Genetics
Back to Denome's Home Page Old Genetics Tests Sample Classical Genetics Problems
Genetics
Biology 202, Spring 2006
Texts
Denome, Synopsis of Genetics (Required).
If you feel the need for another book, let me know. I have a whole shelf of them.
Instructor
Roger Denome, Associate Professor of Biology, rdenome@stonehill.edu
Office hours: T-F 8:30-9:30, T-F 11:15-12:00, and by appointment.
Room 213 Science building. Office phone: 565-1196. Lab phone: 565-1171
Web site: http://faculty.stonehill.edu/rdenome/genetics.htm
Class Times and Rooms:
Lectures
Tuesday, Thursday, 10:00-11:15, in Martin 105
Lab
Tuesday, Wednesday, Thursday 1:30-4:00, room 022 Science building.
Labs will begin in a room other than 022 and move to room 022 after initial discussions.
Lecture Schedule
1/19 Introduction, basic cell biology. Chpt. 1
1/24 Mendelian genetics. Chpt. 2
1/26 Rules of Expression. Chpt. 2
1/31 Genetics crosses. Chpt. 4
2/2 Genetic linkage and Mapping Chpt. 4
2/3 10:00 a.m. First written assignment due.
2/7 Genetic linkage and Mapping Chpt. 4
2/9 Test over chapters 1,2,4
2/14 Pedigrees and abnormalities. Chpt. 5
2/16 Chromosome abnormalities. Chpt. 5
2/21 Population genetics. Types of variation. Chpt. 15
2/23 Hardy/Weinberg equilibrium, Inbreeding. Chpt. 15
2/24 10:00 a.m. Second written assignment due.
2/28 Mutation and Genetic drift. Chpt. 15
3/3 Test over chapters 5,15
Spring break 3/4-3/12
3/14 DNA, DNA replication. Chpt. 6
3/16 DNA replication. Chpt. 6
3/21 DNA replication, RNA. Chpt. 7
3/23 Transcription. Chpt. 7
3/28 Transcription. Chpt. 8
3/28 10:00 a.m. Third written assignment due.
3/30 Translation. Chpt. 8
4/4 Test over chapters 6-8
4/6 Prokaryotes: Operons, Lac and Trp. Chpt. 9
4/11 Prokaryotes: Operons, Lac and Trp. Chpt. 9
4/18 Attenuation, eukaryotes: DNA packaging. Chpt. 9,10
4/20 Eukaryotic gene regulation. Chpt. 10
4/25 Eukaryotic gene regulation. Chpt. 10
4/25 10:00 a.m. Last written assignment due.
4/27 Eukaryotic gene regulation. Chpt. 10
5/2 Test over chapters 9,10
5/4 Review for Final
5/6 8:30 Final exam
Expectations
You are expected to know and be able to work with the material presented in the Synopsis of Genetics. This is 20-40 pages of text (containing many figures) per test. I will follow the format and information in the Synopsis very closely and will use the figures from that text in lecture. Bring the Synopsis to class with you. Also note that the Synopsis is not a normal text. There isn't a great deal of material in it that you are not expected to know. When in doubt, assume that you need to know everything that is in the Synopsis.
Tests and Written Assignments: There will be 4 lecture tests (each worth 100 points), a cumulative final (200 points) and 4 papers (worth 25 points each). Lecture tests will emphasize the material covered in the previous section of the course (5-6 lectures), plus lab. It is impossible to keep these tests from being cumulative; there is always overlap in the material from section to section and from lecture to lab. The format for the tests will depend on the material. The last 7 pages of the syllabus are old tests.
Written assignments are intended to make you use information presented in class to address a series of topics. These papers are take-home exams. You should use information from both lecture and lab for the written assignments. The content and format for the papers will vary with the subject material. These papers should be word-processed (or typed) and conform to standard rules of English spelling and grammar. Significant deviations from this may result in up to 20% deduction in points.
Grading: There are two ways to get a grade in this class. I figure out grades both ways and award whichever grade is higher.
Grade based on Total points.
The first grading method is based on the sum of all points for the semester, using the following scale:
Total points Grade
609-700 points (87-100%) A
546-608.5 points (78-86.9%) B
483-545.5 points (69-77.9%) C
420-482.5 points (60-68.9%) D
< 420 points (below 60%) F
This scale is for the total points for the class. The average score for the class for the past 10 semesters has been about 72.5%. The scale given here is based on this average. In general, there are few students who score above 90%. This is why I slide the scale down a bit at the top.
The tests and papers in this class have different levels of difficulty. In general, the first test is the hardest (mean=~ 65%), the third test is the easiest (mean=~78-80%). The average score on the papers is about 20 out of 25 points (80%). This means that the grading scale given above does not work for individual assignments; it only works for the final sum of all points at the end of the class.
Because students want to know their grades during the semester, given below is a set of grading scales for 4 points in the semester. These scales are approximate. The only scale that is absolute is the one for the total points given above.
Approximate Midterm Grading Scales
After first test After second test After third test After fourth test
84-100% =A 86-100% =A 88-100% =A 87-100% =A
73-84% =B 75-86% =B 79-88% =B 78-87% =B
60-72% =C 64-75% =C 70-79% =C 69-78% =C
50-60% =D 55-64% =D 60-70% =D 60-69% =D
<50% =F <55% =F <60% =F <60% =F
What this is saying is this. If you have 78% of the possible points after the second test, you are doing “low B” work. If you still have 78% of the possible points after the third test, you are doing “high C” work. The reason for this is that the third test is pretty easy, and to maintain your “B” grade, your score on the third test should raise your average.
Grade based on final exam. The second way of getting a grade in this class is to base your entire grade on the comprehensive final exam. If you do well on the final exam, that shows that you are competent in genetics. However, it cannot show that you are superior in genetics. Consequently, I will not award “A's” based on the final exam alone. This way of assigning grades must be more conservative; it is based on “straight scale”.
