Gene Correction in Human Embryonic and Induced Pluripotent Stem Cells: Promises and Challenges Ahead

The sequencing of the human genome has greatly facilitated our ability to identify the gene candidates that are critical for developmental regulation and disease progression. However, investigators seeking to use such data to gain a better understanding of the developmental biology and therapeutic potential of pluripotent stem cell types may be frustrated by the limitations of current genetic engineering techniques. Targeted genomic manipulation of mammalian cells is inefficient, and unintended effects of such manipulation on the cells’ developmental and proliferative potential remain uninvestigated. To this end, the recent demonstration of targeted gene correction in human induced pluripotent stem cells (hiPSCs) before and after reprogramming, as reported in this issue of Molecular Therapy, warrants special mention.

The successful isolation of human embryonic stem cells (hESCs) has enabled the implementation of renewable generation of a wide variety of cell types in regenerative medicine approaches.The prospect of implementing cell replacement therapies for diverse pathologies, including neurodegeneration, cardiovascular disease, and diabetes, is now conceivable owing to the ability to generate large quantities of functional differentiated cell types. However, the use of hESCs remains controversial, and cell transplantation therapy as a field still faces major obstacles such as acute donor cell death, low engraftment rates, immune rejection, and tumorigenicity,which have impeded efforts toward clinical translation.

The recent development of hiPSCs has offered exciting solutions to allay the ethical concerns and mitigate the possibility of immune rejection by providing a mechanism for the creation of “patient-specific” pluripotent stem cells from somatic cell types.It is possible to generate hiPSCs without destroying human embryos, and autologous hiPSC grafts have the potential benefit of being less likely to provoke an immune response. Nevertheless, challenges such as low derivation efficiency and exposure to viral factors during the reprogramming process remain to be solved, but new advances have recently been made.

Although the utility of gene targeting in hESCs has been confirmed by several reports, these techniques remain unimplemented on a large scale. By contrast, relatively facile reprogramming of somatic cell types from virtually any genetic background into hiPSCs has enabled the generation of myriad cell lines containing disease-conferring mutations. The use of these hiPSCs and their derivatives for disease modeling is expected to significantly enhance investigators’ ability to probe the mechanics of a multitude of pathologies. As the repertoire of genetically diverse hiPSC lines expands, the ability to site-specifically modify the genome in order to alter genes of interest will become increasingly important. To realize the full potential of hESCs and hiPSCs, efficient methods for genetic modifications are needed.

Techniques for precise targeting of defined modifications to the genome rely on activation of the cell’s homologous recombination (HR) machinery (Figure 1). In the presence of a donor DNA sequence substituting for the sister chromatid, HR can yield targeted clones with an appropriate experimental system. However, the donor DNA often enters other pathways of DNA repair, most notably undergoing nonhomologous end joining at a higher frequency than HR. Therefore, targeted HR is a relatively inefficient process, occurring at a frequency between 10−5 and 10−7 in treated mammalian cells.This process may be enhanced by some orders of magnitude with the introduction of a DNA double-strand break at the genomic locus of interest. Other strategies for enhancing HR involve augmenting the activity of nucleofilament proteins RAD51 and RAD52, either by transient overexpression or via small molecules, although some of these studies have yielded conflicting results. Further investigation into the cellular determinants of the choice between the intended (i.e., HR) versus unintended (e.g., nonhomologous end joining, single-strand annealing, microhomology-mediated end joining, and random integration) target is required to yield satisfying insight into potential methods for improving the relative frequency of HR.

Gene-correction efficiency may vary considerably, depending on the genomic context of the target locus, so comparing the results of various studies is challenging. Fortunately, the most frequently targeted gene in hESC studies is theHPRT1 locus encoding hypoxanthine-guanine phosphoribosyl transferase (HPRT). HPRT is a key enzyme in the purine salvage pathway, the deficiency of which is manifested as Lesch–Nyhan syndrome in patients. Also, because HPRT1 is located on the X chromosome, a single gene targeting event can lead to a complete loss of function and 6-thioguanine resistance in XY cells, providing a convenient culture system for screening correctly targeted cells from random integrants. Initial attempts at HR in hESCs at the HPRT1 and POU5F1 loci yielded successful gene targeting at frequencies on the order of 10–6. The percentage of selection marker–resistant clones that actually contained the intended gene correction as confirmed by Southern blotting, referred to as the targeting efficiency, varied between 27% and 40%, depending on the targeting construct used.

Several studies have now validated the use of zinc-finger nucleases (ZFNs) to induce a chromosomal double-strand break at the site of the targeted gene, which greatly enhances the frequency of the homology-directed repair. This approach requires the engineering of zinc-finger DNA-binding domains to recognize a DNA sequence of interest. The DNA-binding domains are then fused to the nuclease domain of the FokI endonuclease to enable cleavage at a specific genomic locus. However, the potential remains for off-target cleavage events at chromosomal sequences bearing a similarity to the ZFN target site. One group has demonstrated the use of ZFNs in conjunction with integrase-defective lentiviral (IDLV) vectors to achieve high gene targeting efficiencies. Increasing concentrations of IDLVs delivering ZFNs that target the CCR5 locus resulted in 0.5–5.3% of transduced cells expressing the green fluorescent protein (GFP) transgene. However, this particular study lacked Southern blot analysis, which would have confirmed whether gene targeting had generated a single-copyGFPinsertion without any random integration elsewhere. Also, the increase in viral titer concomitantly increased the frequency of mismatch mutations from 4% to 28% in the GFP+ clones.

