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Genetics Exam Study Guide

By jgonz516 Sep 10, 2014 2765 Words
Final Exam Genetics PCB3063 Form 1 Fall 2013

1. Gregor Mendel selected traits which could be easily and unambiguously sorted into two classes. Each trait such as seed shape was first bred into true breeding lines or Parental Cross
F1 Phenotype
F2 Phenotypic Ratio
F2 Ratio
Round x Wrinkled Seed
Round
5474 Round:1850 Wrinkled
2.96:1
Yellow x Green Seeds
Yellow
6022 Yellow:2001 Green
3.01:1
Red x White Flowers
Red
705 Red:224 White
3.15:1
Tall x Dwarf Plants
Tall
l787 Tall:227 Dwarf
2.84:1
varieties; these were crossed to produce the first filial or F1 generation. The F1s were, then, intercrossed to produce the data in this table. Mendel’s results were very different from the predicted results under the common theory of inheritance in the nineteenth century. For example the F1 of a cross between red flowers and white flowers should have been pink. The F1 were all red, but white flowers appeared in the F2, proving that ______________. a. the genetic material was particulate. b. the red allele is dominant. c. the white allele is recessive. d. the red and white alleles are syntenic. e. the blending theory is correct. 2. The western honey bee or European honey bee (Apis mellifera) is a species of honey bee. The genus Apis is Latin for "bee", and mellifera comes from Latin melli- "honey" and ferre "to bear"—hence the scientific name means "honey-bearing bee". The somatic cells of haploid males contain 16 chromosomes; the somatic cells of diploid females contain _______ DNA double helices. a. 16 b. 24 c. 32 d. 82 e. 164 3. In an individual who is a Hh heterozygote the daughter cells after the first meiotic division will be a. HH & hh b. Hh & hH c. hH & Hh d. H & h e. h & H 4. Mendelian segregation of alleles depends on the pairing and subsequent disjunction of homologous chromosomes at _________________.

a. anaphase I b. iotaphase I c. prephase I d. interphase e. metaphase I 5. Mendelian independent assortment means that nonhomolgous chromosomes segregate independently, because nonhomologous chromosomes align _______ at metaphase I. a. vertically b. horizontally c. randomly d. precisely e. modally 6. Brachyury is a transcription factor within the T-box complex of genes. It has been found in all bilaterian animals that have been screened, and is also present in cnidarians. In tunicates it is a transcription factor involved in notochord development. In dogs and other mammals mutant alleles produce short tails, and Tbx5 acts in limb (arm, wing, etc) development by binding to __________________.

a. miRNA b. sets of operators c. hnRNA d. ribosomes e. telomeres 7. At metaphase 1 an homologous pair of chromosome contains ___ chromatids and ___ DNA double helices. a. 2 2 b. 4 4 c. 1 1 d. 3 3 e. none of these is correct -2- 8. Heterozygous parents may produce two types of haploid gametes, and it is expected that they will be produced in equal frequency. This expectation will be very closely approximated when large samples are examined. In the real world of, say families, we may see data which is not the result of equal haploid gamete frequency. For example Ciriaco and his lovely wife Carmina have five daughters and one son, instead of three sons and three daughters as expected. This is because real world data is composed of expectations and random sampling. What is the probability of having five daughters and one son? a. (1/2)6 b. 6(1/2)5(1/2) c. (1/2)9(1/2) d. 15(1/2)4(1/2)2 e.  Professor Vanetrot has been collecting data from couples known to be carriers of the STR allele faulty. The data collected look like this

CHILDREN's PHENOTYPES FF Ff ff 222 498 280 1000 The expected number of heterozygotes is

a. 250 b. 500 c.125 d. 249 e. 15 10. Using his one locus two allele hypothesis the expected number of ff homozygotes is__, a. 18 b. 360 c. 41.25 d. 250 e. 125 11. The chi-square value for goodness of fit of this data is [ Chi-square = SUM (obs-exp)2/exp ] a. 0.0 b. 3.84 c. 0.66 d. 6.7 e. 0.198 PROBABILITY df 0.95 0.90 0.50 0.25 0.05 1 .004 .016 .455 1.32 3.84 2 .103 .211 1.39 2.77 5.99 3 .352 .584 2.37 4.11 7.81 12. Professor Vanetrots’ one locus two allele hypothesis

a. should be rejected at the 5% level of significance because it does not fit the data. b. is consistent with his data, because deviations from expectation will occur at this level more than 100% of the time.

