These videos are to support students in BIOL 220 Introductory Biochemistry (Principles of Biochemistry, Fifth Edition, Moran, Horton, Scrimgeour, Perry. Pearson Publishing, 2012) and BIOL 223 Genetics (iGenetics: A Molecular Approach, Third Edition/New International Edition, Russell, Pearson Publishing, 2010/2014, the difference being 11/12 Mendelian Genetics 12/11 Chromosomal Basis of Inheritance, respectively). Some employ standard textbook approaches, and some use my own. They are hosted on YouTube. Several are in parts because the durations were too long or the file sizes were too large.
BIOL 220 how to use line structures
About “line structures” the preferred notation of organic and biochemists.
BIOL 220 chemical equilibria and Gibbs free energy part 1 of 3
An explanation of chemical equilibria, part 1 of 3.
BIOL 220 chemical equilibria and Gibbs free energy part 2 of 3
An explanation of chemical equilibria, part 2 of 3.
BIOL 220 chemical equilibria and Gibbs free energy part 3 of 3
An explanation of chemical equilibria, part 3 of 3.
BIOL 220 Horton 5th 2.7 weak acid equilibria part 1 of 1
Problem 7 of chapter 7 in Principles of Biochemistry, Fifth Edition. What is the concentration of a lactic acid buffer (pKa=3.9) that contains 0.25 M lactic acid and 0.15 M lactate? What is the pH of this buffer?
BIOL 220 Moran, Horton, et al. 5th problem 3.9
Problem 9 of chapter 3 (charge of a peptide at different pH) from Principles of Biochemistry, Fifth Edition
BIOL 220 Moran, Horton, et al. 5th problem 3.12
Problem 12 of chapter 3 (base titration of histidine) from Principles of Biochemistry, Fifth Edition.
BIOL 220 Moran, Horton, et al. 5th problem 3.18 part 1 of 2
Problem 18 of chapter 3 (concentration of neutral, non-zwitterionic form of alanine) from Principles of Biochemistry, Fifth Edition.
BIOL 220 Moran, Horton, et al. 5th problem 3.18 part 2 of 2
Problem 18 of chapter 3 (concentration of neutral, non-zwitterionic form of alanine) from Principles of Biochemistry, Fifth Edition.
BIOL 322 dissociation constants
This is to explain dissociation constants. If the unlabeled partner is not in excess, one has a titration, and the concentration at which bound and free are equal is not the dissociation constant.
BIOL 223 bases and base pairs
An explanation of the chemical structure of DNA and RNA bases. What is meant by Watson-Crick geometry.
BIOL 223 Ames test
Exam 21 Question 6 answered. How one could construct an Ames test.
BIOL 223 Russell Chromosomal Basis of Inheritance p19, elaborated
BIOL 223 Russell Chromosomal Basis of Inheritance problem 19, elaborated with aspects from the chapters on Extensions and Deviations from Mendelian Genetic Principles and Regulation of Gene Expression in Eukaryotes.
BIOL 223 Russell Mendelian Genetics p31
Mendelian Genetics problem 31, an example from Hymenoptera that students often find confusing despite being analogous to human X-chromosome inheritance.
BIOL 223 Russell Mendelian Genetics p37
Mendelian Genetics problem 37 explained to illustrate the problem-solving strategy in a relatively simple metabolic pathway without partial or co-dominance. A short cut taken in the book’s answer is done a generalizable way.
BIOL 223 brief and problem: A bisexually reproducing all-triploid vertebrate
BIOL 223 brief presentation and sample problem about A bisexually reproducing all-triploid vertebrate, Stöck et al. March 2002. Nature Genetics. Volume 30 p325.
BIOL 223: sample traditional pedigree problems
BIOL 223: sample traditional pedigree problems (of the if-this-person-displays-the-trait,-what-is-the-likelihood-a-future-child-will-also? sort).
BIOL 223 brief and problem part 1/2: Evolution of flower color pattern through selection on regulatory small RNAs
BIOL 223 brief and problem part 1/2: Evolution of flower color pattern through selection on regulatory small RNAs. Bradley D, Xu P, Mohorianu II, Whibley A, Field D, Tavares H, Couchman M, Copsey L, Carpenter R, Li M, Li Q, Xue Y, Dalmay T, Coen E. Science. 2017 Nov 17;358(6365):925-928. doi: 10.1126/science.aao3526.
BIOL 223 brief and problem part 2/2: Evolution of flower color pattern through selection on regulatory small RNAs
BIOL 223 brief and problem part 2/2: Evolution of flower color pattern through selection on regulatory small RNAs. Bradley D, Xu P, Mohorianu II, Whibley A, Field D, Tavares H, Couchman M, Copsey L, Carpenter R, Li M, Li Q, Xue Y, Dalmay T, Coen E. Science. 2017 Nov 17;358(6365):925-928. doi: 10.1126/science.aao3526.
BIOL 223-200820-25q6-1-a-A
Question 6 of exam 25, Mars breeding program pedigree, single dominant, autosomal allele, individual in third generation.
BIOL 223-200820-25q6-2-bc-A
Question 6 of exam 25, Mars breeding program pedigree, single dominant, autosomal allele, individuals in fourth and fifth generations.
