Exam 1 W 26 in class bring cheat

Exam #1 W 2/6 in class (bring cheat sheet) Review T 2/5 5 -7 pm in PAI 3. 02

Genotype Phenotype Genes code for proteins (or RNA). These gene products give rise to traits… It is rarely this simple. Figs 1. 15 -17

CB 14. 2 Human blood types Codominance: both alleles are fully expressed AA or AO BB or BO AB (Codominance) OO

CB 14. 10 Incomplete dominance gives rise to blending.

Is this all of your DNA?

Mitochondria have their own DNA. CB pg 289 -290

Both Mitochondria and Chloroplast have DNA CB pg 289 -290

Endosymbiotic Theory - proposed origin of mitochondria and chloroplasts from free-living bacteria to cellular organelles CB 26. 13

Human Life Cycle Combination of two individuals DNA

Only the egg provides mitochondria to the offspring. CB 47. 6 mitochondria

Human Life Cycle In females Mom provides 50. 000275% and Dad provides 49. 999825% of DNA to offspring. … because Mom provides 100% of mitochondrial DNA

A few diseases are caused by mutations in mt. DNA

Pedigree of a mitochondrial disease: (black signifies affected individual) Which shape represents females?

Pedigree of a mitochondrial disease: Males and females may be affected by a disease coded on mt. DNA, but only females pass it on.

Mitochondrial DNA comparisons can be used to trace ancestry:

During the Bolshevik revolution, the Tsar’s family was captured and executed. Tsar’s Family

There are many stories about what happened to their youngest daughter Anastasia Tsar’s Family

Anna Anderson, claimed she was Anastasia

Anna Anderson claimed she was Anastasia, but tests of her mt. DNA and one of Anastasia’s maternal relatives did not match.

For more info check out: http: //en. wikipedia. org/wiki/Grand_Duchess_Anastasia_Nikolaevna_of_Russia Or the book “Seven Daughters of Eve” by Bryan Sykes

CB 13. 9 Crossingover Meiosis: In humans, crossing -over and (Ind. Assort. ) independent assortment lead to over 1 trillion possible unique gametes. (1, 000, 000) Meiosis II 4 Haploid cells, each unique

CB 14. 8 Tracking two separate genes, for two separate traits, each with two alleles. Ratio of 9: 3: 3: 1

Approximate position of seed color and shape genes in peas Y y Gene for seed color r Chrom. 1/7 R Chrom. 7/7 Gene for seed shape

CB 15. 5 Some crosses do not give the expected results

CB 15. 5 Heterozygous wild type Homozygous wild type gray w/ normal wings black w/vestigial wings b+ b vg+ vg b b vg vg

CB 15. 5 =25% 42% 41% 9% 8%

CB 15. 6 These two genes are on the same chromosome

CB 15. 6

By comparing recombination frequencies, a linkage map can be constructed = 17 m. u.

Homologous pair of chromosomes

CB 15. 8 Linkage map of Drosophila chromosome 2

Recombination is not completely random. physical distance Yeast chromosome 3 linkage map

Try this: Does this pedigree show recombination? dominant alleles grandparents offspring recessive alleles

Try this: Does this pedigree show recombination? dominant alleles grandparents offspring arental ecombinant recessive alleles