Final Exam Grading scale
90-100%= A- 80-100%= B 70-79.9% = C 60-69.9% = D <60% = F
The intent of this second grading method is to supply a “safety net” for students who have a disaster sometime during the semester. If you can do well on the final, you can get a decent grade.
Finally, I offer this little bit of incentive to do well during the semester. Anyone who finishes the semester with greater than 87% of the points (an “A-“) does not have to take the final exam.
Labs meet on Wednesday and Thursday. Depending on how you do in the first two weeks of class, we might want to schedule additional problem solving time before the first exam. I will be as flexible as I can in scheduling; the material on the first test absolutely requires that you have practice solving genetics problems. That practice time is the lab periods for the first few weeks of class, plus other times that we schedule.
Week
1/19-1/25 Classical genetics basics and Meiosis.
1/26-2/1 Classical genetics problem solving. Basic approach.
2/2-2/8 Problem solving and mapping.
2/9-2/15 Pedigrees
2/16-2/22 Nucleic acid isolation.
2/23-3/1 Nucleic acid isolation.
3/14-3/16 Nuclei acid identification and Electrophoresis.
3/21-3/23 PCR.
3/28-3/30 Basic gene cloning techniques. Bacterial transformation.
4/4-4/6 Basic gene cloning techniques. Sequencing part I.
4/18-4/20 Sequencing part II.
4/25-4/27 Undefined.
5/2-5/4 Review for final
Notes on plagiarism.
You may discuss written assignments with each other. However, it is essential that the final product of the written assignments be individual projects. Please note the following rules for plagiarism.
1. Make sure that you are responsible for your wording. It is your responsibility to make sure that you put concepts in your own words, not those of a common study group or text. Similar wording and sentence structure are indicators of plagiarism. Copying, then altering a passage slightly so that the wording is not exactly the same as the original is still plagiarism.
2. Give credit to sources. If you do use wording or a significant body of ideas from another source, you must give credit to that source. This means including correct references and page numbers for the source. Using someone's work without giving credit is plagiarism.
In this class, you can't use the instructor, another student or the textbook as a referenced source. This means that you must come up with ways of saying things that are very different from the text. I realize that this is a departure from requirements for other classes, but it is necessary in this case. Written assignments are to see if you understand the material. If all you do is quote from outside sources, you haven't shown me that you understand the material. Work to express the concepts and facts of genetics in your own words.
There are only so many ways to describe many phenomena. The use of jargon words and phrases in genetics is not plagiarism. Plagiarism is the use of significant sections (a sentence or more) from another source without giving credit to the person who produced the original.
3. Response to plagiarism. If I encounter written assignments that seem to be plagiarism, the student(s) involved will be given the opportunity to explain. If the explanation is not satisfactory to all parties involved, an independent member of the science faculty will review the material. If the material is deemed plagiarism, then all of the papers involved will be given a score of 0 (zero). In addition, college rules for dealing with plagiarism will be enforced. This involves reporting the infraction to the Academic Review Board. A second offense will result in a grade of “F” for the course.
Expectations
Classical genetics.
A. You should understand and be able to use (correctly) the terminology and symbols used in classical genetic analysis.
B. You should understand the concepts and cellular mechanisms associated with gene transmission and expression.
1. Gene transmission. Mitosis, cell cycle, meiosis; segregation, independent assortment, recombination; chromosomal structure, chromosomal aberrations, ploidy, euploid and aneuploid variants.
2. Gene expression. Dominance, recessiveness, codominance, incomplete dominance, epistasis, pleiotropy; polygenic inheritance, lethals, variable expressivity, incomplete penetrance; sex-linked, X-linked, sex-limited, holandric inheritance; pseudodominance, pseudolinkage, interference.
3. You should know how those concepts listed in B.1. and B.2. above are manifest in classical genetic analysis.
C. You should be able to use the analytical techniques of classical genetics.
1. Techniques: recombination mapping, somatic cell hybrid mapping, pedigree analysis.
2. Statistics and probability used to make predictions and analyze data.
D. You should be able to develop hypotheses based on data given to you and be able to design experiments (and make predictions) to test your hypotheses.
Population genetics.
A. You should understand and be able to use the terminology and concepts of basic population genetics.
B. You should understand how Hardy/Weinberg equilibrium works, and the effects of violating the conditions required for Hardy/Weinberg equilibrium. This includes being able to deal with each assumption conceptually and mathematically.
C. You should understand how real populations change over time and the relative importance of various aspects of population biology (mating and movement patterns, selection, etc.) in causing this change.
Biochemistry of the gene.
A. You should understand and be able to use the terminology, symbols and techniques used in the biochemistry/cell biology of genetics. These includes chemical formulae, reactions, macromolecular components such as ribosomes and replication complexes, and cellular/subcellular compartments, plus the basic lab techniques of molecular biology.
B. You should understand the basic biochemistry of gene expression, both in terms of the components involved and the functions that they serve.
1. DNA structure. Nucleotides, polarity, chemistry.
2. DNA replication. The enzymology and processes involved in bidirectional and rolling circle DNA synthesis, plus proofreading and DNA repair.
3. Chromatin structure in eukaryotes. Histones, nucleosomes, supercoiling, higher order structure.
4. RNA structure. Nucleotides, polarity, chemistry.
5. Transcription. The enzymology and processes involved in transcription.
6. RNA processing. Processing steps and purposes.
7. Protein structure. Amino acids, peptide bonds, polarity, primary structure, secondary structure, tertiary structure, quaternary structure, allosteric proteins, protein function, native and denatured configurations.
8. Translation. The enzymology and processes involved in protein synthesis; other processes necessary for protein synthesis to take place such as charging of tRNAs with amino acids.
9. Protein processing. Proteolytic cleavage, modification by the addition of adducts. Stabilization of tertiary/quaternary structure by -S-S- bridges.