BIOL 223-200820-25q6-3-d-A
Question 6 of exam 25, Mars breeding program pedigree, single X-linked recessive allele, individual in fifth generation.
BIOL 223-200820-25q6-4-a-X
Question 6 of exam 25, Mars breeding program pedigree, single X-linked recessive allele, individual in third generation.
BIOL 223-200820-25q6-5-bc_X
Question 6 of exam 25, Mars breeding program pedigree, single X-linked recessive allele, individuals in fourth and fifth generations.
BIOL 223-200820-25q6-6-d-X
Question 6 of exam 25, Mars breeding program pedigree, single X-linked recessive allele, individual in fifth generation.
BIOL 223 Russell 14.14 part 1/2
BIOL 223 Russell Genetic Mapping in Eukaryotes problem 14 part 1/2. Note occasional misstatements: In Drosophila, only females (not males) undergo crossing over during meiosis, which means mapping requires multiply heterozygous females. This problem would make a very difficult exam question, yet it is good as an exercise because it recapitulates the process of conceiving and testing possibilities.
BIOL 223 Russell 14.14 part 2/2
BIOL 223 Russell Genetic Mapping in Eukaryotes problem 14 part 2/2. Note occasional misstatements: In Drosophila, only females (not males) undergo crossing over during meiosis, which means mapping requires multiply heterozygous females. This problem would make a very difficult exam question, yet it is good as an exercise because it recapitulates the process of conceiving and testing possibilities.
BIOL 223 Russell’s iGenetics 3rd ed problem 14.31 part 1 of 3
Problem 31 of chapter 14, part 1 of 3, linkage mapping, solved a different method (conceptual not math) iGenetics A Molecular Approach.
BIOL 223 Russell’s iGenetics 3rd ed problem 14.31 part 2 of 3
Problem 31 of chapter 14, part 2 of 3, linkage mapping, solved a different method (conceptual not math) iGenetics A Molecular Approach.
BIOL 223 Russell’s iGenetics 3rd ed problem 14.31 part 3 of 3
Problem 31 of chapter 14, part 3 of 3, linkage mapping, solved a different method (conceptual not math) iGenetics A Molecular Approach.
BIOL 223: Russell Genetic Mapping in Eukaryotes p31 backwards part 1/3
BIOL 223: Russell Genetic Mapping in Eukaryotes problem 31 backwards part 1/3 This relates to problem 26 as well.
BIOL 223: Russell Genetic Mapping in Eukaryotes p31 backwards part 2/3
BIOL 223: Russell Genetic Mapping in Eukaryotes problem 31 backwards part 2/3 This relates to problem 26 as well.
BIOL 223: Russell Genetic Mapping in Eukaryotes p31 backwards part 3/3
BIOL 223: Russell Genetic Mapping in Eukaryotes problem 31 backwards part 3/3 This relates to problem 26 as well.
BIOL 223-202020-HW3abc
BIOL 223-202020-HW3abc: solving the easier part of HW3.
BIOL 223-202020-HW3d
BIOL 223-202020-HW3d. This is about the limit of what is worth the time it takes to solve on an exam: there is one only recombination.
BIOL 223-202020-HW3e part 1/2
BIOL 223-202020-HW3e part 1/2. Starting from the end of d, this needs two recombinations and is here as an exercise to practice and demonstrate the method. I would not consider such a problem to be worth the time needed on an exam: your effort would be better spent elsewhere. This problem, especially in part 2/2, illustrates the common observation that most mistakes are made on simple matters (for example, 9+27 does not equal 32), and that there are roles for computers.
BIOL 223-202020-HW3e part 2/2
BIOL 223-202020-HW3e part 2/2. Starting from the end of part 1/2. This needs two recombinations and is here as an exercise to practice and demonstrate the method. I would not consider such a problem to be worth the time needed on an exam: your effort would be better spent elsewhere. This problem, especially in part 2/2, illustrates the common observation that most mistakes are made on simple matters (for example, 9+27 does not equal 32), and that there are roles for computers.
BIOL 223 problem: Pedigree with recombination – method of tracking gamete pools as fractions
BIOL 223 problem: Pedigree with recombination. This is to demonstrate how to solve inheritance problems from ancestors to descendants by tracking gamete pools with fraction of whole numbers and replaces a previous video that used decimals and fractions and was a bit messy.
BIOL 223 chiasma resolution
How to resolve chiasmata of ordered tetrads such as Neurospora crassa.
BIOL 223: Stella safeguards the oocyte methylome by preventing de novo methylation mediated by DNMT1
For BIOL 223 Genetics, a sample brief and how to approach an abstract (or here, a first paragraph) on Stella safeguards the oocyte methylome by preventing de novo methylation mediated by DNMT1. Li Y, Zhang Z, Chen J, Liu W, Lai W, Liu B, Li X, Liu L, Xu S, Dong Q, Wang M, Duan X, Tan J, Zheng Y, Zhang P, Fan G, Wong J, Xu GL, Wang Z, Wang H, Gao S, Zhu B. Nature. 2018 Dec;564(7734):136-140. doi: 10.1038/s41586-018-0751-5.