遗传学周测题目含参考答案.docx

1、遗传学周测题目含参考答案注意：开始时间2014-10-18 的16:00 ，结束时间2014-10-19的24:00周测一1. Which of the following diseases results from faulty DNA repair? (6.0分)A. neurofibromatosis type 1 B. xeroderma pigmentosum C. retinoblastoma D. osteogenesis imperfecta E. hemophilia A 2. 在基因组的间隔序列和内含子等非编码区内，广泛存在着与小卫星DNA相似的一类小重复单位，重复序列为（

2、），称为微卫星DNA或STR，如(A)n/(T)n、(CA)n/(TG)n、(CT)n、(AG)n等。 (6.0分)A. 26kb B. 15100kb C. 15kb D. 1060kb E. 170300kb 3. If the sequence of nucleotides in a gene is T-T-A-C-G-A-G, the sequence of nucleotides in mRNA synthesized by it is (6.0分)A. A-A-U-G-C-U-C B. T-T-A-C-G-A-G C. A-A-T-G-C-T-C D. A-A-T-G-C-T-G

3、E. T-T-U-G-C-U-G 4. Gene is split, alternate with the exons and the introns. (6.0分)A. The numbers of exons and introns in a gene are N+1 and N respectively. Exons are variable and introns are conserved. B. Introns are the non-encoding sequences and the corresponding RNA sequence will be removed from

4、 mRNA. Their mutations do not influence the splicing of RNA and affect the production of mRNA C. Exons are the encoding sequences corresponding to the sequence of mRNA and their mutations will influence the formation of proteins. D. Splice site is located in the junction region of intron and exon, w

5、here intron always starts at AT and ends at CG. 5. A nucleotide substitution cannot cause which of the following? (6.0分)A. nonsense B. Frameshift C. missense D. same sense 6. Which of the following statement about polymorphysim is not correct? (6.0分)A. Single nucleotide polymorphisms (SNPs) occur ab

6、out once every 500-1000 bp in the human genome. B. Minisatellites and Microsatellites are not the repeat DNA sequence in the human genome and both can not be used as polymorphysim marker. C. RFLP is the 1st generation of markers used in humans D. Restriction RFLP Mapping can be combined with Souther

7、n Blot Analysis to detect the differences in DNA samples. 7. The hereditary characteristics are transmitted as units was discovered by (6.0分)A. Fleming B. Morgan C. Mendel D. Sutton 8. Cross-over is a familiar term to (6.0分)A. Hybridiztion B. Diffusion from one side of a membrane to the other C. Mig

8、ration across difficult geographical barriers D. Interchange of sections of chromatids E. Pollination involving separate flowers, male and female 9. The hypothetical Southern blot show below illustrates a DNA “fingerprinting” analysis to examine paternity, where maternal (M), child (C), and paternal

9、 (F) DNA samples have been restricted, blotted, and hybridized simultaneously to the probe D1S220 and D7S123. (DNA marker is the standard ladder). The distributions of restriction fragment alleles suggest (6.0分)A. Correct maternity and paternity B. False maternity (i.e. baby switched in the nursery)

10、 C. False paternity D. The child is adopted E. None of the above 10. The presence of genes in the same chromosome is an example of (6.0分)A. Segregations B. Coordination C. Free assortment D. Linkage E. Fusion 11. 外显子与内含子接头序列为_。 (6.0分)A. 5 GT-AT 3B. 5 GT-AG 3C. 5 GG-AT 3 D. 5AG-GT 3 12. 发病率最高的遗传病是（ ）

11、 (7.0分)A. 单基因病 B. 染色体病 C. 多基因病 D. 体细胞遗传病 13. 目前已知人类最大的基因，大约由（）和相应的内含子组成，cDNA全长为11 000bp，编码分子量为427 000的蛋白质，从dystrophin转录形成一条完整的mRNA分子需要16h。 (7.0分)A. 60个外显子 B. 40个外显子 C. 70个外显子 D. 80个外显子 E. 79个外显子 14. According to present thinking, which of the following statements about codons is incorrect? (7.0分)A.

12、There is evidence to support the theory that the cell reads the codons in sequence from a given point B. The codon are nonoverlapping C. Two or more codons often code for the same amino acid D. The codon consisits of four nucleotides E. Nucleotide sequence (genes) are probably separated by spacers t

13、hat are nonsense codons 15. Meiosis 1 is known as the (6.0分)A. reduction division B. diploid division C. sexual division D. equilibration division 16. Genetic diseases caused by which of the following types of mutation would be most amenable to replacement therapy with the relevant gene product(prov

14、ided you could get it where it needed to go)? (7.0分)A. haploinsufficient B. gene duplication C. dominat negative D. any of the above周测二1. Thus far, two genes have been found that can cause autosomal dominant breast cancer (one on chromosome 13 and one on chromosome 17). This is best described as an

15、example of: (4.0分)A. Allelic heterogeneity B. Linkage C. Linkage disequilibrium D. Locus heterogeneity 2. Contrasting genes at the same locus are called (4.0分)A. Difactors B. Alleles C. Allies D. Homologus E. Associates 3. 某人的父亲为Huntington舞蹈病患者，他现已30岁还未发病，如果30岁时此病的外显率为30%，则他将来发病的可能性为（）。 (4.0分)A.35%B

16、.15%C.0%D.100%E.30%4. Your patient is a 5-year-old girl who appears to have Duchenne muscular dystrophy. What is the most likely genetic explanation for this disease in a girl? (4.0分)A. She has a dominant negative mutation. B. She has skewed X inactivation. C. She has two independent DMD mutations.