10. Regulation of gene expression. DNA rearrangement, promoter utilization in prokaryotes; promoter utilization in Eukaryotes; RNA processing, rRNA transport, RNA stability; translation, Protein stability.
Early papers.
I will look at first drafts of papers up until 48 hours before the paper is due. The closer to the due date, the less information you will get back from me. This is a simple matter of time. As the due date gets closer, the number of students asking for input increases; I have less time to review each paper. In any event, I will point out where there are problems with a paper and possibly suggest where to go to get the solutions to these problems. I will try my best to catch all of the problems, but given time constraints, I may miss some. Don't assume that I have caught everything. It's your paper, not mine. It's your responsibility to make sure everything is in it that you want in it.
Late papers.
My standard policy for late written assignments is 20% of the maximum point total for the paper will be deducted from the score for each day that the paper is late. (A paper is one day late if it is from 1 minute to 24 hours late.) Obviously, this suggests that getting the paper in on time makes sense. The assignments for papers and the date on which they are due are announced long in advance (syllabus, page 6). It is difficult to imagine an excuse that would justify turning in a paper late. It is conceivable that such an excuse is out there. If you think that you have such an excuse, let me know and we'll discuss it. I might waive or reduce the 20% deduction.
Late tests.
The only sure excuse for taking a test late is death in self or a member of your immediate family. Everything else you need to discuss with me. In all cases I need some sort of independent confirmation of the excuse. This may be a doctor's appointment slip or some other “paper” confirmation. If the issue causing you to miss a test is too private to discuss with me, then I will depend on another member of the faculty or staff at Stonehill to confirm your excuse. Just let me know who to talk to. Various deans, members of the counseling center and other faculty members have been helpful in the past.
A makeup test will cover the same material as the original test, although the format may change. The test must be given as close to the date of the original as possible. When I grade a makeup test, I try to keep the grading on a par with the original. However, since the formats of the test may be different, it may be impossible to prove that the grading is comparable. This is just one of the problems with makeup tests.
Corrections in Synopsis of Genetics.
page 26, left column, three lines below “F2” should read
“50% white eyed females”. (text reads red eyed females)
page 29, right column, the line below “Parental generation” should read
“AAbbcc X aaBBCC” (Switch capital and lower case “c's”.)
page 37, right column, figure title should read.
“pericentric inversion heterozygote” (reads “paracentric” now.)
This figure goes with the discussion on the following page (left column text “pericentric inversions”), and the figure on page 38 goes with the discussion in the right column of page 37 (right column text “paracentric inversions”).
page 52, left column, DNA Chemistry A.5.a. In parentheses should read
“(A=T, G=C)”.
page 75, right column, Prokaryotic gene regulation, A.7.a., third line change
“eukaryotes” to “prokaryotes”.
page 91, right column, figure, top right cell (over the big ”+”) should read
“laca- lacb- “. (The superscript on the first “lac” is wrong.)
page 121, left column, Deviations from Hardy/Weinberg equilibrium conditions,
C. “Non random mating”.
Study hints.
1. Study outline. At the end of each chapter in the synopsis is a study outline that breaks the material into logical sections. This is a good place to start defining what you need to work on.
2. Index. Because Genetics is so “terminology rich”, knowing the definitions of words goes a long way towards learning the material. The synopsis has an alphabetical listing of terms (siting where they are first used or defined), plus a list of terminology in the order that it first appears. This second list is a good place to start reviewing for tests. Go through and make sure that you know what each term means.
3. Write it out. If you can write out or draw the process in detail, starting from a blank sheet of paper, without looking at your notes, you are probably going to be able to use that process to answer questions. Beware of just being “familiar” with the material. I want you to be able to use the material to deal with new situations. Simple familiarity won't help.
4. Work in groups. If you can teach the material, then you will know it better. Also, different students will have different strengths and so are able to help each other with different problems.
5. Write test questions. Most of the test questions that I write are pretty obvious. Go through and figure out what the questions are going to be and answer them. Think like a test writer.
6. Do the problems. Especially in classical genetics, the toughest and most time consuming questions are mapping/pedigree problems. Do practice problems without looking at the answers!!
7. Get help. Don't put off getting help, even for small problems. All of this material is interconnected. That little problem will come back to haunt you later. Fix it now. ASK!!!!!!!First Genetics Paper, Spring 2005
First Genetics Paper, Spring 2006
25 points total. Paper is due at 10:00 a.m. on February 3, 2006.
1. Draw and describe clearly meiosis in a male mammal. The haploid number of chromosomes in this species is 5 (i.e. 1N = 5 chromosomes). The only organelles that you need to pay attention to are the spindle, the chromosomes, the nucleus and nuclear membrane, and the plasma membrane. Make sure that you clearly distinguish:
A. which chromosomes are homologous, and whether they are maternal or paternal in origin.
B. which chromosomes are autosomes and which are sex chromosomes.
C. sites of recombination and the products of recombination.
D. at what point cells become haploid, and at what point they attain the right number of chromatids to function as gametes.
E. how the meiosis that you have drawn would differ from meiosis in other cells in the same male.
Label everything clearly the first time you draw it. I can't read your mind.
The basic drawing is worth 10 points. A-E are worth 3 points each.
Hint: draw clear, large (whole page) diagrams and use them as the starting point for your descriptions. Tiny diagrams are almost always ambiguous and they are very difficult to label well. Use colored pencils to help make the distinctions requested above.
Second Genetics Paper, Spring 2006
25 points total. Paper is due at 10:00 a.m. on February 24, 2006.
Answer completely and concisely. Use diagrams, if you think that will make your answer clearer.
1. 15 points. Please describe the biology of an individual that is heterozygous for a pericentric inversion. Include descriptions of the effects on the heterozygote's own viability and its ability to make gametes. How could such a translocation be created?
2. 10 points. Please describe Hardy/Weinberg equilibrium, the conditions required to maintain it, and the value of this concept to population genetics. Describe the effects on allele and genotype frequencies when each of the assumptions is violated (individually) in a population that is in Hardy/Weinberg equilibrium in all other ways.