Subsequent studies have demonstrated the use of ZFNs to successfully target the POU5F1PIGAPITX3, and AAVS1 loci in hESCs and hiPSCs. Because techniques employing ZFNs require concomitant delivery of three DNA constructs (two to encode the ZFN pair and one donor DNA sequence), reported efficiencies may be lower for these studies, which employed electroporation in place of viral vectors. Using a nonviral electroporation system, these studies demonstrated successful gene targeting for a variety of loci in hESC and hiPSC lines on the order of 10−5. Targeting efficiency varied from 8% to 100%, depending on the ZFN pair used and genomic locus targeted. Notably, these studies utilized ZFNs designed by different groups using entirely different methods of validation. However, a ZFN-based gene-correction approach requires extensive design and validation of new zinc-finger binding domains for each target locus. In addition to verifying proper targeting when using ZFNs, it is of paramount importance to limit cells’ exposure to ZFN-expressing plasmids to prevent off-target cleavage and even potential integration of ZFN-expressing DNA into the genome.

Viral vector delivery systems have also been implemented in order to overcome the low transient transfection efficiencies observed using electroporation in hESCs. Adenoviral and lentiviral vector gene delivery systems are well established in their ability to transduce a broad range of cell types at high efficiency, although safety concerns regarding random chromosomal integration have prevented their clinical use. Modified viral vectors that have been developed to allay such concerns include AAV, helper-dependent adenovirus, and IDLV. Helper-dependent adenoviruses, for example, have been modified by deletion of all viral genes from the vector genome, resulting in reduced cytotoxicity and an expanded cloning capacity to allow for insertion of larger targeting constructs. These properties have been exploited by one group to achieve successful gene correction at the HPRT1 locus in hESCs at a frequency of ~10–6, with a targeting efficiency of ~40%.

It is within this context that Khan et al., in this issue of Molecular Therapy, make a significant step forward. Remarkably, the investigators observed successful insertion of a neomycin-resistance cassette into the HPRT1locus in ~10–5 of hESCs and hiPSCs after transduction with AAV gene targeting vectors. Importantly, their work is also the first demonstration of successful gene targeting in differentiated somatic cells (fibroblasts) before their reprogramming into hiPSCs. A 4–base pair (bp) segment was successfully inserted into the HPRT1 locus of transduced fibroblasts, and one of these fibroblast cell lines was then successfully reprogrammed into hiPSCs using lentiviral vectors, albeit at a reduced derivation efficiency when compared with an unmodified fibroblast cell line. Once derived, these genetically modified hiPSCs were also subsequently corrected to delete the original 4-bp insertion. Despite these multiple genomic modifications, karyotypic abnormalities were observed only rarely (i.e., in one hiPSC line at later passages, as is commonly observed in hESC culture).

However, unintended mutations are a common result of gene correction attempts using any approach, including the one used in the current article. Improving the proportion of site-specific versus nonspecific integration events has remained a major challenge to the gene therapy field. Althoughin vitro culture allows for the careful selection and clonal expansion of the targeted cell population of interest, clinical protocols may involve transduction of hundreds of millions of cells. Currently, the approach demonstrated by Khan et al. remains limited to use in cultured cells that can subsequently be expanded to achieve the cell numbers required for downstream applications. Also, despite potential uniformity of HLA subtypes across hiPSC derivatives and their recipients, it remains to be seen whether the short exposure of transplanted cells to viral capsids is sufficient to induce an immune response in the transplanted host. It is remarkable, however, that differentiated fibroblasts, having undergone AAV transduction, retain the proliferative capacity for subsequent reprogramming into hiPSCs. Differentiated somatic cells senesce over time, and senescence is known to inhibit successful reprogramming. The low derivation efficiency of hiPSCs is already a well-recognized and prominent impediment in the reprogramming field. The further reduction in efficiency observed when reprogramming genetically modified fibroblasts raises further questions. Will additional increases in the degree and number of genomic modifications in fibroblasts lead to further reductions in reprogramming efficiency? Or is the reduction in reprogramming efficiency correlated only with the increased number of plating and passaging steps required for fibroblast culture prior to transduction with reprogramming factors?

Khan et al. have convincingly demonstrated the feasibility of AAV-mediated gene correction in hESCs and hiPSCs. Overall, their results open many new avenues for investigators to site-specifically modify candidate genes before and after the derivation of pluripotent stem cell types. After perturbing a gene of interest, the derived iPSC lines can be differentiated into the relevant cell population to measure the gene’s effect on cell function. Importantly, the ability to site-specifically modify the genome mitigates the risk of malignant transformation induced by random retroviral or lentiviral transgene insertion. Safe, reliable genomic modification would greatly assist the therapeutic cell transplantation field to overcome this significant obstacle to clinical realization. In any case, these advances have significantly streamlined the genetic manipulation of pluripotent stem cells, and have brought a small step closer the realization of the full potential of pluripotent stem cells for investigating disease and treating patients.


نوشته شده توسط در شنبه 23 مرداد1389 ساعت 8:0 بعد از ظهر | لینک ثابت |

21. Why is a high-moor bog an extreme habitat?
1 because it is oligotrophic and permanently cold
2 because it allows extreme situations concerning climate on the surface.
3 because it is oligotrophic and has a low pH value
[  ] A.  All three explanations are correct
[  ] B.  Only explanation 1 is correct
[  ] C.  Only explanations 1 and 2 are correct
[  ] D.  Only explanations 2 and 3 are correct
[  ] E.  Only explanation 3 is correct.
22. Which of the following situations is in operation when blood is being pumped into
the aorta?
[  ] A.  left ventricle contracted, bicuspid valve open, semilunar valve shut
[  ] B.  left ventricle contracted, bicuspid valve shut, semilunar valve open
[  ] C.  left ventricle contracted, bicuspid valve open, semilunar valve open
[  ] D.  left ventricle relaxed, bicuspid valve shut, semilunar valve open
[  ] E.  left ventricle relaxed, bicuspid valve open, semilunar valve shut.
23. Only one of the following features of the phylum of the Chordata also is present in
adult Tunicata (=Urochordata). Which feature?
[  ] A.  possession of a chorda
[  ] B.  possession of visceral slits ( = pharyngeal slits)
[  ] C.  possession of a tail
[  ] D.  possession of a dorsal tubular nervous system
24. Which of the following is true for transport in xylem? 
[  ] A.  The primary force involved is osmotic pressure 
[  ] B.  Xylem is the primary site for transport of sucrose 
[  ] C.  Movement through xylem depends mainly on transpiration
[  ] D.  All of the above
25. Which of the given actions are part of the light stage of photosynthesis? 
[  ] A.  1, 3, 6
[  ] B.  1, 4, 6
[  ] C.  2, 3, 6
[  ] D.  2, 4, 5
[  ] E.  1, 3, 5
26. Animal behaviour patterns, in which an individual endangers its life to benefit
other members of the group, are called altruistic. It is believed that altruistic
behaviour was favoured by kin selection. Which if the examples given below
CANNOT be explained as kin-selection-favoured?
[  ] A.  suicidal attack by a worker bee guarding its hive
[  ] B.  protection of the queen of an ant species by "soldier ants"
[  ] C.  protection of lion cubs by a lioness NOT being their mother
[  ] D.  warning cries of a bird warning other individuals about approaching danger

