio of a female true-breeding wild-type fly for red eyes crossed with a white-eyed male? XR XR XRXr XRXr Xr Phenotypic Ratio Red-eye female:Red-eye male 2:2 reduced to 1:1 Genotypic Ratio XRXr:XRY 2:2 reduced to 1:1 XRY XRY Y

  7. Sample Problems Example 2: What would the genotypes and phenotypes be of the F2 generation? XR Xr XRXR XRXr XR Genotypic Ratio XRXR : XRXr : XRY : XrY 1 : 1 : 1 : 1 XRY XrY Y Phenotypic Ratio Red-eye female : Red-eye male : White-eyed male 2 : 1 : 1

  8. X-linked disorders Definition: diseases or disorders whose genes are found on the X-chromosome, but not on the Y. Ex: hemophelia (Xh), color blindness (Xb), muscular distrophy (Xm) If the disorder is recessive, more males than females will tend to have the disorder. Why?

  9. Take, for example, colorblindness (Xb) XBXB XBXb If you have a normal female, what is her possible genotype(s)? _____, or _____ If you have a colorblind female, what is her genotype? ______ If you have a colorblind male, what is his genotype? ______ How many colorblind genes do males need to inherit to be colorblind? _____ Females? _____ Who does the male inherit the colorblind gene from? _____________________________________ XbXb XbY 1 2 His mother, who donates the X chromosome

  10. Other traits and alleles of Drosophila melanogaster Body-color and wing-type are NOT located on the sex chromosome, so they are considered autosomal Things to think about independently….How would you confirm or test that these mutant traits are recessive?

  11. Example 3 In flies, grey bodies (G+) and normal-wing size (N+) are dominant to black bodies (g) and small wing size (n). Predict a cross between G+gN+n and ggnn.

  12. Surprising Results! Predicted Cross G+gN+n x ggnn G+N G+n gN+ gn gn gn gn gn 25% G+gN+n 25% G+gnn 25% ggN+n 25% ggnn

  13. Actual Results 8.5% 8.5% 41.5% 41.5% 41.5% 41.5% Why did this happen???

  14. Linked Genes The genes for body color and wing size are linked, meaning they are found on the same chromosome. They will most likely be inherited together and will not undergo Mendel’s Law of . cross over segregates the linked genes g g G+ G+ Independent Assortment N+ n N+ n unless Homologous Chromosomes

  15. What are recombinants? Recombinants are offspring that have different phenotypes from those of the parents. Let’s look back at our original cross X

  16. X C B D 8.5% 8.5% 41.5% 41.5% 41.5% 41.5% A Which offspring (A-D) from this cross are the recombinants?

  17. How do we determine if two genes are linked or if two genes are located on different chromosomes? Calculate the recombination frequency! If the frequency is less than 50%, it is assumed that the two genes are on same chromosomes. Answer

  18. X C B D 185 206 944 965 A Total Flies: 965 + 206 + 185 + 944 = 2500 flies Calculate the recombination frequency for this cross! Recombination Frequency = recombinants / total flies = 391/2500 = .156 = 16% Total number of recombinants: 206 + 185 = 391

  19. Gene Mapping • Genes are mapped on a chromosome based upon the recombination frequency. • For ex. the distance between the genes for body color and wing type is therefore 16 “map-units” apart (16% frequency) Grey Body Black Body Normal wings Small wings

  20. Using recombination frequencies, create a linkage map for the following a - c:between genes b and a = 10.5%; between genes c and a = 48%; between genes c and b = 37.5% 10.5% 37.5% 48% a b c

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