Third Genetics Paper, Spring 2006
25 points total. Paper is due at 10:00 a.m. on March 28, 2006.
1. 10 points. Please draw a double stranded DNA molecule that has the sequence 5' AAGCT 3'
on one strand. (This is the entire sequence of one DNA strand. It consists of only these 5 nucleotides. You figure out the sequence of the other strand.) Include all atoms (including hydrogens) and covalent bonds. Indicate roughly where hydrogen bonds are between strands.
2. 15 points. Describe tRNA synthesis and processing in eukaryotes. Include the basic reactions that are common to all RNA synthesis in prokaryotes. How does it differ from the same process in eukaryotes?
Last Genetics Paper, Spring 2006
25 points total. Paper is due at 10:00 a.m. on April 25, 2006.
1. Describe the regulatory mechanisms used in the Tryptophan operon. Please describe the organization of the DNA, the purpose of each section of DNA, and all regulatory phenomena and mechanisms. Make sure you describe the purpose of the whole operon, and why that purpose is important to the organism.
Genetics. Sample First test
Sample test answers start on page 14 of the syllabus.
Write your name at the top of each page. Name
This test consists of 7 short answer questions (4 points each), 11 multiple choice questions (4 points each), and 2 problems (14 points each).
For the short answer questions, you must answer the question in 25 words or less. Make it clear, concise, and legible. There is partial credit on short answer questions.
For the multiple choice questions, choose the best answer from those available. Write your answer in the space provided. Only unambiguous answers written in the correct place will count. There is no partial credit on multiple choice questions.
For the problems, you may use a calculator, but make sure that you write out your calculations so that I can see what math you did to get to your answer. Make sure that you answer the questions that are asked clearly, concisely, and completely. Partial credit will be given for problems. However, a correct answer in which there is additional incorrect material will not receive full credit.
If you need clarification of a question or problem, raise your hand. I will clarify standard English, but I will not define genetics terms for you.
Short answer questions. 4 points each.
1. At what stage of meiosis would you see single chromatid chromosomes for the first time?
2. A sixth grade student mates her female black mouse (Josephine) with her male black mouse (Napoleon). Josephine and Napoleon make lots of baby mice over a 12 month period. When all of the offspring of this mating are added up, the results are:
40 female black mice
20 male black mice
How do you explain this result?
3. What is the difference between maternal effect and maternal inheritance?
4. Recently, developmental biologists have created genetically identical copies of some mammals. (“Cloning”). In one case, they produced two calves that were both genetically identical to their “mother”. These black and white calves did not have identical patterns of black and white spots, nor was their coloration identical to their mother. Given that the calves are genetically identical to their mother, explain this lack of identical coloration. I need an explanation, not just a name.
5. The fD allele of the F gene causes fatal familial insomnia (FFI). FFI is a disease in which people slowly become fatigued, lethargic, and eventually enter a coma and die. The fD allele is recessive, incompletely penetrant and has variable expressivity. Describe the phenotype that you would expect to see in an fDfD individual.
6. In classical genetics, what is used to measure the distance between two genes? (Hint: “map units” is not enough.)
Multiple choice questions. Choose the best answer from those given.
7. In butterflies, females are the heterogametic sex. Dosage compensation is achieved by X inactivation in butterflies. How will this effect phenotypes in normal butterflies.
A. Females that are heterozygous for X-linked genes that have cell-autonomous expression will have mosaic phenotypes.
B. Males that are heterozygous for X-linked genes that have cell-autonomous expression will have mosaic phenotypes.
C. All males and females will have mosaic phenotypes.
D. Neither males nor females will show mosaic phenotypes.
E. Males that are heterozygous for X-linked genes that have cell-autonomous expression will have mosaic phenotypes, but only for genes involved in exclusively male phenotypes.
8. A primate researcher crosses a purebred, female, shaggy coat, Rh+ rhesus monkey with a purebred, male, short coat, Rh- male monkey. (Rh is a blood type.) This mating is repeated to produce 20 offspring. The offspring are:
10 short coat Rh+ males
10 short coat Rh+ females.
One gene controls coat length and another controls Rh; which of the following is correct?
A. The allele causing short coat is dominant to the allele causing shaggy coat.
The allele causing Rh+ is dominant to the allele causing Rh-.
The gene controlling coat length is autosomal; the gene controlling Rh blood type could be
either X-linked or autosomal.
B. The allele causing short coat is dominant to the allele causing shaggy coat.
The allele causing Rh+ is dominant to the allele causing Rh-.
The genes controlling coat length and Rh blood type are X-linked.
C. The allele causing short coat is dominant to the allele causing shaggy coat.
The allele causing Rh+ is dominant to the allele causing Rh-.
The genes controlling coat length and Rh blood type could be autosomal or X-linked.
D. The allele causing short coat is dominant to the allele causing shaggy coat.
The alleles causing Rh+ and Rh- are codominant.
The genes controlling coat length and Rh blood type could be autosomal or X-linked.
E. The allele causing short coat and the allele causing shaggy coat show incomplete dominance.
The allele causing Rh+ is dominant to the allele causing Rh-.
The genes controlling coat length and Rh blood type could be autosomal or X-linked.
9. In humans, the Rb+/Rb- gene is on the third autosome. Individuals carrying the Rb- allele develop retinal cancer at an early age. Females carrying this allele also have an increased risk of ovarian, uterine, and breast cancer. Males carrying Rb- have an increased risk of testicular and prostate cancer. The expression of the Rb- allele is:
A. Sex-limited B. Sex-linked C. Sex-influenced
D. Dosage compensated E. Lyonized
Use these data for questions 10 and 11. In guppies (little fish), males have flamboyant coloration, long fins, and extravagant behavior patterns. Females never show these characteristics. The genes controlling these phenotypes are all on the second autosome, and combinations of alleles of these genes are transmitted as a unit from one generation to the next.