27. In a grazed field the net primary production is less than the gross primary
production. Which of following could account for this difference?
1. Plants use energy in translocation.
2. Plants use energy in actively transporting ions across membranes.
3. Hydrolytic enzymes in plants use energy in breaking down food reserves.
4. Plants use energy when water vapour is lost via the stomata.
5. Grazing by herbivores.
[  ] A.  1 & 2
[  ] B.  2 & 3
[  ] C.  3 & 4
[  ] D.  1, 2 & 3
[  ] E.  2, 3 & 4
28. Which of the following hormones is most elevated after a large, carbohydrate-rich
[  ] A.  glucagon
[  ] B.  glucose
[  ] C.  GnRH
[  ] D.  insulin  
29. Which of the following animal taxa only occur in the sea?
1. starfish and sea urchins (Stellaroidea, Echinoidea)
2. jellyfish (Cnidaria)
3. sponges (Porifera)
4. squids (Cephalopoda)
5. highly developed crustacea (Malacostraca ñ Crustacea, Astacus)
[  ] A.  1 & 4
[  ] B.  2 & 3
[  ] C.  only 5
[  ] D.  1, 2 & 3
[  ] E.  2, 4 & 5
30. How many different phenotypes can be expected in the F2 of the crossing: AA BB *
aa bb when:
I the genes are completely coupled and,
II the genes inherit independently:
[_] A 3 4
[_] B 3 9
[_] C   4   9  
[_] D 4 16
[_] E 9 16
 31. Which hypothesis seeks to explain why a plant auxin produces different effects on
the growth of a root and of a shoot?
[  ] A.  Gravity affects the action of the auxin.
[  ] B.  The growth rates of roots and shoots differ.
[  ] C.  The auxin travels faster downwards towards the root.
[  ] D.  The root and shoot respond differently to similar auxin concentrations.
[  ] E.  Light affects the action of the auxin.
32. Potatoes are stored for one week in pure air, then for one week in pure nitrogen
and then in pure air again. During the experiment the excretion of CO2 is
measured. The diagram shows the results.
The extra amount of CO2 produced and excreted during the third week probably came from
[  ] A.  ethanol
[  ] B.  ethanal
[  ] C.  lactic acid
[  ] D.  NADH
[  ] E.  NADPH
33. Who presented a manuscript of a theory of evolution identical to Darwinís one
year before the publication of The Origin of Species?
[  ] A.  Hutton
[  ] B.  Gould
[  ] C.  Malthus
[  ] D.  Cuvier
[  ] E.  Wallace
34. Generally, there are more species per square kilometer in which of the following
types of ecosystem?  
[  ] A.  Temperate grassland  
[  ] B.  Arctic tundra  
[  ] C.  Boreal forest  
[  ] D.  Coral reef
35. Which of the following contribute to phenotypic variation in a population?
I. Mutations
II. Crossing over
III. Constancy of the environment
IV. Independent assortment.
[  ] A.  I & II only
[  ] B.  II & III only
[  ] C.  III & IV only

[  ] D.  I, II & IV 

36. Which of the following is NOT a characteristic of all chordates?  
[  ] A.  vertebrae
[  ] B.  notochord  
[  ] C.  pharyngeal slits  
[  ] D.  postanal tail
37. In a particular breed of cattle, the allele for the polled condition (no horns) is
dominant to the horned condition. Coat colour is determined by another gene
which has two alleles so that the animals can be homozygous red and homozygous
white, whilst the heterozygous condition is called roan (red with white patches). A
cross was made between a horned red cow and a horned white bull. Assuming that
neither of these characteristics is sex-linked, which of the following statements
about the offspring from this cross are true?
I. The offspring will show equal numbers of horned roan and polled roan individuals.
II. All offspring will be horned, but coat colour will have red, white and roan individuals.
III.  All off spring will be homozygous for the horned condition and heterozygous for coat
IV. All the offspring will be horned and roan
V. None of the offspring will be polled.
[  ] A.  I & II
[  ] B.  II & III
[  ] C.  III & IV
[  ] D.  I, II & III
[  ] E.  III, IV & V
38. The vitelline sac (=yolk sac) is expected to be very small in one of the following
groups. In which one?
[  ] A.  in groups that fertilize externally
[  ] B.  in groups with embryos that are fed from maternal blood
[  ] C.  in groups that fertilize internally
[  ] D.  in groups that have an allantoic membrane
39. C4-plants can start photosynthesis with a lower concentration of CO2 in the
atmosphere than C3-plants. This is because:
[  ] A.  respiration of C4-plants is higher
[  ] B.  respiration of C4-plants is lower
[  ] C.  C4-plants do not have photorespiration
[  ] D.  C4-plants do have photorespiration
40. Which one of the following is the site of the influx of sodium ions during the
passage of an action potential along a myelinated axon?
[  ] A.  the nodes of Ranvier
[  ] B.  the whole of the axon membrane
[  ] C.  the sodium pump area of the axon membrane
[  ] D.  the chemical synapse
[  ] E.  the myelin sheath


Answer Key:

