BIO 442 MENU
syllabus 
1 - genome
2 - mutate
3 -cell cycle
4 - karyotype
5 - chromoabn
6 -sex-determ
7 -prenatal
8 - mendelian
9 - complex
10 - non-trad
11 - clinical
12 - newborn
13 - teratog
14 - linkage
15 - DNA prof
16 - quanti
17 - links
18 - quizzes
(full title of lecture appears in status bar on the top or at the bottom of your window)
Biology 442 - Human Genetics
Quizzes Spring 2005
Quiz 1
1. (5 pts) Describe in some detail, the structure of a single eukaryotic chromosome and relate it to the function. Begin with DNA; include histones, and the DNA sequences necessary to the function of all eukaryotic chromosomes.
The eukaryotic chromosome consists of a single double-stranded DNA molecule wrapped around histone proteins to form the nucleosomes. The structural necessities of a chromosome are telomeres at each end of the chromosome, centromeres where the spindle will attach, and origins of replication on each arm where the DNA synthesis enzyme complex bind to replicate the DNA.
2. (4 pts) Answer > greater than; < less than; or = equal to
The number of DNA molecules per chromosome is __=__one.
The number of base pairs of DNA in a nucleosome is__<___1 kb.
The number of chromosomes in a chimp is _>___the number of chromosomes in a human.
The amount of non-coding DNA in the human genome is __>__ the amount of protein coding.
3. (1 pt) What are Alu sequences? What is the origin of the name “Alu”? In what species are they found?
Alu sequences are a SINE (short interspersed repeats) of about 300 bp with about a million copies which comprise about 10% of our genome. They are unique to humans (and great apes) and can be used to distinguish human cell lines from non-human cell lines if a lab mix-up is suspected. The are named for the restriction enzyme which cleaves the sequences, AluI.
Quiz 2
1. (6 pts) What is the signal for the beginning and ending of each of these processes?
a. replication: Beginning--origin of replication sequences; ending is when the replication "bubbles" run into one another.
b. transcription: Beginning--promotor sequences; ending at some ill-defined stop sequence past the 3" end where the polymerase falls off.
c. translation: Begriming is at the AUG start codon (which may later be cleaved off); ending is one of the 3 termination codons.
2. (3 pts) Explain the result or possible results of each of the following single base mutations?
a. deletion: will cause a "frameshift" in the reading frame which usually results in a truncated protein because of encountering a premature stop codon.
b. addition: same as for deletion.
c. substitution: when one base is substituted for another you may still have the same amnio acid coded (silent), a similar amino acid coded for (quiet), a very different amino acid coded for (missense), an amnio acid codon changed to a termination codon or vice versa.
3. (1 pt) What is meant by a null mutation? A mutation which causes the gene product not to be formed or which causes the loss of the protein function.
Quiz 3
1. (2 pts) Explain what is meant by allelic heterogeneity. A gene can have a mutation anywhere within it. The presence of these different alleles in the population is referred to as allelic heterogeneity.
2. (2 pts) The student I told you about who had later onset Tay-Sachs Disease was a compound heterozygote. Explain the use of this term. He had two different mutant alleles for the gene, hexoseaminidase A, responsible for Tay-Sachs disease. His mother and father were heterozygous for these two different alleles. One was a null allele, the other gene coded for an hex A which had some residual activity.
3. (2 pt) Which of the mutations we have discussed give rise to the normal variation we see among humans? The missense mutations which cause a minor change in the gene product without affecting its function.
4. (4 pts) Cystic Fibrosis is an autosomal recessive disorder. The disorder is caused by mutations in the cystic fibrosis conductance regulator gene (CFTR), located on chromosome 7.
The following are some of the numerous mutations found in the gene. Explain what the abbreviations represent.
• CFTR, PHE508DELThere is a deletion of the codon for the amnio acid phenylalanine at position 508 in the protein
• CFTR, GLN493TER The codon at 493 which coded for glutamine had a base substitution which changed to a termination codon.
• CFTR, ASP110HIS The codon for the amino acid at position 110 in the protein had a base substitution which changed the codon for aspartic acid to histidine,
• CFTR, IVS10, G-A, -1 Base substation of adenine for guanine at the acceptor site of intron 10
• CFTR, 2-BP INS, 2566AT A 2 base pair insertion of adenine and thymidine at nucleotide position 2566
• CFTR, 1-BP DEL, 3659C A 1 bp deletion of cytosine at nucleotide position 3659
• CFTR, 5-BP DUP, NT3320 A 5 bp duplication of the nucleotides at position 3320 of the nucleotide chain
• CFTR, IVS20, G-A, +1 A base substitution of adenine for guanine at the donor site of intron 20.
