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Announcements Exam 1 next week, 9/19, 9/20 in testing center. Covers chapters 1 through 4, with emphasis on material: from lectures through 9/13, from.

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Presentation on theme: "Announcements Exam 1 next week, 9/19, 9/20 in testing center. Covers chapters 1 through 4, with emphasis on material: from lectures through 9/13, from."— Presentation transcript:

1 Announcements Exam 1 next week, 9/19, 9/20 in testing center. Covers chapters 1 through 4, with emphasis on material: from lectures through 9/13, from “Monk in the garden”, and from lab. Part multiple-choice, part short answer - emphasis on problem-solving. No time limit, but must finish that day so choose a 2-3 hr. time block. Closed book; bring calculator, #2 pencils and BLUE BOOK. **Review sessions next week in lecture and in lab. Bring your questions! “Problem set 2” answers due Friday, 9/13 at start of class Also practice Ch.4 problems this week (but do not turn in): 1, 7, 16, 27, 31. 3. re. printing power point slide files: when in computer room of Brooks, please choose “handout” when asked print what and print 6 slides/page. Do not print 1 slide per page.

2 Problem Set 2: due Friday 9/13 in class - show all your work.
The LM and LN alleles at the MN blood group locus exhibit codominance. Give the expected genotypes and phenotypes (with their ratios) of progeny from the following crosses: a) LMLM x LMLN b) LMLN x LNLN A woman of blood group AB marries a man of blood group A whose father was group O. What is the probability that: a) their 2 children will both be group A? b) one child will be group B and the other child group O? 3. In snapdragons, red flower color (R1) is incompletely dominant to white (R2); the R1/R2 heterozygotes are pink. A red-flowered snapdragon is crossed to a white colored one. Determine the ratios of the flower colors in the progeny from a cross of an F1 with the red parent. 2 points each question part for 10 points total

3 Review of last lecture I. Chi-square revisited: small deviation from expected yields small X2 value; this correlates with high probability that deviation is due to chance and you should NOT reject your hypothesis II. Pedigree analysis- recessive vs. dominant traits - solving pedigree problems

4 Solving Pedigree Problems
Inspect the pedigree: If trait is dominant, it will not skip generations nor be passed on to offspring unless parents have it. If trait is recessive, it will skip generations and will exist in carriers. Form a hypothesis, e.g. autosomal recessive. Deduce the genotypes. Check that genotypes are consistent with phenotypes. Revise hypothesis if necessary, e.g. autosomal dominant.

5 Pedigree Example 2: p. 71, #26

6 Outline of Lecture 7 In all crosses discussed so far, one of two traits for a character has been dominant to the other. ie. according to Mendel’s second postulate of dominance/ recessiveness. Does the expression of all genes occur in this way? ex. Are there only two colors of hair for humans with one clearly dominant to the other? NO I. Alleles alter phenotypes in different ways; a variety of symbols are used for alleles II. Incomplete dominance - where neither allele is dominant III. Codominance - both alleles in a heterozygote are expressed IV. Multiple alleles of a gene are studied in a population V. Lethal alleles - recessive or dominant VI. Modification of the 9:3:3:1 ratio

7 I. Alleles - alternate forms of the same gene
Wild-type allele - allele (form of gene) most frequently found in nature (“normal”); specifies normal phenotype and is usually dominant. Mutant allele specifies an altered phenotype. Mutation creates new alleles.

8 Gene Symbol Conventions
ebony body color mutation in Drosophila: e Normal (wildtype) color is gray: e+ e+/e+ or +/+ is homozygous wildtype e/e is homozygous ebony e+/e or +/e is heterozygous Other systems are also used, but symbol usually reflects the function of the gene, e.g. cdc, leu-, BRCA1

9 II. Incomplete Dominance
Neither of two alleles is dominant, e.g. snapdragon flower color: R1 is red R2 is white Heterozygotes give an intermediate (blended) phenotype: P1 cross gives pink flowers in F1 F1 cross gives 1:2:1 red:pink:white in F2

10 III. Codominance When two alleles of a gene specify two distinct, detectable gene products MN blood group in humans: LM, LN alleles MN locus codes for surface glycoprotein on red blood cells; can detect immunochemically. LM LM gives M phenotype LM LN gives MN phenotype LN LN gives N phenotype LM LN X LM LN produces 1/4 LM LM, 1/2 LM LN, 1/4 LN LN

11 IV. Multiple Alleles: ABO Blood Groups
When 3 or more alleles present (allelic series); can only be studied in populations. A and B alleles code for glycoproteins on red blood cells which can be detected immunochemically: mix blood sample with type A or type B antibodies look for clumping of RBC’s O allele carries neither antigen

12 ABO Blood Groups A - A antigen only B - B antigen only
AB - Both A and B antigens O - Neither antigen

13 ABO Genotypes and Phenotypes

14 ABO, continued IA and IB are codominant
Both IA and IB are dominant to IO All possible matings shown in Table 4.1 Applications: testing compatibility of blood transfusions disproving parentage of a child forensic science

15 Biochemical Basis of ABO
A and B antigens are on carbohydrate groups bound to fatty acids (glycolipid) on RBC membrane The A and B alleles code for enzymes that differentially process the carbohydrate during synthesis of the glycolipid: A enzyme adds N-acetylgalactosamine B enzyme adds galactose

16 Complexity with ABO blood groups: The Bombay Phenotype

17 Biochemical Basis of Bombay Phenotype
h mutation prevents addition of fucose to form H substance. A and B enzymes no longer recognize structure, don’t add A and B antigens. So individual is phenotypically O but can be genotypically A_, B_ or AB. H/h acts upstream of A and B in the pathway.

18 Bombay Phenotype: hh masks the expression of ABO (Epistasis)

19 The Secretor Locus also affects the expression of the ABO blood type
About 80% of human population have the A and B antigens present in various body secretions - not only in blood. Genetics - dominant allele, Se (Se/Se or Se/se) In what societal application would the secretor locus have significance? Forensic science - ABO blood typing can be performed on tissue samples other than blood.

20 Multiple alleles - a second example
White locus in Drosophila - over 100 alleles may occupy this locus. This results in an allelic series of eye colors ranging from pure white, to light buff to yellowish pink to deep ruby. (Table 4.3 in text)

21 V. Lethal Alleles Recessive lethal if heterozygote is viable, homozygous mutants die; common, e.g. Yellow (AY) in mice Manx (M) in cats Curly wing (Cy) in Drosophila Dominant lethal if either heterozygotes or homozygous mutants die; rare, e.g. Huntington disease

22 Mouse coat colors Agouti x agouti Yellow x yellow Agouti x yellow
All agouti 2/3 yellow, 1/3 agouti 1/2 yellow, 1/2 agouti

23 Effect of Dominant Yellow, Recessive Lethal in F2
AY is dominant to A AAY is yellow, but AYAY is lethal Results in 2:1 monohybrid ratio

24 VI. Modified Dihybrid Cross: First Consider Each Trait on its Own
Ex. 2 humans heterozygous for albinism and are blood type AB

25 Modified Dihybrid Cross: Next Consider Both Traits Together

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