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The Chromosomal Basis of Inheritance Chapter 15. Review Mitosis Meiosis Chromosome Genotype and Phenotype Mendelian Genetics.

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Presentation on theme: "The Chromosomal Basis of Inheritance Chapter 15. Review Mitosis Meiosis Chromosome Genotype and Phenotype Mendelian Genetics."— Presentation transcript:

1 The Chromosomal Basis of Inheritance Chapter 15

2 Review Mitosis Meiosis Chromosome Genotype and Phenotype Mendelian Genetics

3 Thomas Hunt Morgan Studied Drosophila melanogaster – Large number of offspring – Small, easy to care for – 4 chromosomes with easily observable phenotypes

4 Drosophila melanogaster Phenotypes Wild type: red eyes Variations: eye color, body color, wing shape Males and females are easy to tell apart

5 Morgan’s Cross Red eyed female x White eyed male (w+)(w) F1  100% red eyed (wild type) F2  3:1 ratio BUT only males had white eyes

6 Morgan’s Conclusion The gene for eye color must be carried on the X chromosome and NOT an autosome Higher probability of a male having the recessive phenotype

7 Sex Linked Traits

8 Color Blindness X b = No color blindness X B = color blindness Determine the crosses (both phenotype and genotype) for the following crosses 1.Color blind father x normal mother yields one color blind son. 2.Normal father, carrier mother. 3.Normal father, color blind mother. 4.What cross will yield a color blind daughter?

9 Sex Linked Traits Called hemizygous Do occur in females but males have a higher probability of inheriting the trait Duchenne muscular dystrophy, hemophilia, color blindness

10 X Inactivation in Females Having two X chromosomes is a lot of genes! One X chromosome will be turned off – Barr body: X chromosome condenses and will be near the nuclear envelope – Ovaries – Barr body will be duplicated for viable egg cells – In development, different X chromosomes could be turned off About ½ of the cells display the mother’s traits and half display the father’s

11 Review of Pedigrees and Blood Typing

12 Linked Genes Genes on the same chromosome that tend to be inherited together Morgan believed body color and wing shape were inherited together

13 Genetic Recombination Recombinant: Offspring show combinations of traits not found in the parents How does this happen?

14 Recombinants Mendel’s peas YyRr x yyrr – Complete the punnett square for the above cross. Which genotype and phenotypes are recombinants? Which are parental types?

15 Crossing Over During Meiosis I (Prophase I) Homologous chromosome pairs come together forming a tetrad – Crossing over Each chromosome will cross with the other in the pair Parts of the chromosome will be exchanged

16 Recombination Frequency How likely is it that the two genes will be linked? Based on how close they are on the chromosome – Closer they are, more likely they will be linked Linkage map: genetic map based on recombination frequency

17 Linkage Mapping Problem Mating: AaBb x aabb A = Long antennae a = Short antennae B = Green eyebrows, and b = Blue eyebrows. Say you make this mating, and your actual results look like this: Long Green - 850 Long Blue - 150 Short Green - 150 Short Blue - 850

18 Calculate Linkage Map Distance One linkage map unit (LMU) is 1% recombination. Thus, the linkage map distance between two genes is the percentage recombination between those genes. In this case, we have a total of 300 recombinant offspring, out of 2000 total offspring. Map distance is calculated as (# Recombinants)/(Total offspring) X 100. What is the LMU of the two genes?

19 Linkage Map Summary 1.How can you recognize when genes are linked? 2.How do you calculate linkage map distance? 3.What does the linkage map distance tell you? 4.How does a linkage map relate to independent assortment of genes?

20 Linkage Map for Drosophila melanogaster

21 Dihybrid Cross In summer squash, white fruit color (W) is dominant over yellow fruit color (w) and disk- shaped fruit (D) is dominant over sphere- shaped fruit (d). If a squash plant true- breeding for white, disk-shaped fruit is crossed with a plant true-breeding for yellow, sphere-shaped fruit, what will the phenotypic and genotypic ratios be for the F1 generation and the F2 generation?

22 F1

23 F2

24 Summary of Dihybrid Cross White color (W) is dominant over yellow fruit color (w) and disk-shaped fruit (D) is dominant over sphere-shaped fruit (d)

25 Chi Square (X 2 ) Test Determines statistical significance of a set of data Use in genetics: if genes sort INDEPENDENTLY they will follow the expected ratios of domintant : recessive – If not, they will not follow expected ratios and we would reject the null hypothesis

26 Chi Square Set Up PhenotypeGenotype#observed#expectedobserved – expected (o-e) (o-e) 2 (o-e) 2 / e

27 Monohybrid Cross Practice Problem A student makes a monohybrid cross with Drosophila. She crosses two heterozygotes for the white eye. Ww x Ww. She expects to see a 3:1 phenotypic ratio of Red eyes (WW and Ww) to white eyes (ww), her null hypothesis. She rears the next generation through to adult flies and counts the following numbers: – White eyes 210 – Wild type 680 What change occurs in the allelic frequencies between generations 1 and 2? Perform a chi square analysis on these results and find out if it is close enough to 3:1 to fail to reject her null hypothesis. Make sure to show all work and explain your conclusions.

28 Dihybrid Chi Square Problem In the garden pea, yellow cotyledon color is dominant to green, and inflated pod shape is dominant to the constricted form. Considering both of these traits jointly in self-fertilized dihybrids, the progeny appeared in the following numbers: – 193 green, inflated – 184 yellow constricted – 556 yellow, inflated – 61 green, constricted Do these genes assort independently? Support your answer using Chi-square analysis.

29 Baby chicks are pecking at grains of cracked corn of different colors. A researcher is attempting to determine if they have a preference for one color over another. He counts the number of grains chosen in an arena that has equal numbers of red, blue, green and yellow corn chunks. Here are his results: – red: 250 – blue: 220 – green 230 – yellow 240 Perform a Chi Square analysis on these results. Can he reject the null hypothesis of no preference? Explain your answer. Exit Ticket

30 15.4 Alterations of chromosome number or structure cause some genetic disorders

31 Abnormal Chromosome Number Nondisjunction: chromosomes do not separate correctly – Meiosis I or II

32 Aneuploidy Gamete with abnormal number of chromosomes unites with a normal gamete Offspring will have abnormal number of chromosomes Monosomic (2n – 1) Trisomic (2n + 1) Polyploidy: triploidy (3n), tetraploidy (4n)

33 Chromosomal Mutations Involve changes in the number or structure of chromosomes 4 types: – Deletion – Duplication – Inversion – Translocation

34 Deletion and Duplication Deletion: – Loss of all or part of a chromosome Duplication: – Produce extra copies of the chromosome

35 Inversion and Translocation Inversion: – Reverse direction of parts of the chromosome Translocation: – Part of one chromosome breaks off and attaches to another chromosome

36 Human Disorders caused by Chromosomal Alterations Turner Syndrome Down syndrome Cri du chat Prader-Willi Syndrome Huntington’s Disease Angleman Syndrome

37 Inheritance Patterns Not all fall into set patterns of equal inheritance Genomic Imprinting – Effect of the allele for a certain trait depends on which parent passed on the trait – Could be expressed in different strengths

38 Organelle Genes Extranuclear genes (found in cytoplasm) Mitochondrial genes Chloroplast genes NOT distributed to the offspring in the Mendelian fashion Come from mother – why?

39 Mitochondrial Genes Most make up the ETC and ATP synthase Defects in these genes will affect energy production – Most severely nervous system and muscular system

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