21. D
22. B
23. B
24. C
25. B
26. D
27. D
28. D
29. A

30. B

31. D
32. B
33. E
34. D
35. D
36. A
37. E
38. B
39. C
40. A

نوشته شده توسط در یکشنبه 3 آذر1387 ساعت 3:10 بعد از ظهر | لینک ثابت |

1. Protein synthesis occurs in/on the
[  ] A.  nucleus
[  ] B.  ribosome
[  ] C.  smooth endoplasmic reticulum
[  ] D.  lysosome
2. Which of the following biomes has the greatest species diversity?
[  ] A.  Temperate Rain Forest
[  ] B.  Savannah
[  ] C.  Grassland
[  ] D.  Temperate deciduous forest
[  ] E.  Coral Reef
4. In a food chain isolated from others, which of the following (measured in kJ m
 ) is
numerically the greatest?
[  ] A.  Net primary production in plants
[  ] B.  First carnivore consumption
[  ] C.  Herbivore assimilation
[  ] D.  Herbivore respiration
[  ] E.  Plant biomass increase.
5. Tissues that form long, tough strands, as in the leaf stalk of celery, are:
[  ] A.  epidermis
[  ] B.  collenchyma
[  ] C.  sclerenchyma
[  ] D.  parenchyma
6. During migrations, some birds use which of the following as reference
I.  The sun 
II.  Constellations 
III.  Earth's magnetic field 
[  ] A.  I only 
[  ] B.  I and II only 
[  ] C.  I and III only 
[  ] D.  I, II, and III
7. When the base composition of DNA from bacterium Mycobacterium tuberculosis
was determined, 18% of the bases were found to be adenine. What is the [G] + [C]
[  ] A.  32%
[  ] B.  36%
[  ] C.  64%
[  ] D.  18%

8. The pH of the lysosome is most nearly
[  ] A.  1.0
[  ] B.  5.0
[  ] C.  7.0
[  ] D.  11.0
9. Which of the following is an example of habituation? 
[  ] A.  Chickadees learning new songs when they shift to living in large winter flocks from small
family groups. 
[  ] B.  Stalking and attacking litter mates by lion cubs. 
[  ] C.  Hydra initially contract when gently touched, but soon stop responding.
[  ] D.  Yearly migration of golden plovers from Arctic breeding grounds to southeastern South
10. Energy is transferred between trophic levels in a food chain. Which of the
following stages represents the least efficient transfer of energy?
[  ] A.  Sun to primary producer
[  ] B.  Primary producer to primary consumer
[  ] C.  Primary consumer to secondary consumer
[  ] D.  Secondary consumer to tertiary consumer
[  ] E.  Tertiary consumer to decomposer.

11. The final sere during succession is termed the:
[  ] A.  niche
[  ] B.  biotic potential
[  ] C.  carrying capacity
[  ] D.  climax
[  ] E.  seral community
12. Brown algae differ from the green algae and red algae in having:
[  ] A.  chlorophyll a
[  ] B.  differentiated cells
[  ] C.  phycocyanin within their cells
[  ] D.  Fucoxanthin within their cells
13. Which two functional groups are always found in amino acids?  
[  ] A.  Amine and sulfhydryl  
[  ] B.  Sulfhydryl and carboxyl  
[  ] C.  Carboxyl and amine
[  ] D.  Alcohol and aldehyde
14. Woodlice normally live in damp habitats. This is to prevent desiccation and to
facilitate gaseous exchange. In regions of low humidity they move about quite
rapidly but in a random manner. When they find an area of high humidity they
slow down and eventually become stationary. This is an example of:
[  ] A.  a kinesis
[  ] B.  a taxis
[  ] C.  a reflex action
[  ] D.  habituation
[  ] E.  operant learning
15. Sex-linked recessive alleles are usually carried on:
[  ] A.  The homologous part of the X chromosome
[  ] B.  An extra Y chromosome
[  ] C.  The non-homologous part of the X chromosome
[  ] D.  The homologous part of the Y chromosome
[  ] E.  An extra X chromosome
16. Populations that are likely to be living at a density near the limit imposed by their
resources, are characterized as  
[  ] A.  K-selected  
[  ] B.  p-selected  
[  ] C.  Q-selected  
[  ] D.  r-selected
17. Which of the following is responsible for transcription?
[  ] A.  spliceosome
[  ] B.  RNA polymerase
[  ] C.  DNA polymerase
[  ] D.  Ribosome 
18. Dichlorophenol indophenol (DCPIP) is a blue dye that is decolorized when
reduced. After being mixed with DCPIP which one of the following would show the
greatest change in color?
[  ] A.  Isolated chloroplasts in the dark;
[  ] B.  Isolated chloroplasts in the light;
[  ] C.  Chlorophyll extract in the dark;
[  ] D.  Boiled chloroplasts in the dark;
[  ] E.  Boiled chloroplasts in the light.
19. In situation that conflict between attack and flight animals have been noted to
behave in a most peculiar fashion e.g. the behaviour of fighting cocks interrupting
their fight to peck at the ground for food. What is this kind of behavior called?
[  ] A.  Feeding
[  ] B.  Ritualisation
[  ] C.  Displacement Activity

[  ] D.  Aggressive Behaviour

Answer Key:

1. B
2. E
3. B
4. A
5. C
6. D
7. C
8. B
9. C

10. A

11. D

12. D
13. C
14. A
15. C
16. A
17. B
18. B
19. C
20. E 
نوشته شده توسط در جمعه 3 آبان1387 ساعت 2:51 بعد از ظهر | لینک ثابت |

Questions 8-10:

Much attention is currently focused on the potential long-term environmental dangers of producing transgenic organisms. In many countries in Western Europe consumer action has resulted in the specific labeling of food obtained from genetically engineered crop plants. Initially, scientists were unmoved by this groundswell of public opinion. Now the tide is beginning to turn, and potential risks are being assessed or at least discussed by the biotechnology community.


8. An introduced gene could:
A. Cause silencing of endogenous genes.
B. Not cause high levels of a toxic compound to accumulate.
C. Not cause any alteration in endogenous gene expression or regulation.
D. Cause mutations to occur with increased frequency in the host plant.