How do you suppose this one differs from the others in its effect?
• CFTR POLYMORPHISM
• CFTR, MET470VAL Polymorphisms refer to harmless base changes which account for normal variation in a population (see question #3 above). In this case the codon for methionine at position 470 (not at the beginning) was changed by a substitution mutation to the codon for valine. In this case it did not cause a problem.
Quiz 4
Circle all answers that are correct.
1. (2 pts) Which is true of oogenesis but not spermatogenesis.
a. it begins in fetal life
b. the result is one functional gamete for each meiotic event
c. there is a greater chance of non-disjunction
d. the origin of most new mutations (sporadics)
2. (1 pts) Answer > greater than; < less than <; or = equal to
The amount of recombination that occurs in oogenesis is__>___the amount of recombination that occurs in spermatogenesis.
The number of non disjunctional events that occurs in Meiosis I of oogenesis is__>__the number of non disjunctional events that occur in Meiosis II of oogenesis.
3. (5 pts) Matching. Answer a and/or b. More than one may be correct
A. mitosis B. meiosis
____A____cell division that occurs during formation of the fetus
____B____cell division in which homologous chromosomes pair up
___A and B_____cell division in which non disjunction can occur
___A__cell division in which mosaicism can arise
____A and B____DNA and histone synthesis occur before prophase
____A and B____chromatids are formed prior to cell division
4. (2 pts) Short Answer.
When is Meiosis I completed in the human egg?
At ovulation
When is Meiosis II completed in the human egg?
At fertilization
Quiz 5
1. (4pts) In the process of preparing lymphocytes for karyotyping, what is the function of each of the following:
a. PHA (phytohemaglutinin or pokeweed mitogen:They are both stimulators of mitosis. They are used primarily for stimulating lymphocytes to divide. This will increase the percent of cells in mitosis since the number is usually low otherwise.
b. methotrexate: This is a DNA synthesis inhibitor. It is used to synchronize the cells by stopping the cells in G1. When the methotrexate is removed the cells move into the S period and through G2 and into mitosis. Colchicine is added at the time the cells enter prometaphase (this time has been determined previously), Chromosomes in prometaphase allow higher resolution banding and the ability to see smaller deletions.
c. colcemide (colchicine): This drug binds to microtubules and thereby prevents mitotic spindle assembly. This prevents cells from exiting the metaphase condition and collects mitotic cells for analysis.
d. hypotonic saline: This causes the cells to swell (not burst) so that when they are dropped onto the slide the chromosomes will be more spread out....and not overlap as much.
2. (3 pts) Which of the chromosomes are acrocentric? The D and G groups chromosomes (13,14,15 and 21,22)
What is similar about their p arms? They contain only rDNA...the DNA that codes for ribosomal RNA.
What intra nuclear structure do they form? The nucleolus (the site of ribosome synthesis)
3. (1 pt) One of the common causes of early miscarriages is chromosome abnormalities. Within those triploidy is common. What is the origin of triploidy? Dispermy
4. (2 pts) FISH is an acronym. What do the letters stand for? Give one example of how FISH can be used. Fluorescence In Situ Hybridization. It can be used to detect aneuploidies by probes made for the centromeres for specific chromosomes (21, 18, 13, X and Y) It can be used to detect microdeletion syndromes.
Quiz 6
1. (4 pts) Pericentric inversions
a. involve two breaks in a chromosome
b. involve a break on each side of the centromere
c. carriers have more abnormal offspring than paracentric inversion carriers
d. if de novo always cause problems in the inversion carrier
2. (4 pts) A person with 45 chromosomes and a robertsonian translocation between his/her two number 21 chromosomes
a. has a balanced chromosome constitution
b. will have all Down syndrome children
c. will have some normal children
d. will produce some gametes with 22 chromosomes and no #21
3. (2 pts) For what reason would a geneticist order chromosome analysis for the parents of a child with an unbalanced translocation?
To provide recurrence risks for the parents, it must be determined whether the child inherited the unbalanced translocation from a parent with a balanced translocation or whether it arose de novo. There is less of a risk if the condition is de novo.