17、D. She has a 46, XY karyotype with sex reversal. E. This diagnosis is impossible. 5. 父亲为A血型，母亲为B血型，生育了1个B血型的孩子，如再生育，孩子可能的血型为（）。 (4.0分)A. A、B和AB B. B和AB C. A和B D. A和AB E. A、B、AB和O 6. Many biochemical defects, which are inherited in a single-gene Mendelian fashion, have multiple effects upon the indiv

18、idual affected. This phenomenon is known as: (4.0分)A. pleiotropy B. phenocopying C. incomplete penetrance D. codominance 7. Huntington舞蹈病患者重复扩增的三联体是（）。 (4.0分)A. GCG B. CAG C. CTG D. GAA E. CGG 8. Mutt and Jeff are two brothers who were born with congenital deafness. Their parents had normal hearing.

19、 Mutt married Mabel. Mabel was also congenitally deaf and her parents had normal hearing. Mutt and Mabel had 10 children (5 boys and 5 girls) all of whom were born congenitally deaf. Jeff married Jane. Jane is not related to Mabel. Jane was congenitally deaf although her parents had normal hearing.

20、Janes sister, Myrna, was also congenitally deaf. Jeff and Jane had 10 children (5 boys and 5 girls) none of whom were congenitally deaf. The probable explanation of the above pedigree is that: (4.0分)A. there is a single genetic locus with at least 2 different abnormal recessive alleles (a and b) whi

21、ch will result in congenital deafness when homozygous as either recessive aa, ab or bb. B. there are at least 2 different genetic loci which have abnormal recessive alleles which will result in congenital deafness when either locus is homozygous for its abnormal allele. C. there are at least 2 diffe

22、rent genetic loci which have abnormal recessive alleles which will result in congenital deafness when both loci are homozygous for their abnormal alleles. D. Jeffs congenital deafness resulted from a spontaneous mutation and hence cannot be passed on. 9. 一个男孩是甲型血友病（XR）的患者，其父母和祖父母均正常，其亲属中不可能患此病的人是（）

23、(4.0分)A. 姑姑 B. 姨表兄弟 C. 外祖父或舅父 D. 同胞兄弟 10. Match the descriptions below with the appropriate term. A grandson and paternal grandfather have ectrodactyly (aoutosomal dominant disorder with absent middle fingers), but the father has normal hands (4.0分)A. Variable expressivity B. Germinal mosaicism C. N

24、onrandom X iinactivation D. Genetic heterogeneity E. Incomplete penetrance 11. Match the characteristics to the mode of inheritance. Elevated paternal age is characteristic (4.0分)A. Autosomal dominant B. Chromosomal C. X-linked recessive D. Autosomal recessive E. Polygenic 12. A young couple both af

25、fected with classical achondroplasia come to you for genetic counseling. There is no other family history of achondroplasia. You should inform them that if they have a tull term liveborn, the probabi1ity that the newborn wi11 not have achondroplas if about: (4.0分)A. 33% B. 25% C. 1% D. 50% 13. 多指症为常

26、染色体显性遗传病，如果其外显率为60%，两个杂合型患者婚后所生子女表型正常的概率为（）。 (4.0分)A.20%B.50%C. 30% D.55%E.15%14. One characteristic of a mutation is that it is (4.0分)A. Genetically transmitted to future generations B. Almost an improvement C. Always a dominant characteristic D. Prevented by using colchicine 15. Match the characte

27、ristics to the mode of inheritance. When male-to-male transmission is observed, this mode is unlikely (4.0分)A. Autosomal recessive B. Autosomal dominant C. X-linked recessive D. Chromosomal E. Polygenic 16. A 34-year-old woman presents to the emergency room with burning, stabbing, epigastric abdomin

28、al pain. She reports nausea and vomiting over the past 24 hours and states that she thinks that she may be pregnant. Urinalysis is positive for ketones, WBCs, RBCs, and bacteria. Pregnancy test is positive. An abdominal CT reveals thromboses in the splenic, superior mesenteric, and left renal and po

29、rtal veins. Further testing reveals a factor V Leiden mutation in one allele. There is no family history of thrombosis disorders in her family or her husbands family. Further analysis reveals a mutation in the mothers factor V Leiden gene; the father does not have the mutation. The patient is concer

30、ned about her pregnancy in light of these findings. What is the probability her child will be affected by factor V Leiden thrombophila? (4.0分)A. Virtually 0% B.75%C.50%D. 25% E.100%17. Match the descriptions below with the appropriate term. A 90-year-old man with autosomal dominant neurofibromatosis

31、 has a son and grandson who died in their twenties from neural tumors (4.0分)A. Genetic heterogeneity B. Nonrandom X iinactivation C. Germinal mosaicism D. Variable expressivity E. Incomplete penetrance 18. Phenylketonuria (PKU) is an autosomal recessive disease that causes severe mental retardation if it is undetected. Two normal parents are told by their state neonatal screening program that their third child has PKU. Assume the initial screening is accurate and answer the questions below. What is the risk for their next child to have PKU? (4.0分)A. 100 % B. 50% C. 67% D. 25%