10. The group of alleles of these genes that is passed as a unit is called:
A. a chromatid B. a haploid C. a sex chromosome
D. a haplotype E. pleiotropy
11. The expression of these genes in guppies is:
A. sex-limited B. sex-linked C. sex-influenced
D. dosage compensated E. Lyonized
12. A geneticist has been studying sheep fertility for years. She has found 6 different genes with alleles that affect fertility. These loci are on various chromosomes in the sheep genome. None of these loci act as the sole determinant of the level of fertility. This is a case of:
A. pleiotropy B. additive inheritance C. epistasis
D. sex-linked inheritance E. variable expressivity.
13. John and Jane both have type AB blood. Their firstborn (Judy) has type O blood, but all other tests indicate that Judy is in fact the biological offspring of John and Jane. This is a case of:
A. pleiotropy B. phenocopy C. epistasis
D. incomplete penetrance E. variable expressivity.
14. Cats that carry the CBlue allele are white, they have blue eyes, and they are deaf.
This is an example of:
A. pleiotropy B. additive inheritance C. epistasis
D. sex-linked inheritance E. variable expressivity.
15. The R/r, S/s, G/g and T/t genes are all on different autosomes. An individual who is heterozygous for all of these genes makes gametes. What proportion of the gametes will have the genotype RsTG?
A. 0.5 B. 0.52 C. 0.54 D. 0.58
E. There is not enough information here to answer this question.
16. The gene controlling Red/green color sight is on the X chromosome in humans; the allele that causes red green color blindness is recessive to the allele causing color sight.
A gene controlling hearing is on the second autosome; the allele causing deafness is dominant to the allele causing normal hearing.
Mary's father is color blind and can hear. Mary can see color and is deaf. Mary's husband can see color and hear. What proportion of Mary's children will be color blind and deaf?
A. 1/2 B. 1/4 C. 1/8 D. 1/16 E. 1/32
Use the following information for questions 17 and 18. The A/a and B/b genes are autosomal and 15 map units apart. A purebred individual showing the A and b phenotypes is crossed with a purebred individual with the a and B phenotypes. The offspring make gametes.
17. What proportion of the gametes will have the genotype AB?
A. 50% B. 25% C. 15% D. 7.5% E. None
18. very short answer question. The offspring of the mating above are allowed to mate among themselves. What proportion of the F2 generation will have the same genotypes as the original purebred parents. Just show the calculation.
Problems. 14 points each.
Problem #1
Purebred female black cats with long fur and green eyes are crossed with purebred ginger (= orange) cats with short fur and yellow eyes. The offspring are:
males black, long fur, yellow eyes.
females ginger, medium fur, yellow eyes.
The offspring are allowed to mate among themselves and produce the following kittens.
Males
135 black long fur yellow eyes 45 black long fur green eyes
135 ginger short fur yellow eyes 45 ginger short fur green eyes
15 black short fur yellow eyes 5 black short fur green eyes
15 ginger long fur yellow eyes 5 ginger long fur green eyes
females
135 black long fur yellow eyes 45 black long fur green eyes
135 ginger medium fur yellow eyes 45 ginger medium fur green eyes
15 black medium fur yellow eyes 5 black medium fur green eyes
15 ginger long fur yellow eyes 5 ginger long fur green eyes
A. Determine rules of expression for the genes controlling fur color, fur length, and eye color.
B. Determine mode of transmission for each gene. (autosomal, X linkage
C. Are any of these genes linked to each other? If so, determine the map distances and draw the map.
Problem #2.
In garden peas, genes controlling height, flower color, and seed size are on the third autosome.
height 10 m.u. color 20 m.u. seed size
A purebred tall, red flowered, large seeded plant is crossed the a purebred short white flowered, small seeded plant. All of the offspring are medium height, pink and small seeded.
A. What are the rules of expression for these three genes?
B. What proportion of the pollen produced by the F1 will carry the tall, red flower and large seed alleles?
C. If the F1 is allowed to produce an F2, what proportion of the F2 will be tall, red flowered and large seeded?
Second Example Test
Sample test answers start on page 14 of the syllabus.
Short answer questions. 4 points each.
1. What is a reciprocal translocation?
2. A reciprocal translocation heterozygote will have what characteristics (both phenotypic and genetic)? Just give a list.
3. A geneticist has found a pericentric inversion that has no phenotypes when the inversion is heterozygous with a normal chromosome. The inversion is lethal when homozygous? Please explain this.
4. The fifth autosome is about the same size as the X chromosome in humans. Why is monosomy for the X chromosome so much less harmful than monosomy for the fifth chromosome?
5. A researcher has found a strain of mice that always has some type of non-disjunction of the sex chromosomes in males. Autosomes act normally. Given this information, what would you expect for karyotypes for sperm in these mice? In mice, 1N = 22.
6. While looking at meiosis in lilies, a geneticist notices a loop in one member of the largest homologous pair in a prophase I cell. What are two possible explanations for this phenomenon, and how can the geneticist distinguish between these possibilities?
7. In polar bears, a single gene with two incompletely dominant alleles (PBig and Plittle), controls paw size. In the Greenland population of polar bears, the following is observed.
78 Big pawed bears
180 Medium pawed bears
522 Little pawed bears
Please give genotype and allele frequencies for this population.
8. Is the population given in question 7 randomly mating? Explain your answer.
9. What is positive assortitive mating, and what effects does it have on allele and genotype frequencies?
10. A young woman shows up in a doctor's office. She is 17 years old and has yet to enter puberty. She is short in stature, but otherwise normal. Which of the following genetic conditions could explain her condition?