c.is consistent with his data, because deviations from expectation will occur at this level between 50 and 90 % of the time.
d. is untestable.
13. The very simple genotype-phenotype relationships of dominance, codominance and recessiveness have been complicated by issues of penetrance and epistatsis. Thousands of simple single gene disorders have been described in various human populations. Now we are interested in identifying the gene(s) responsible for complex phenotypes such as eye color and intelligence. The two approaches to this problem are known as a. the candidate gene and genome wide association studies. b. the multiple allele and the linkage approaches. c. the Rayleigh and Strassburger approaches. d. the D2 and Duarte conjunctions. e. crossover and recombination mapping studies.

-3-
14. Methylation of the upstream cytosines blocks transcription so that the inheritance of a mutant IGF2 allele from the _____ does not produce the dwarf phenotype, because only the_____ IGF2 allele is transcribed/expressed.

a. aunt tetraallelic b. father maternal c. mother paternal d. brother abnormal e. mother abnormal 15. What fraction of the children in the Punnett Square for two blue-eyed parents will have blue eyes? a. ½ b. ¾ c. ¼ d. 1/8 e. 0 16. E. coliK cells are infected simultaneously with rII T29 and rII R836; neither of these bacteriophage is capable of growing on E. coliK when infected by itself. The plates which were simultaneously infected have 27 plaques at a 10-8 dilution within twenty-four hours. The best explanation of these observations is that ………………….. a. rII T29 and rII R836 harbor separate mutations which are very close together in the same gene.

b. both are wild-type with respect to E. coli K.
c. rII T29 and rII R836 harbor separate mutations which are in different genes.
d. rII T29 has been methylated and rII R836 has been demethylated. e. rII T29 is very closely related evolutionarily to rII Q51.

-4- 17. E. coli B is simultaneously infected with two rII mutant strains unable to grow on E. coli K, but capable of growth on E.coli B. The E. coli B culture was diluted 10-9 and 11 plaques were counted. For the E. coli K simultaneous inoculation culture 4 plaques were found at a 10-3 dilution. The recombination rate between the two mutants is …. a. estimated as 4/11. b. estimated as 2(4 x 103)/(11 x 109) c. estimated as (4 x 103)/(11 x 109)

d. estimated as 2(4 x 109)/(11 x 109) e. estimated as 2(4 x 103)/(11 x 103) 18. A cross between flies with wild-eyes and normal wings and a fly with purple eyes and vestigial wings produces F1 dihybrids which were then testcrossed to a homozygous purple, vestigial fly yielding 1339 red-eye, full wing 1195 purple-eye, vestigial wing 251 red-eye, vestigial wing and 264 purple-eye full wing progeny.

The total number of parental or nonrecombinant progeny is …… a. 1339 b. 1195 c. 2534 d. 305 e. 67
19. The map distance between the purple locus and the vestigial locus is a. 0.169 b. 0.0011 c. 0.5 d. 0.107 e. unknown 20.

3,8 4,6

3,4 8,6 8,4 3,4 8,4 3,4 3,6 The father and the first and fourth child in this pedigree suffer from Pitzer Syndrome a dominant condition leading to sustained lethargy and overindulgence. Based on a trial recombination rate of r=0.2 [(1-r)=0.8] the _______ child is probably a recombinant. a. 1st b. 3rd c. 9th d. 5th e. 7th

21. The denominator for the odds ratio is
a. (1/4)6 b. (1/4)1 c. (1/4)7 d. (1/2)7 e. unknowable 22. The numerator for the odds ratio is
a. unknowable b. (0.8)6(0.2) c. (0.4)5(0.6)2 d. (0.4)6(0.1) e. (0.4)(0.1)6 23. Pitzer Syndrome is probably_______________ to the Short Tandem Repeat marker locus. a. linked b. not linked c. unknowable 24. The odds ratio would be ______ if r were assumed equal to 0.3. a. greater b. less c. equal

25. Mutations in the lactose I gene which produce polypeptides unable to bind lactose are ____________ mutants, because they are always _______ . a. loss of function off b. repressible off c. repressible on d. constitutive on e. lagomorph furry 26. In prokaryotic control of gene expression inducible operons are on when the inducer binds to the repressor. Represssible operons are turned off when the corepressor binds to the repressor. There are additional controls. In -5- lactose metabolism where ________ keeps the operon off, and in tryptophan synthesis _______ slows down synthesis.