Quiz 7
1. Regarding the process of X inactivation
a. Activations is random , inactivating either the maternal X or the paternal X in the cells of the 20-30 cell embryo.
b. Both X’s are inactivated in oogenesis.
c. The same X is inactivated in the progeny of the cell once inactivation occurs.
d. Not all of the genes on the inactive X are inactivated.
e. Does not occur in the Turner female.
f. The single X in the male is inactivated in spermatogenesis.
g. The product of the XIST gene is a methylase enzyme.
2. What does “PAR” mean?
Pseudo autosomal region
Where is it?
At the Xpter, Xqter, Ypter and Yqter
What is its function?
The region(s) of pairing of the X and Y chromosomes in spermatogenesis (male meiosis I prophase)
Quiz 8
1. (2 pts) Since half of Klinefelter males get their extra chromosomes from dad explain at which meiotic division the non-disjunctional would have had to occur. (Work it out by starting with the chromosomes dad contributed.)
The father would have contributed an X and a Y. This would be a meiosis I failure of the X and Y "homologs" to separate.
2. (2 pts) Why are the genes on the Y chromosome considered a (perfect) haplotype? In what circumstances would this be useful in paternity testing? In what circumstances would it not be useful?
Since there is no recombination of the Y with another chromosome the Y linked genes will stay together. This provides a way to trace the relatedness of males in a family since each male received his Y from his father whose male sibs received the same Y from their father and so on. I can be used in paternity testing only for male children. It might be a problem if there is a possibility that another male in the putative father's family was also a possible father.
3. (2 pts) What are the functions of the “primary” genes for “maleness” on the Y chromosome (those in the male specific region, MSR, i.e., do not include the PAR)? On which arms are they?
The primary "maleness" genes on the Y chromosome are the SRY on the Yp and the fertility genes on the Yq.
4. (2 pts) Why are individuals with more than 2 sex chromosomes taller than those with just 2? Why are Turner females who are 45,X short?
The SHOX gene is in the Xp and Yp PARs. This gene influences height so the more X's and Y's you have the taller you are! The Turner female who is 45,X has only one SHOX and thus is shorter than females with two X's
5. (2 pts) Many Turner females have been found to be mosaic. Why is an iXp not found as a second cell line although iXq has?
An X chromosome cannot be inactivated unless it has the XIST gene which is on the Xq. Therefore if the normal X is inactivated the cell would be monosomic for the active genes on the Xp and nullosomic for the genes on the Xq and if the normal X is not inactivated, the cell world be making too much gene product for the genes on the Xp. With this lack of dosage compensation and gene imbalance, the cells with an iXp would not survive.
Quiz 9
1. (7 pts) Choose the ONE best answer for each example.
Genetic terms
a. Multiple alleles
b. Variable expressivity/reduced penetrance
c. Holandric trait
d. Co dominance
e. Sex influenced trait
f. Genetic lethal
g. X-linked trait
Examples of the above.
__d__Inheritance of STRs used in DNA fingerprinting
__ f__Progeria
___b__Polydactyly
___e__Pattern baldness
___a__FGFR mutation disorders
___c__Presence of testes
___g__Presence of androgen receptor
2. (3 pts) Two baby girls were born the same day and got mixed up in the nursery (THL…. always have your babies at home!) Knowing just the blood types (phenotypes) of the babies and the parents can you match up the babies with the correct parents?
EXPLAIN YOUR ANSWER FULLY. Give all genotypes as fully as possible.
Parents A and B are Type O and Type AB
Parents C and D are Type A and Type B
Baby E is Type O
Baby F is Type A
Baby E must be the child of parents C and D since parents A and B can only have type A or B babies. If baby E belongs to C and D then baby F belongs to parents A and B.
QUIZ 10
1.(5 pts)Osteogenesis Imperfecta is
a. caused by a mutation in a FGFR gene
b. always a dominant trait
c. is divided into 4 types which are due to 4 different mutations
d. is caused by a mutation in a procollagen gene
e. due to mutations in 4 different collagen genes
2.(3 pts) Which of the following autosomal dominant disorders is more severe in the homozygous state than in the heterozygous state?
a. Achondroplasia
b. Polydactyly
c. Familial hypercholesterolemia
3. What is meant by the loss of heterozygosity? Define it and then give a disorder as an example and explain how it works.