9. Selection markers are commonly used to identify transgenic organisms in the initial stages of their production.  These markers are thought to pose a potential risk if they are allowed to remain in transgenic crop plants when they are released for agricultural use.   Selection markers could:
A. Silence endogenous gene expression.
B. Stimulate cancerous growth.
C. Impart antibiotic resistance.
D. Do no foreseeable harm.


10. The next generation of genetically modified crops may include "functional foods" such as plants with increased vitamin levels.  Safety trials will be implemented for these super-vitamin producers.  Plants that produce increased amounts of a particular vitamin:
A. will produce less other essential vitamins.
B. will have reduced rates of photosynthetic carbon fixation.
C. may produce the vitamin at a toxic level since production rates within the plant cannot yet be controlled.
D. will not pose any substantial risk to consumers.


Questions 11-16:

Your doctor, a researcher in medical genetics, tells you that you have a rare genetic condition, which is controlled by one gene alone, in which a protein present in your brain cells makes you think faster than your biological brother. You want to get more information about your great intellectual advantage. The doctor has two samples of DNA, one a copy of the gene from your body and the other from your brother's. She tells you that each sample encodes one form of the brain protein in question. The DNA sequences have not been made public yet, and the doctor is sure that her discovery will make her famous, so she is unwilling to give you access to the DNA samples. A friendly technician in the doctor's group, however, carries out in vitro transcription and translation of these two DNA samples and slips you some of each of the two translation products.


11. What method could you use to get information about these translation products?
A. Nucleotide sequencing
C. N-terminal amino acid sequencing
D. RNase protection assays


12. The process by which you can use information gleaned from examination of the translation products to learn about the two DNA sequences is called:
A. Reverse genetics
B. Database mining
C. Computational biology
D. Classical genetics


13. Which molecular property of the brain protein might differ between you and your brother if the protein in question were an enzyme?
A. Lariat structure
B. Differences in sequence of active sites
C. The presence of thymine-thymine dimers
D. RNase activity


14. What difference might you find in the samples if the brain protein were a membrane receptor?
A. Proteolytic activity of more efficient protein would be higher than the other form
B. Proteolytic activity of the less efficient protein would be higher than the other form
C. Differences in binding properties for signaling molecule
D. Participation in DNA repair


15. What difference between your brain protein gene and that of your brother would explain the observed difference between the 2 proteins?
A. Altered promoter activities
B. Stability of transcripts
C. Differences in 3'-untranslated sequences
D. Differences in open reading frame


16. The first step in analysis of the two samples of brain proteins leads directly to:
A. A partial knowledge of the sequence of the open reading frames
B. A knowledge of the genomic sequences
C. A knowledge of site of binding of transcription factors
D. An understanding of the functional differences between the 2 proteins


Questions 17-18:

Your brother is about to start a family. You decide that this advantage of yours should be shared by your brother's children. You ask your doctor about the possibility of carrying out gene therapy to ensure that each of the children born to your brother and his wife have the more efficient brain protein gene. The doctor immediately refuses for a non-scientific reason.


17. She does so because:
A. The procedure might fail.
B. The procedure might damage the developing embryo.
C. The procedure is illegal germline therapy and she would lose her US license.
D. The procedure would cost a lot of money.


18. Objections have been raised against using the potential power of biotechnology to fight genetically-based diseases.  These objections are based on:
A. Ethical and/or religious considerations
B. Economic considerations
C. A knowledge of the limitations of available techniques
D. A desire to relieve human suffering


19. Two patients suffering from a genetic disorder were successfully treated by somatic gene therapy in the US.  Somatic gene therapy involves:
A. Introducing a functional gene into an affected tissue or group of cells
B. Homologous recombination
C. The gene gun
D. Introducing a functional gene into embryo stem cells


20. The gene prolactin in turkeys is associated with maternal behavior.  It has its name due to its resemblance to the mammalian prolactin gene, which is also associated with maternal behavior.  The relationship between the turkey and the mammalian genes is likely to be one of:
A. Identity over the entire genomic sequence.
B. Homology over the entire genomic sequence.
C. Homology among introns.
D. Homology among exons.


21. Regions of greatest similarity between genes with homologous functions mostly do not occur:
A. in conserved domains.
B. at the 3' untranslated end.
C. in the case of enzymes, at the sequence encoding the active site.
D. in the case of receptor proteins, at the binding site.


22. Levels of turkey prolactin are correlated with the egg laying and incubation phases of the hen's reproductive cycle.  Based on its resemblance to the mammalian gene, where it is associated with milk production, do you expect:
A. high levels of prolactin at the time of egg laying?
B. low prolactin levels at the broodiness phase when the hen sits on the developing eggs?
C. high prolactin levels at the time of broodiness?
D. no change in prolactin levels in the transition from egg laying to broodiness?


23. Efforts to control malaria caused by parasitic protozoa harbored by the genus Anopheles stephensi (mosquito) has shifted from antibiotic and insecticide development to a molecular genetic approach.  This change of approach is most likely due to:
A. The appearance of antibiotic-resistant parasites and insecticide -resistant mosquitoes.
B. The greater fame of molecular geneticists compared to pharmacologists.
C. A shortage of drugs that are used to treat malaria.
D. The reluctance of the drug companies to market expensive products.


24. Antibiotic-resistant parasites have most likely appeared because of:
A. overuse of antibiotics leading to selection for resistant strains of the parasite.
B. the appearance of stronger mosquitoes.
C. the appearance of new pathways for antibiotic degradation in the mosquito.
D. antibiotic biosynthesis in the parasite.


25. It has been found that the first line of immune response mounted by the mosquito to invasion by the parasite is the production of nitric oxide, which also functions as a defense molecule in vertebrates.   One possible strategy that is being explored for malaria control is to engineer mosquitoes with a superior immune response that will have the capacity for a successful defense against the invading parasite.  In order for this strategy to succeed, these transgenic mosquitoes must:
A. overrun the natural mosquito population in areas where malaria is a threat.
B. succumb to the threat of the natural mosquito population in areas where malaria is a threat.
C. bite malaria patients.
D. bite people at risk for malaria.