a. galactose leucine b. glucose attenuation c. lactose ATP d. leucine galactose e. tryptophan glucose
27. ____________ neuronal signaling and synapse proteins are _________ among humans, flatworms and flies; and this increases our confidence in the applicability of fly and worm research findings to humans. a. some correlated b. nearly all different c. most conserved d. a few conserved e. twenty-seven orthologs 28. Using cells in culture it was shown that _______ follows circadian cycles. a. the X chromosome b. mRNA c. DNA synthesis d. courtship e. feeding behavior

29. Assume that we are studying a character controlled by five loci where the alleles have additive effects and capitol alleles contribute 2 units while lower case alleles contribute 1 unit. A female AABBCCDDEE is crossed to an aabbccddee. The progeny will have a 15 phenotype. An intercross of these progeny with a 15 phenotype will produce _______ phenotypic classes.

a. 2 b. 5 c. 7 d. 11 e. 14
30. The phenotypes for the range of phenotypic classes produced by the intercross described above will range from ______ to _______.
a. 10 to 20 b. 4 to 8 c. 12 to 16 d. 5 to 10 e. 3.14 to  31. In nonhumans h2 is measured by the rate of response to artificial selection. For example, a farmer wishing to increase the size of his ralbdots only allows the largest(22) ralbdots to breed, those with a phenotypic score of 22. If the progeny all have phenotypic scores of 22, the h2 value is _____in the parental generation and _____ in the progeny generation.

a. 1.0 0.0 b. 22 22 c. 0.5 0.25 d. indeterminate e. 27 27 32. When two trihybrid folks have children they can have children with the full range of predispositional phenotypes. (1 A1A1 : 2 A1A2 : 1 A2A2) * (1 B1B1 : 2 B1B2 : 1 B2B2) * (1 C1C1 : 2 C1C2 : 1 C2C2) Assume that the A1, -6- B1 and C1 alleles contribute one phenotypic unit to the predispositional score. The two alleles contribute two units. The range of predispositional scores for this population is __ to ___.

a. 6 18 b. 2 8 c. 6 12 d. 9 27 e. 94 118 33. In 1908 Hardy and Weinberg published papers showing that the binomial or multinomial could be applied in genetics to predict the frequencies of genotypes when allele frequencies where known or at least estimated. The formal algebraic proof of this Hardy-Weinberg Equilibrium Law assumes that the population is infinite in size. This assumption is necessary, because real populations a. reproduce sexually and litter or family sizes vary.

b. will exhibit random sampling changes in allele frequency. c. are not infinite and the Hardy-Weinberg proof is only a very crude approximation when natural selection is acting. d. reproduce both sexually and asexually so an infinite size population is required to preserve genetic variation. e. vary in size over time and infinity is the mathematicians way of controlling variation. 34. In an area of the Central African Republic which has a high frequency of malaria, the following genotype frequencies were observed at birth in a large sample. TT Tt tt 0.59 0.32 0.09 What is the expected frequency of tt homozygotes in the next generation ? a. 0.6444 b. 0.4225 c. 0.4000 d. 0.09 e. 0.0625 35. The tt homozygotes has a low fitness because they may die from anemia, and the TT homozygotes have a low fitness, because they may contract malaria. Under these conditions we expect to see______________

a. a balanced equilibrium set by the severity of malaria and anemia. b. an increased frequency of malaria.
c. an increased frequency of anemia.
d. spectacular mutation rate increases.
e. no change.
36. Tachahiro and his cousin Maria del Marty have conceived a child. They are very concerned about the baby, because they have heard from friends taking Dr. Mill’s genetics class that children of close relatives have a higher incidence of birth defects and/or inherited recessive disorders. They know that their maternal grandmother was a carrier of the Wapatuknak syndrome allele, and they are both carriers. Assuming that this is the only source of genetic problems for Tachahiro and Maria del Marty what is the chance that they will have an affected child ?
a. 1 in 16 b. 1 in 4 c. 1 in 8 d. 1 in 106 e. 1 in 64 37. MM Mm mm Phenotype normal normal sterile Frequency at gen 0 0.83 0.16 0.01 0.84 0.16 0 normalized = 1.0 -7- gen 11 0.85 0.15 0.01 0.85 0.15 0

gen 100 0.98 0.02 0
gen 1000 0.998 0.002 0 gen 10,000 0.9998 0.000198 0 The sterile allele is still in the population after 10,000 generations because a. sterile homozygotes are not as bad in selective terms as lethals. b. the sterile allele is in heterozygotes at a frequency of 0.000099. c. sterile homozygotes are selected against with a fitness of zero. d. only 1 in a million will be mm in generation 1000. e. pigs can’t fly.