A. 47, XXX B. 45, X C. 47, XXY D. 46, X E. Down syndrome
11. Which of the following would always have a Barr body?
A. Turner syndrome B. Klinefelter syndrome
C. XYY male syndrome D. Edward syndrome E. Down syndrome
12. As part of mapping studies, a geneticist has crossed a purebred fruit fly stock from Detroit with a purebred stock from Daytona. When she tries to map 4 genes on the X chromosome in these hybrids, all 4 genes map to the same position. Using other crosses, these 4 genes map to 4 separate loci spread over half of the X chromosome. Which of the following could explain these results?
A. One group of flies (either Detroit or Daytona) carries a reciprocal translocation of part of the X chromosome.
B. One group of flies (either Detroit or Daytona) carries a reciprocal translocation of part of an autosome.
C. One group of flies (either Detroit or Daytona) carries an inversion of part of the X chromosome.
D. One group of flies (either Detroit or Daytona) carries a deletion of part of the X chromosome.
E. More than one of the above could explain these data.
13. Pseudolinkage is:
A. a phenomenon associated with deletions.
B. the conversion of all alleles to recessives in deletion heterozygotes
C. the reduction in map distances associated with paracentric inversions.
D. the preferential association of unlinked genes in gametes from translocation heterozygotes.
E. a phenomenon associated with duplications that makes linked genes look unlinked.
14. Recombination in the area of the inversion in paracentric inversion homozygotes produces:
A. a dicentric chromatid. B. an acentric chromatid.
C. an inverted chromatid. D. a non-inverted chromatid.
E. more than one of the above.
15. Which of the following is a good definition of selection?
A. Selection is a measure of reproductive failure.
B. Selection is a change in genotype frequency caused by mate choice.
C. Selection is a change in allele frequency caused by differences in fitness.
D. Selection is a change in allele frequency caused by any process.
E. More than one of the above is a good definition of selection.
16. Genetic Drift and Founder effect:
A. have exactly the same effects (in all regards).
B. result in similar effects, but drift occurs in all populations. Founder effect occurs in all small populations.
C. result in similar effects, but drift occurs in all populations. Founder effect only occurs in the first generation of a new population.
D. result in similar effects. Drift only occurs in small populations. Founder effect only occurs in the first generation of a new population.
E. have very different effects.
Essay. 14 points.
Please discuss how differences in reproductive success and changes in population size might interact to cause rapid changes in allele frequencies. Limit your answer to these two factors. Before you discuss the interaction between these phenomena, please describe how each can alter allele frequencies in the absence of other factors.
Genetics Third Test Practice.
Sample test answers start on page 14 of the syllabus.
1. Which of the following never occurs in eukaryotes?
A. removal of extra piece of RNA from the end of a precursor RNA.
B. removal of extra piece of RNA from an internal area of a precursor RNA.
C. binding of a ribosomal subunit at the very end of an mRNA.
D. synthesis of two different types of protein from the same mRNA molecule.
E. RNA is processed in the nucleus.
2. Which of the following removes primers in prokaryotes?
A. RNA polymerase III B. DNA polymerase III C. RNase H
D. DNase H E. DNA polymerase I
3. Okazaki fragments are:
A. Products of leading strand synthesis in prokaryotes.
B. Products of leading strand synthesis in eukaryotes.
C. Products of lagging strand synthesis in prokaryotes.
D. Products of lagging strand synthesis in eukaryotes.
E. Two of the above.
4. Which of the following fixes only UV damage?
A. Suicide enzymes B. DNA photolyase. C. Thymidine dimerase.
D. Primer removal enzymes. E. DNA methylase.
5. Which of the following is a specialized mechanism that just removes bulky adducts from DNA?
A. Suicide enzymes B. DNA photolyase. C. adductase.
D. SOS repair. E. DNA methylase.
6. Which of the following is not part of the normal DNA synthetic machinery in prokaryotes?
A. An RNA polymerase. B. A DNA polymerase.
C. An exonuclease. D. An endonuclease. E. A ligase.
7. Prokaryotic mRNA processing usually involves:
A. addition of the 5' cap. B. endonuclease function.
C. exonuclease function. D. addition of the poly(A) tail.
E. none of the above.
8. During translation in prokaryotes, the small ribosomal subunit:
A. initially binds to the mRNA at an AUG codon.
B. associates with the mRNA after binding to a protein bound to the mRNA.
C. binds to the mRNA through basepairing between the medium rRNA and the mRNA.
D. initially binds to the mRNA through basepairing between the tRNA and the mRNAs.
E. binds to the mRNA through basepairing between the large rRNA and the mRNA.
9. A researcher has found a very sick mutant. She isolates proteins produced by this mutant. Although the proteins are the right length and have the right primary sequence, many of them are still associated with ribosomes, and all of them have a tRNA attached to the carboxy terminus. What is wrong with this mutant?
A. peptidyl transferase is non-functional.
B. release factor (RF) is non-functional.
C. aminoacyl tRNA synthetases are non-functional.
D. polyadenylation machinery is non-functional.
E. elongation factor G is non-functional.
10. Which of the following is not a semi-conservative synthetic mechanism?
A. Double stranded rolling circle DNA replication. B. Eukaryotic DNA synthesis.
C. Prokaryotic DNA synthesis. D. Adenovirus DNA synthesis.
E. Two of the above are not semi-conservative synthetic mechanisms.
#11-20. Short answer. Keep the answers SHORT. You do not need to use full sentences.
11. What do topoisomerases do? (Don't worry about the different classes.)
12. What do aminoacyl tRNA synthetases do, and why is it so important?
13. If you were to inactivate DNA polymerase a (alpha), what effect would it have and in what sort of organism?
14. What is the wobble hypothesis, and what basic chemical phenomenon is it based on?
15. List the steps in Rho-dependent termination. In what sort of organism does it occur?
16. Why don't eukaryotes have polycistronic mRNAs?
17. What are the three “rules” that govern the addition of nucleotides onto DNA and RNA?
18. List two differences between eukaryotic and prokaryotic rRNA synthesis and processing.
Transfer RNA
Messenger RNA
Ribosomal RNA
snRNA
20. Name this compound and the indicated components. Be as specific as you can.
At this point in the test there would be a molecular structure of a nucleotide. Hint: learn how to name them.
