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Pedigree Chart Symbols Male Female Person with trait.

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Presentation on theme: "Pedigree Chart Symbols Male Female Person with trait."— Presentation transcript:

1 Pedigree Chart Symbols Male Female Person with trait

2 Sample Pedigree

3 Dominant Trait Recessive Trait

4 Chapter 15 The Chromosomal Basis of Inheritance

5 First Experimental Evidence to connect Mendelism to the chromosome Thomas Morgan (1910) Used fruit flies as model organism Allowed the first tracing of traits to specific chromosomes.

6 Fruit Fly Drosophila melanogaster 3 pairs of Autosomes 1 pair of sex chromosomes

7 Morgan Observed: A male fly with a mutation for white eyes.

8 Morgan crossed The white eye male with a wild type (red eyed) female. Wild type is most common – NOT always DOMINANT Male ww x Female w + w +

9 The F1 offspring: All had red eyes. This suggests that white eyes is a _________? Recessive. F1= w + w What is the predicted phenotypic ratio for the F2 generation?

10 F1 X F1 = F2 Expected F2 ratio - 3:1 of red:white He got this ratio, however, all of the white eyed flies were MALE. Therefore, the eye color trait appeared to be linked to sex.

11 Morgan discovered: Sex linked traits. Genetic traits whose gene are located on a sex chromosome

12 Fruit Fly Chromosomes Female Male XX XY Presence of Y chromosome determines the sex Just like in humans!

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14 Morgan Discovered There are many genes, but only a few chromosomes. Therefore, each chromosome must carry a number of genes together as a “package”.

15 Sex-Linked Problem A man with hemophilia (a recessive, sex- linked, x-chromosome condition) has a daughter of normal phenotype. She marries a man who is normal for the trait. A. What is the probability that a daughter of this mating will be a hemophiliac? B. That a son will be a hemophiliac? C. If the couple has four sons, what is the probability that all four will be born with hemophilia?

16 Original Man - X h Y Daughter - must get the dad’s X chromosome X H X h (normal phenotype, so she’s a carrier) Daughter’s husband X H Y (normal phenotype) A. daughter must get X H from the dad. 0% (50% carrier, 50% homo dom.) B. son must get Y from dad. 50% chance to be hemophiliac C. ½ x ½ x ½ x ½ = 1/16

17 Multiple Genes Parents are two true-breeding pea plants Parent 1 Yellow, round Seeds (YYRR) Parent 2 Green, wrinkled seeds (yyrr) These 2 genes are on different chromosomes (all problems so far have assumed this)

18 F1: YyRr x YyRr What are the predicted phenotypic ratios of the offspring? ¾ yellow ¾ round ¼ green ¼ wrinkled ¼ (green) x ¼ (wrinkled) = 1/16 green, wrinkled 9:3:3:1 phenotypic ratio

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20 Linked Genes Traits that are located on the same chromosome. Result: Failure of Mendel's Law of Independent Assortment. Ratios are different from the expected

21 Example: Body Color - gray dominant/wild b + - Gray b - black Wing Type - normal dominant/wild vg + - normal vg – vestigial (short)

22 Example b + b vg + vg X bb vgvg Predict the phenotypic ratio of the offspring

23 Show at board b + b x bb vg + vg x vgvg ½ gray ½ black ½ normal ½ vestigial ----------------------------------------------------- ¼ gray normal, ¼ gray vestigial, ¼ black normal, ¼ black vestigial 1:1:1:1 phenotypic ratio

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25 Conclusion Most offspring had the parental phenotype. Both genes are on the same chromosome. bbvgvg parent can only pass on b vg b + b vg + vg can pass on b + vg + or b vg

26 b + b vg + vg - Chromosomes (linked genes)

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28 Crossing-Over Breaks up linkages and creates new ones. Recombinant offspring formed that doesn't match the parental types. Higher recombinant frequency (non- parental types) = genes further apart on chromosome

29 If Genes are Linked: Independent Assortment of traits fails. Linkage may be “strong” or “weak”. Strong Linkage means that 2 alleles are often inherited together.

30 Degree of strength related to how close the traits are on the chromosome.

31 Genetic Maps Constructed from crossing-over frequencies. 1 map unit = 1% recombination frequency. Can use recombination rates to ‘map’ chromosomes.

32 Comment - only good for genes that are within 50 map units of each other. Why? Over 50% gives the same phenotypic ratios as genes on separate chromosomes

33 Genetic Maps Have been constructed for many traits in fruit flies, humans and other organisms.

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35 Barr Body Inactive X chromosome observed in the nucleus. Way of determining genetic sex without doing a karyotype.

36 Lyon Hypothesis Which X inactivated is random. Inactivation happens early in embryo development by adding CH 3 groups to the DNA. Result - body cells are a mosaic of X types.

37 Examples Calico Cats. Human examples are known such as a sweat gland disorder.

38 Calico Cats X B = black fur X O = orange fur Calico is heterozygous, X B X O.

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40 Chromosomal Alterations Changes in number. Changes in structure.

41 Number Alterations Aneuploidy - too many or too few chromosomes, but not a whole “set” change. Polyploidy - changes in whole “sets” of chromosomes.

42 Nondisjunction When chromosomes fail to separate during meiosis Result – cells have too many or too few chromosomes which is known as aneuploidy

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44 Meiosis I vs Meiosis II Meiosis I – all 4 cells are abnormal Meiosis II – only 2 cells are abnormal

45 Aneuploidy Caused by nondisjunction, the failure of a pair of chromosomes to separate during meiosis.

46 Types Monosomy: 2N - 1 Trisomy: 2N + 1

47 Question? Why is trisomy more common than monosomy? Fetus can survive an extra copy of a chromosome, but being hemizygous is usually fatal.

48 Structure Alterations Deletions Duplications Inversions Translocations

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51 Result Loss of genetic information. Position effects: a gene's expression is influenced by its location to other genes.

52 Summary Know about linkage and crossing-over. Sex chromosomes and their pattern of inheritance.

53 Summary Be able to work genetics problems for this chapter.


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