26. A common form of immunization against bacterially caused infectious disease involves the use of live vaccine.  Live vaccine is:
A. a low dose of the infectious bacteria administered as prophylactic.
B. a dose of the bacterial strain in a modified form which retains immunogenicity but is not pathogenic.
C. a low dose of the toxin that is produced by the bacterium.
D. a sample of cells from a patient who recently recovered from the disease.


27. One method of disarming bacterial pathogenicity is by transposon mutagenesis. To use this technique in the creation of a vaccine, some prior knowledge is required.
A. The sequence of the transposon must be known.
B. Its site of insertion in the bacterial genome must be known.
C. The gene responsible for bacterial pathogenicity must be known.
D. Transposon restriction sites must be known.


28. In order to have the Food and Drug Administration approve a transgenic bacterium for use in a human vaccine, the exact nature of the mutation must be known and, most importantly, the probability of its reversion to wild type.  Which of the following mutations is the FDA least likely to approve?
A. Point mutations
B. Insertions
C. Deletions
D. Chromosomal rearrangements


29. A genetic predisposition to some colon cancers has been traced to a recessive gene encoding a protein involved in DNA mismatch repair.  Nonetheless, heterozygous individuals are known to be increasingly at risk for the disease as they age.  This is likely due to:
A. somatic mutations in the damaged copy of the gene.
B. germline mutations in the undamaged copy of the gene.
C. an accumulation over time of mutations in a clone of cells with two damaged copies of the gene.
D. eating fatty, high salt, foods.


30. It is possible, at this time, to control the point of insertion of an introduced gene into
A. a bacterial genome
B. a plant genome
C. an animal genome
D. (B) and (C)


31. The human eye has evolved to respond to a broad range of light intensities from moonlight to full sun.  This versatility is made possible by:
A. Rapid adjustment of rhodopsin pool sizes.
B. Positive feedback of signals involving changes in intracellular calcium.
C. Modulation of signal amplification strengths.
D. None of the above.


32. Patients who suffer from phenylketonuria are homozygous for a mutant gene encoding a deficient form of phenylalanine hydroxylase.  Heterozygous individuals ("carriers") show a normal phenotype.   This is because:
A. The recessive gene is not expressed in carriers.
B. The normal gene is haplo-insufficient.
C. The dominant allele produces sufficient protein in carriers to sustain normal function.
D. The gene product of the abnormal allele is preferentially degraded in carriers.


33. Cells isolated from a carrot were grown in suspension and eventually developed into a mature carrot plant.  For this to occur, the original cell must have initially:
A. Undergone meiosis
B. De-differentiated
C. Activated its senescence genes.
D. Undergone photomorphogenesis.


34. Dolly the sheep was cloned by introducing a nucleus from the mammary gland of a mature ewe into an enucleated egg cell.  Dolly, therefore, has:
A. A nuclear genome from one cellular source and an organellar genome from another source.
B. Nuclear and mitochondrial genomes from the same cellular source.
C. A haploid genome.
D. More potential to transmit genetic variation to her offspring than animals produced as a result of sexual reproduction.


35. Sex-linked traits are expressed:
A. Only in males.
B. Only in females.
C. Most commonly in males and rarely in females.
D. Primarily in people of Eastern European descent.


36. Specific targeting sequences direct:
A. mRNAs from the nucleus to their place of expression in the mitochondrion or the chloroplast.
B. Completed polypeptide chains from the cytosol to nuclei, mitochondria and chloroplasts.
C. RNA splicing.
D. Newly degraded protein fragments to lysosomes.


37. During RNA splicing:
A. Exon lariats are formed.
B. Primary transcripts are formed.
C. Specific sequences are recognized at intron-exon borders.
D. Transposons are inserted into primary transcripts.


38. The first stages in the origin of life probably involved:
A. The appearance of autocatalysis.
B. The appearance of DNA.
C. The appearance of proteins.
D. The appearance of molecular oxygen.

39. Drosophila with a wild type white gene have red eyes.  Flies with a mutant white gene have white eyes.  This confusing nomenclature arose because:
A. The geneticist who made the original observation was mistaken.
B. The geneticist who made the original observation was not an English speaker.
C. The gene was initially discovered in its mutant form.
D. The red pigment in the eyes of wild type flies is not synthesized under the conditions of the original observation.
نوشته شده توسط در پنجشنبه 18 مهر1387 ساعت 4:31 بعد از ظهر | لینک ثابت |


Questions 1-3:

A gene encoding growth hormone in rats was introduced into mice. The construct used consisted in an open reading frame encoding rat growth hormone fused to a mouse promoter A. The growth hormone sequence was obtained from a rat cDNA library, the only available source of sequence information. Fertilized mouse egg cells were transformed with the rat growth hormone construct and expression of the rat growth hormone was detected. Transformation of fertilized egg cells with rat growth hormone cDNA alone did not result in expression.


1. Why was it necessary to fuse the rat growth hormone sequence to a promoter to obtain expression?
A. The rat sequence was obtained from a cDNA library and therefore did not include a promoter sequence.
B. No factors that recognized the rat promoter sequence were present in the mouse.
C. The rat promoter was contained within the rat sequence but was inactivated in the mouse.
D.  Rat growth hormone is inactive in mice.


2. When the eggs were placed under conditions conducive to activation of the mouse promoter, one of the resulting offspring grew larger than the expected size.  The appearance of the large mouse is consistent with one of the following statements:
A. That mouse grew bigger because it produced more mouse growth hormone.
B. That mouse grew bigger because it was more aggressive and ate more mouse chow.
C. That mouse grew bigger because the mouse promoter A was activated, resulting in increased production of rat growth hormone.
D. That mouse grew bigger because his nose was longer.