Essay #1. 10 points
Describe and draw prokaryotic DNA synthesis up through the point where the first deoxynucleotide is used. Include all enzymes and processes, but keep the verbiage to a minimum. You don't have to include the rules for synthesis (they're in question #17) or the chemical reaction.
Essay #2. 10 points.
Describe prokaryotic rRNA synthesis. Describe all of the processes and products. You don't have to include the rules for synthesis (they're in question #17) or the chemical reaction.
Last Practice test
Sample test answers start on page 14 of the syllabus.
1. Which of the following lab techniques requires the use of DNA ligase?
A. Southern blots. B. PCR. C. Genomic library construction.
D. DNA sequencing. E. Northern blots.
2. RAPDs are a modified form of:
A. Southern blots. B. PCR. C. Genomic library construction.
D. DNA sequencing. E. Northern blots.
3. The trp repressor is capable of:
A. binding to tryptophan. B. binding to the trp enhancer.
C. binding to the trp silencer. D. binding to the trp attenuator.
E. binding to the trp operator.
4. CAP:
A. is much more abundant in cells with higher cAMP.
B. binds to the operator upstream of lac Z.
C. alters DNA shape near some promoters.
D. binds to the 5' end of prokaryotic mRNA.
E. converts ATP to cAMP.
5. Why don't eukaryotes have operons?
A. Eukaryotic small ribosomal subunits can't bind to mRNA internally.
B. In eukaryotes transcription and translation take place in different areas of the cell.
C. Eukaryotes have too much junk DNA to make operons efficient.
D. Eukaryotes don't need to regulate their genes because they have so many more of them.
E. The initial statement is wrong. Eukaryotes do have operons.
6. A geneticist has changed the 5' AUG 3' start codon in the attenuator region of the trp operon to a tryptophan codon (5' UGG 3'). There are no other 5' AUG 3' codons in the attenuator region. The rest of the operon is normal. How would this alteration affect regulation of the trp operon?
A. It would function normally under all conditions.
B. Very high tryptophan concentrations would reduce transcription of the trpE-A cistrons. However transcription of the trpE-A cistrons would never halt completely.
C. Very high tryptophan concentrations would reduce transcription of the trpE-A cistrons. Transcription of the trpE-A cistrons would only halt completely in high concentrations of methionine.
D. The operon would not function at all.
E. The operon would function at 10% of maximum all the times, regardless of amino acid concentrations.
7. A geneticist has altered the lac I gene so that its transcription rate is twice as high as normal. Everything else in these cells is normal. How would this alteration affect regulation of the lac operon?
A. The regulation of the operon would be absolutely normal.
B. The operon would work, but it would take a lower concentration of lactose to induce synthesis of b-galactosidase.
C. The operon would work, but the amount of b-galactosidase produced in the presence of excess lactose would be half that of normal.
D. The operon would work, but it would take a higher concentration of lactose to induce synthesis of b-galactosidase.
E. The operon would not function at all.
8. What is the difference between gene families and multicopy genes?
A. Gene families are eukaryotic; multicopy genes are prokaryotic.
B. A gene family will have more copies of a gene than a multicopy gene will.
C. Gene families encode mRNAs; multicopy genes only encode non-messenger RNAs.
D. Multicopy genes are identical; genes in a gene family are all very similar but not identical.
E. Gene families encode mRNAs; multicopy genes don't encode anything.
9. Which of the following is least abundant in the mammalian genome?
A. unique sequence genes. B. structural DNA.
C. Repetitive junk DNA found as tandem repeats. D. Pseudogenes.
E. non-functional retroviruses.
10. What is an enhancer? What characteristics distinguish enhancers from other regulatory elements?
11. What does histone H1 do? Would it be found in prokaryotes, eukaryotes, or both?
12. Why is DNA that is packaged in the solonoid inactive?
13. What is attenuation and how does it work? Be brief.
14. What is genetic leakiness and what purpose does leakiness play in the lac operon? Be specific.
15. Why don't eukaryotes use attenuation to regulate their genes?
16. What is structural DNA? Give 3 examples. Would large quantities of it be found in prokaryotes, eukaryotes, or both?
17. List 3 eukaryotic gene regulation mechanisms that don't involve promoter control.
18. What characteristics distinguish processed pseudogenes from pseudogenes created by duplication? (How would you know if you had one or the other just by looking at its sequence?)
Essay #1. 14 points
Compare and contrast gene regulation in eukaryotes and prokaryotes. You may use specific examples to explain your points.
Essay #2. 14 points.
Compare and contrast genome complexity in eukaryotes and prokaryotes.
Key, first practice test
1. Anaphase II 2. Josephine is heterozygous for an X-linked lethal.
3. Maternal Inheritance: inheritance of genes in (egg) cytoplasmic organelles.
Maternal Effect: egg proteins and mRNAs affect embryonic development.
4. Two possible answers.
either Something in calves' environment modulates phenotype. Variable expressivity.
or X-inactivation. Coat color controlled by X-linked locus with cell autonomous expression.
5. Severity of disease in homozygote will vary. Some homozygotes will not get disease.
6. frequency of recombinant gametes.
7. B 8. A 9. C 10. D 11. A 12. B
13. D 14. A 15. C 16. C 17. D
18. (0.85 X 0.5)2 + (0.85 X 0.5)2 = .36125
Problem #1
Fur color: "ginger" allele is dominant to the "black" allele.
Fur length: "long fur" allele and "short fur" allele show incomplete dominance.
Eye color: "yellow eye" allele is dominant to "green eye" allele.
Fur color locus and Fur length locus are X-linked. Eye color locus is autosomal.
This is a test cross (for the X-linked loci). Use all recombinants
(15+15+15+15+5+5+5+5)/800 = 0.1=10%=10 m.u.