38. Frequencies of STR genotypes used in fifteen locus forensic DNA analysis are often in the range of 10-15 to -21. The odds that a particular fifteen locus profile came from a Cuban aboriginal vs an Irish aboriginal are on the order of ________________. a. 1 in 10 b. 1 in a thousand c. 1 in million d. 1 in 1021
39. Double stranded small interfering RNAs and proteins assemble into ribonucleoprotein particles which in the RISC is cleaved to make single stranded siRNAs which bind message leading to ___________.

a. ubiquitination b. methylation c. nucleation d. degradation e. acetylation 40. miRNA and mRNA paired imperfectly, lead to a translation ___________. a. inhibition b. degradation c. transcribed d. ubiquilation e. ased 41.Deletion of 15q11-q13, the PWS/AS region produces different symptoms depending on the parent of origin. If the mother’s copy of the chromosomal area is deleted the child will have Angelman syndrome.If the father’s copy of the chromosomal area is deleted Prader-Willi syndrome results. In addition to deletions, uniparental disomy of chromosome 15 gives rise to the same genetic disorders, indicating that genomic imprinting must occur in this region. Human SNRPN, a gene in the 15q11-15q13 deleted region is monoallelically expressed in -8- normal fetal brain and heart and in adult brain. Analysis of maternal DNA and SNRPN cDNA confirmed that the maternal allele of SNRPN is not expressed in fetal brain and heart. Thus it is reasonable to assume that PWS symptoms result at least in part from the absence of SNRPN. Of course the assumption is a hypothesis and testing is always a good idea in science. Which of the following experimental suggestions might provide evidence to support or refute the hypothesis that SNRPN absence leads to PWS? a. use a -galactosidase reporter to show that SNRPN is paternally imprinted b. use a constitutive promoter construct to turn SNRPN on in the male parent’s deletion c. use siRNA to silence SNRPN in normal male parents d. use a CCK deleted maternal SNRPN to facilitate imprinting e. use PstI to digest the male, deletion parents DNA and probe for SNRPN 42. In children with Angelman syndrome the 15q11-15q13 deletion is inherited from the mother, and people with AS are sometimes known as "angels", both because of the syndrome's name and because of their youthful, happy appearance. There are approximately seven known genes in the deleted region, and the severity and extent of the Angelman symptoms depends on the father’s genetic content in 15q11-15q13, because___. a. the presence of a Y chromosome reduces expression in a male hormone environment. b. the father’s genetic content in 15q11-15q13 is the only genetic content in these children. c. the mother’s genetic content in 15q11-15q13 is imprinted. d. the coprophagic nature of the paternal uncles leads to sterility in the nieces and nephews. e. the presence of X deletions often preceeds 15q11-15q13 deletion. 43. Schizophrenia is a general term for a group of psychotic illness(es) characterized by disturbances in thinking, emotional response and behavior. Decades of research have failed to identify the allele(s) responsible for schizophrenia although it has a very high heritability. A possible explanation of these failures is that the disease is a. primarily psychosomatic. b. a heterogeneous collection of disorders. c. the result of genotype shiftin. d. inherited as a recessive. e. a copy number variant.

44. The Disrupted in Schizophrenia 1 (DISC1) gene is disrupted by a balanced chromosomal translocation (1; 11) (q42; q14.3) in a Scottish family with a high incidence of major depression, schizophrenia, and bipolar disorder. Subsequent studies provided indications that DISC1 plays a role in brain development, because suppression of DISC1 expression reduces neural progenitor proliferation, leading to premature cell cycle exit and differentiation. These results suggest that schizophrenia is a disorder of a. neuronal development. b. oxidative phosphorylation. c. electron transport. d. leptin metabolism. e. demonic intrusion. 45. Cocaine increases the firing potential of neurons of the nucleus accumbens, a brain area involved in pleasure and reward. That is, the efficacy with which transmission occurs between the neurons is increased. Light stimulation of ____ to reduce activity of neurons in the nucleus accumbens reverses the increased activity, called potentiation, caused by cocaine. a. dopamine receptors b. opsins c. alpha-ketoglutarate d. cAMP e. the retina

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