3. The mouse promoter used is controlled by zinc.  It controls the expression of metallothionein, a zinc-binding protein that regulates the level of that metal in mouse cells.  It is important to control intracellular zinc levels because
A. Zinc is toxic at all levels.
B. Intracellular levels of zinc must be kept within certain narrow limits to avoid toxic responses.
C. Although zinc is not required for the regulation of transcription, it is required for translational control.
D. None of the above.


Questions 4-7:

The researchers now want to get information about the corresponding gene in mice. They have identified a strain of true-breeding mice that do not grow to full size (dwarfism). Their thought is that if they could understand this sluggish growth in mice, that perhaps they could begin to understand dwarfism in humans. The research team sets out to isolate and clone the growth hormone gene in mice, starting from the rat gene. They have a mouse cDNA library available to them.


4. The first step in their new project involves:
A. Screening a mouse genomic library.
B. Probing the mouse cDNA library with cDNA encoding the rat growth hormone.
C. Probing a rat genomic library with mouse cDNA encoding mouse growth hormone.
D. Amplifying mouse cDNAs encoding growth hormone.


5. The team has now succeeded in identifying a candidate for the open reading frame encoding the mouse growth hormone.  The final step in the procedure, which resulted in a tentative identification of the sequence as a growth hormone gene involved:
A. Purifying the candidate sequence on an agarose gel.
B. Sequencing the DNA.
C. Using the sequence, once known, to scan GenBank to identify homologous cDNA sequences.
D. Using the primary sequence, once known, to infer the definitive 3-D molecular structure of the protein.


6. Armed with the mouse growth hormone cDNA, the researchers now want to get more information about regulation of the mouse growth hormone gene.   Their first step will be to:
A. Identify and clone the mouse growth hormone genomic sequence.
B. Search for transcription factors that bind to the known promoter sequence of the mouse growth hormone gene.
C. Search for transcription factors that bind to the known promoter sequence of the rat growth hormone gene.
D. Search for the promoter sequence of the rat growth hormone gene.

7. They now propose to use the dwarf mice to get more information concerning mouse growth hormone function.  A first measurement shows that growth hormone levels and body weight in dwarf mice are the same at birth as in normal size mice but that levels do not increase at the same rate in the former as they do in the latter (50 separate matings in each case). This could be due to:
A. Lack of responsiveness in the dwarf mice pups of the growth hormone promoter to activating factors produced during development.
B. Accidental environmental factors affecting the dwarf mouse mothers.
C. Accidental environmental factors affecting the dwarf mouse fathers.
D. None of the above
نوشته شده توسط در پنجشنبه 18 مهر1387 ساعت 4:29 بعد از ظهر | لینک ثابت |



1. Which of the gases in the table below is most polar?

A. carbon dioxide
B. hydrogen sulfide
C. oxygen
D. nitrogen

2. In ice, each water molecule forms hydrogen bonds with four other water molecules, as compared to liquid water in which each water molecule forms hydrogen bonds with 3.4 other water molecules. A consequence of this is that
A. ice is more dense than water.
B. water has a relatively low boiling point.
C. water has a relatively high melting point.
D. water turning into ice is a spontaneous reaction because more hydrogen bonds are involved in ice.


3. Which of the following is true about hydrogen bonds?
A. Hydrogen bonds are longer and stronger than covalent bonds.
B. The geometry of a water molecule results in the equal sharing of electrons between the hydrogen and oxygen. (see p. 48)
C. Hydrogen bonds must involve at least one water molecule.
D. Polar molecules are soluble in water because they can form hydrogen bonds with water molecules.


4. Micelles are characteristic of what kinds of molecules?
A. nonpolar molecules
B. charged molecules
C. amphipathic molecules
D. polar molecules


5. What is the concentration of H+ in a pH 0 solution of acid?
A. 0 M
B. 1 M
C. 1 X 10-7 M
D. 1 X 10-14 M


6. What is the concentration of H+ in a solution of 0.01M NaOH?
A. 0.01M
B. 2 M
C. 10-12 M
D. 10-2 M


7. Osmosis is water movement across a semipermeable membrane. Which of the following is true about water movement across cell membranes?
A. In a hypotonic solution, cells will swell.
B. In an isotonic solution, cells will shrink.
C. In a hypertonic solution, cells will stay the same.
D. Cells can neither shrink nor swell because water cannot penetrate the plasma membrane


8. Which of the following is true about acids and bases?
A. When bases ionize, they donate protons. (see p. 63)
B. Strong acids and bases are completely ionized in dilute aqueous solutions.
C. The dissociation constant of a strong acid is lower than that for a weak acid.
D. The pKa of a strong acid will be higher than that for a weak acid


9. Cells need to be buffered because
A. they need to be able to increase or decrease their cytosolic pH to adapt to various environmental conditions.
B. their proteins work best at low pH.
C. they need to maintain a specific cytosolic pH to keep biomolecules from being ionized.
D. they need to maintain a specific cytosolic pH to keep biomolecules at their optimal ionic state.


10. Which of the following is true about condensation reactions?
A. Condensation reactions are invariably exergonic.
B. Condensation reactions involve the depolymerizion of macromolecules.
C. Condensation reactions involve the loss of the elements of water.
D. A typical condensation reaction is the formation of ADP from ATP and inorganic phosphate


11. Hydrophobic interactions account for
A. why biomolecules are amphipathic.
B. why the nonpolar regions of molecules cluster together in water.
C. the tendency of lipids to disperse in water.
D. why the polar regions of molecules are associated with water


12. Based on the Henderson-Hasselbalch equation (shown below), calculate the pH when the ratio of acetic acid to acetate is 10 to 1 (the pKa of acetic acid is 4.76).

A. 1.00
B. 3.76
C. 4.76
D. 5.76


13. Which of the following pairs of groups cannot form a hydrogen bond with each other (the proposed hydrogen bond is indicated by the blue rectangle)?