X chromosome an autosome
fur color 10 m.u. fur length eye color
Problem #2
Height locus: "tall" allele and "short" allele show incomplete dominance.
Color locus: "red" allele and "white" allele show incomplete dominance.
Seed size locus: "small" allele is dominant to "large" allele.
frequency of F1 gamete containing tall, red and large seed alleles (completely parental)
= 0.9 X 0.8 X 0.5 =0.36.
Frequency of F2 individuals showing the tall, red and large seed phenotype
= ( 0.9 X 0.8 X 0.5)2 = 0.362 = 0.1296.
Key, second practice test
1. Chromosome rearrangement where pieces of non-homologous chromosomes are switched.
2. A. Cruciform structure in synapsis. B. Pseudolinkage
C. 50% reduction in fertility. D. Phenotypes vary, minor. Sometimes break a gene when translocation occurs, but in a heterozygote this is likely to be unnoticed.
3. During creation of the inversion, at least one essential gene was inactivated (by breakage). When heterozygous this is minor. When homozygous, this is lethal.
4. Dosage compensation makes all mammals “functionally monosomic” for the X, so only having one is trivial. No such mechanism exists for autosomes.
5. 22, X 22, Y 21, 0 23, XY
6. Could be a deletion or duplication heterozygote. (Inversion would involve both chromosomes.) If this is a deletion, all prophase I cells in this plant will have the loop in the same spot. If it is a duplication, there are two possible positions for the loop. Half the prophase I cells will have it is one position, half in the other.
7. [PBigPBig] = 78/780 = .1 [PBigPLittle] = 180/780 = .23
[PLittlePLittle] = 522/780 = .67 [PBig]= [(72 X 2) +180]/(780 X 2) = .22
[PLittle]= [(522 X 2) + 180]/(780 X 2) = .78
8. expected heterozygosity = .22 X .78 X 2 = .3432 observed heterozygosity = .23
Many fewer heterozygotes in the population than expected. Population is not randomly mating.
9. Positive assortitive mating is mate choice based on similarity. Like mates with like. This increases homozygosity for those alleles that determine the phenotypes used for mate choice.
10. B 11. B 12. C 13. D 14. C 15. A 16. C
Essay. There were lots of way to approach this, but this is roughly what I was after.
1. Differences in reproductive success will move the population to the point of highest fitness. This is because alleles that confer greater reproductive success get transmitted to the next generation more frequently. This can either remove alleles from the population, or stabilize the population at a point where multiple alleles are present.
2. Population size affects the ability to gain and maintain genetic diversity. In small populations, rare alleles can be lost easily. Those same rare alleles may also rapidly become more common, just because of random choice of gametes (genetic drift) or when a new population becomes established (founder effect).
3. The interactions between population size and fitness come in many forms. You only needed one. A small population may have fewer alleles/locus, so it doesn't have the “right allele” to survive a change in environment or pathogen. A rare recessive allele may become much more common (because of drift) in a small population. This makes it easier for homozygosity to occur for this allele, which means that it can affect fitness quicker. There are other options.
Pedigree
1. The gene controlling color is Y linked. No. Ic can't have IId. Females show variation.
2. The gene controlling color is X linked. Allele causing black is dominant to the allele causing white. No. IIIa couldn't have IV b,c. I d couldn't have IIe.
3. The gene controlling color is X linked. Allele causing black is recessive to the allele causing white. No. Ib couldn't have IIa. IIc couldn't have IIIc,d.
4. The gene controlling color is autosomal. Allele causing black is dominant to the allele causing white. Works.
5. The gene controlling color is autosomal. Allele causing black is recessive to the allele causing white. Works.
Key, third practice test
1. D 2. E 3. E 4. B 5. A 6. D 7. E 8. C 9. B 10. A
11. Stop positive supercoiling and untangle DNA
12. Charge tRNAs with right amino acids. If this isn't done correctly, codons won't specify a specific amino acid during translation.
13. extends primers in eukaryotes by about 200 bases.
14. Sloppy basepairing in third position of codon. Both G-U pairing and non-canonical bases in anticodon are responsible for this.
15. a. Sequence transcribed that Rho binds to.
b. Rho binds to it.
c. Rho winds RNA around itself until it hits RNA polymerase.
d. Transcription stops.
16. Ribosomes only bind to cap binding protein in eukaryotes.
17. Strands must be antiparallel, you must have correct base pairing, you can only add nucleotides onto 3' ends.
18. Prokaryotes use spacer RNA as pre-tRNA. Eukaryotes don't. Prokaryotes have short tandem arrays of rRNA genes, eukaryotes have large tandem arrays.
19. tRNA-RNA pol III
mRNA- RNA pol II
rRNA- RNA pol I
snRNA- RNA pol III or pol III, depending on the snRNA.
Essays.
See sections of the text.
Key, last practice test
1. C 2. B 3. E 4. C 5. A 6. B 7. D 8. D 9. A
10. Enhancers are distant independant, orientation independant transcription regulatory sequence that may affect multiple genes. Promoter elements are both distance and orientation dependant and only affect one gene. Both have many sequences and trans factors in common.
11. Eukaryotic. Responsible for higher order packaging and gene regulation. Binds to DNA between core particles.
12. RNA polymerase can't get through all the protein to see the DNA
13.
14. Need permease to get the lactose into the cell to start with, and need beta gal to make allolactose to bind to repressor to derepress the operon. If no permease or beta gal, can't start operon.
15. translation occurs in cytoplasm, transcription occurs in nucleus. Attenuation requires that they happen on the same RNA molecule at the same time.
16. telomeres, matrix attachment points, centromeres, origins. All but origins are eukaryotes. Origins are both.
17. Regulated RNA processing, transport, translation, editing
18. Processed pseudogenes do not have sequences corresponding to promoters or introns, but do have a "poly(A) tail". Duplicational pseudogenes have promoter and intron like sequences, but lack poly(A) tails.