The loss of heterozygosity refers to the loss of the normal allele in a heterozygote who has inherited a mutant tumor suppressor gene. The loss of the normal allele results in a tumor. Examples of these disorders are: Neurofibromatosis 1, retinoblastoma, breast cancers. The loss occurs due to somatic mutations, deletion of the normal allele, isodisomy of the chromosome with the mutant allele or recombination during mitosis (we usually only think of this occurring during prophase I of meiosis) whereby one of the 2 resulting cells takes 2 normal alleles and the other cell receives the two mutant alleles.
QUIZ 11
1. (2 pts) Why are enzyme disorders (usually) recessive?
Since enzymes are very efficient in catalyzing their reactions the 50% of enzyme made by one allele is sufficient for normalcy. It takes the knocking out of both alleles to have an effect on the phenotype.
2, (3 pts) What are the three genetic disorders in Newborn Screening in California?
Phenylketonuria (PKU), Galactosemia, and Sickle cell disease
3. (1 pt) What is their inheritance pattern?
All are recessive
4. (2 pts) If congenital adrenal hyperplasia occurs in 1/5000 in California, what is the carrier frequency? Show your work…you can get partial credit.
1/5000 is the prevalence of the disease and is the frequency of homozygotes for CAH. This is q2 and the gene frequency for the mutant allele is q. Since the square root of 1/5000 is close to 1/70 (1/70 squared is 1/4900) we can use this as the gene frequency. The Hardy-Weinberg equation says that 2pq is the heterozygote or carrier frequency. The gene frequency of the normal allele is p and it is near 1 (1 - 1/70 is still near 1) Therefore, the carrier frequency is 2 x 1 x 1/70 = 1/35.
5. (2 pts) In what genes/proteins are the mutations in lysosomal storage diseases?
The lysosomal storage diseases are caused by mutations in the genes that code for the various lysosomal hydrolases.
Give one example of a lysosomal storage disease and its effect.
There are many but we discussed the mucopolysaccharidoses and Tay Sachs disease. The first set results in accumulation of glycosaminoglycans in the joints and bones since the enzymes involved normally break these down and when they are not there, the substrates accumulate. In TSD a brain lipid, a ganglioside, accumulates in the lysosomes which cannot be degraded and the cells of the brain become engorged with the substrate and eventually die.
QUIZ 12
1. (3 pts) Why is alpha thalassemia lethal in fetal life while beta thalassemia is not? Alpha thalassemia deletes alpha globin chains that are needed for fetal hemoglobin as well as adult hemoglobin. Beta thalassemia does not cause a problem until birth when adult hemoglobin begins to be made using beta chains.
2. (4 pts) Answer the following with regard to mt DNA mutations
a. what gene products are affected?
The mtDNA codes for mitochondrial ribosomal RNA, transfer RNA and several of the proteins of the ox-phos system.
b. why do they show a maternal inheritance pattern? Only the mother gives the embryo mitochondria in the egg. Any of the father's mitochondria that enter the egg are eliminated.
c. what is meant by heteroplasmy and homoplasmy?Heteroplasmy refers to the situation whereby an individual has both mutant and non mutant mitochondria in their cells. Homoplasmy refers to the situation whereby all the mitochondria are either mutant or normal but there is not a mixture.
3. (3 pts)
Using the nomenclature zeta, alpha, epsilon, gamma, delta, beta for the various alpha and beta chain genes, show the composition of the hemoglobin (Hb) for the following: (For example, the second adult hemoglobin which is about 3% or our normal adult Hb is represented as alpha 2 beta 2)
An early embryonic Hb: zeta 2 epsilon 2
Fetal Hb: alpha 2 gamma 2
Adult Hb: alpha 2 beta 2 and also a small amount of alpha 2 delta 2
QUIZ X
Prader Willi syndrome can be caused by
a. a microdeletion
b. a mutation of the SNRPN gene
c. a loss of the maternal allele on chromosome 15
d. unimaternal disomy of chromosome 15
Microdeletion syndromes
a. include Williams syndrome
b. DiGeorge syndrome
d. can usually be detected by FISH probes
e. usually involve several genes
Trinucleotide repeats
a. are usually inherited in an AD pattern of inheritance
b. can be in 5’ untranslated regions
c. can be in 3’ untranslated regions
d. can be in introns
e. can be in exons
f. the disorders they cause often show anticipation
g. usually show a parent of origin effect