14. Which of the following is true about how crystalline salts, such as NaCl, behave in water?
A. Water can form hydrogen bonds with NaCl.
B. NaCl does not spontaneously dissolve in water because the Na+ and Cl- ions are in the form of a stable crystalline lattice.
C. Crystalline salts dissolve in water because it results in an increase in entropy.
D. Crystalline salts dissolve in water because water adds to the electrostatic attractions of Na+ and Cl- ions


15. Based on the Henderson-Hasselbalch equation (shown below), calculate the pH when the ratio of acetic acid is half dissociated to acetate (the pKa of acetic acid is 4.76).

A. 1.00
B. 3.76
C. 4.76
D. 5.76


16. A buffer system consists of
A. a weak acid and its conjugate base.
B. a weak acid and a proton donor.
C. a weak acid and a proton.
D. a weak base and a proton acceptor

17. Which of the following is true about the titration curves of solutions of weak acids?
A. The pH for optimal buffering power of a weak acid is 7.00.
B. You can calculate the pKa of an acid, given the pH and the molar ratio of the acid and its conjugate base.
C. The pKa of a weak acid is the pH at which the acid is completely dissociated.
D. At a pH below the pKa of a weak acid, its conjugate base will predominate
نوشته شده توسط در پنجشنبه 18 مهر1387 ساعت 4:25 بعد از ظهر | لینک ثابت |
المپیاد زیست
آزمون المپیاد زیست مرحله اول توسط سایت المپیاد ایران و گروه المپیاد زیست در تاریخ 19 الی 25 مهر 1387 برگزار خواهد شد. جهت ثبت نام به این سایت مراجعه نمایید. سه روز پس از آزمون برای هر دانش آموز یک کارنامه صادر می گردد. لطفا جهت اطلاع رسانی به دیگران نیز اطلاع دهید.
نوشته شده توسط در پنجشنبه 18 مهر1387 ساعت 4:20 بعد از ظهر | لینک ثابت |




The IBO theoretical examination should concentrate on biological concepts applied to the majority of organisms of the same group. It should not contain specific facts, exceptions or knowledge about local organisms that require special or local experiences


The majority of questions should test students' understanding, science process skills and application of their biological knowledge. The questions testing only knowledge should be as few as possible and they should not exceed 25 % of the total points


After approval of the test questions by the International Jury the maximum obtainable points for correct answers of each particular question have to be stated in the examination papers


Questions concerning Principles of Scientific Reasoning and Principles of Biological Methods should be included in the Theoretical test.


The IBO practical examination should concentrate on the evaluation of competitors for their ability to solve given biological problems using the following skills.
In the IBO tasks the names of organisms will be the national names (no description) together with the scientific names (Latin) in brackets. Any description instead of name is prohibited. The organizers should construct the questions so that the name of the organism is not a key element for answering; otherwise they should use very well-known organisms (general representatives of a group) mentioned in the list for biosystematics

I Science Process skills

Grouping or classification
Relationship finding
Data organization and presentation: graphs, tables, charts, diagrams, photographs
Prediction / projection
Hypothesis formulation
Operational definition: scope, condition, assumption
Variable identification and control
Experimentation: experimental design, experimenting, result/data recording, result interpretation and drawing conclusions
Representing numerical results with appropriate accuracy

II Basic biological skills

Observation of biological objects using magnifying glasses


Work with a microscope (objective max. 45 x)


Work with a stereomicroscope

Drawing of preparations (from a microscope, etc)
Exact description of a biological drawing using tables of biological terms marked with a numerical code

III Biological methods

Competitors in the IBO should know the following methods and be able to use them. If any method requires extra specific information concerning procedures that depend on special technical equipment, instruction will have to be provided.
A Cytological methods
Maceration and squash technique
Smear method
Staining of cells and slide preparation
B Methods to study plant anatomy and physiology
Dissection of plant flower and deduction of flower formula
Dissection of other plant parts: roots, stems, leaves, fruits
Free - hand sectioning of stems, leaves, roots
Staining (for example lignin) and slide preparation of plant tissues
Elementary measurement of photosynthesis
Measurement of transpiration
C Methods to study animal anatomy and physiology
Dissection of invertebrates.
Dissection of parts or organs from vertebrates bred for the consumption is allowed, too
Whole - mount slide preparation of small invertebrates
Elementary measurement of respiration
D Ethological methods
    Determination and interpretation of animal behavior
E Ecological and environmental methods
Estimation of population density
Estimation of biomass
Elementary estimation of water quality
Elementary estimation of air quality
F Taxonomic methods
Use of dichotomous keys
Construction of simple dichotomous keys
Identification of the most common flowering-plant families
Identification of insect orders
Identification of phyla and classes of other organisms

IV Physical and chemical methods

Separation techniques: chromatography, filtration, centrifugation
Standard tests for monosaccharides, polysaccharides, lipids, protein (Fehling, I2 in KI(aq), biuret
Measuring quantities by drip and strip methods
Dilution methods
Pipetting, including use of micropipettes
Microscopy, including use of counting chambers
Determination of absorption of light
Gel electrophoresis

V Microbiological Methods

Preparing nutrient media
Aseptic techniques (flaming and heating glass material
Inoculation techniques

VI Statistical methods

Probability and probability distributions
Application of mean, median, percentage, variance, standard deviation, standard error, T test, chi-square test.

VII Handling equipment

Due to differences in the equipment between participating countries, these skills can only be evaluated if the competitors have been informed beforehand about the algorithm, how to use the equipment, how to proceed with a particular experiment, ...etc.

نوشته شده توسط در جمعه 1 شهریور1387 ساعت 3:16 بعد از ظهر | لینک ثابت |
سال كشور
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6 1995 Thailand
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8 1997 Turkmenistan
9 1998 Germany
10 1999 Sweden
11 2000 Turkey
12 2001 Belgium
13 2002 Latvia
14 2003 Belarus
15 2004 Australia
16 2005 China
17 2006 Argentina
18 2007 Canada
19 2008 India
نوشته شده توسط در جمعه 1 شهریور1387 ساعت 3:9 بعد از ظهر | لینک ثابت |

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نوشته شده توسط در جمعه 1 شهریور1387 ساعت 3:4 بعد از ظهر | لینک ثابت |