遗传学双语教学试卷 9

GENETICS FINAL EXAMINATION 8 1 Meiotic cell division produces haploid gametic cells by DNA replication and cell division s a 1 chimps have 24 pairs of homologous chromosomes Assume that a man and a male chimp make a bet about which of them can produce more genetically different sperm at least in theory Further assume that both are heterozygous at one locus on each chromosome The will win a gorilla b chimp c man d fish 6 Ten years after the Avery et al experiment which identified DNA as the genetic material was reported Watson and Crick used model building and the data of other workers to propose a double helical model for DNA The data of Erwin Chargaff showed that the purine pyrimidine ratio was in almost all cases approximately 1 0 This suggested that a phosphodiester linkages are covalent b A T the dominant allele polled eliminates horns If a PP x pp mating is made of the kids will have horns a 100 b 75 c 25 d 11 78 e 0 9 ABO blood groups in humans may be viewed as a three allele one locus genetic system If an AB individual and an A individual have four children who are A B AB and AB the parents are a both heterozygotes b AA and BB c AB and AB d Rh e AA and AB 10 In crosses between true breeding black and white guinea pigs all of the F1 are black crossing F1s produces black or white offspring in a ratio a 1 1 b 6 1 c 3 1 d 0 1 e 9 3 3 1 11 In a cross of black feathered Andalusian chickens with a strain having splashed white feathers all the F1 chicks were slate blue Black X white blue F1s When the F1s were intercrossed the chicks were 193 black 409 blue and 198 white Assuming that prior information had suggested single locus codominance for feather color we would calculate expected values a 3 1 b 1 1 c 9 1 d 1 2 1 e 8 3 12 The expected number of blue chicks is a 600 b 400 c 198 d 409 e 193 13 The expected number of white chicks is a 600 b 400 c 200 d 198 e 193 14 The expected number of black chicks is a 600 b 400 c 200 d 198 e 193 15 The chi square value for this test is a 468 b 3 84 c 9 74 d 0 002 e 9 4 Chi square SUM obs exp 2 exp 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 16 The appropriate number of degrees of freedom for this test is a 1 b 2 c 1 5 d 3 e 0 17 The percentage of times one would expect to detect observed minus expected discrepancies as large as these is a between 25 and 5 b less than 5 c more than 95 d between 90 and 50 e between 50 and 25 18 In a cross between a male Horribilis herrerasaurus who is HhUu and a homozygous female 1000 progeny are produced 245 of the progeny are HU 257 are hu 250 are Hu and 248 are hU These loci are most probably a linked b on different chromosomes 19 The method of scores compares hypotheses of linkage and independent assortment a homozygous b heterozygous c prokaryotic d linked e lod 20 The affected father I2 has a disorder Sweetwater syndrome which is thought to be inherited as a single locus dominant A first hypothesis might be that the marker locus with 2 7 12 and 14 alleles was seen in the affected father s chromosomes in and phase configurations a 2 s and 14s b 7 s and 12 s c 2 s and 14s d 2 s and 7 s 21 Under this hypothesized phase configuration there are noncrossover and crossover progeny a 5 and 0 b 5 and 1 c 4 and 2 d 4 and 1 22 The probability of observing this particular family under the hypothesis of independent assortment is a b c 5 d 5 e 0 23 If we assume that the rate of recombination theta is 0 25 the probability of observing this particular family under the hypothesis of linkage is a 0 375 4 0 125 b 4 0 125 c 1 10 5 d 0 125 e 0 24 The ratio of the probability under linkage and independent assortment is equal to approximately 2 5 This means that the hypothesis of is approximately two and one half times more likely a independent assortment b linkage 25 Mutations in the lac I gene which produce polypeptides unable to bind to the operator are termed mutants because they are always a constitutive off b repressible off c repressible on d constitutive on e lagomorph furry 26 Gene control in eukaryotes allows many more control points than in prokaryotes For example bone morphogenetic proteins are synthesized as large molecules with an amino terminal signal sequence of from 15 to 25 amino acids a poorly conserved pro domain that has between 50 to 375 amino acids and a carboxy terminal domain between 110 to 140 amino acids The active form is released by proteolytic cleavage a polynucleotide b polypeptide c precursor d disulfide e polyrazmataz 27 Eukaryotic genes are for the most part split by introns so that the bases which actually encode polypeptides exons are separated This allows of transcribed mRNA to produce different polypeptides a operonic processing b translational processing c differential splicing d continuous looping e transmembrane portage 28 In a quantitative single locus model where A 2 a 1 B 2 and b 1 the Aabb genotype will have a score of a 2 b 5 c 4 d 1 e 8 29 In a three locus model A 2 a 1 B 2 b 1 C 2 and c 1 In a cross between two triple heterozygotes progeny with scores ranging from 12 to 6 are seen The genotypes producing these extreme scores are and a AABBCC and aabbcc b AaBbCc and AabbCC c AAbbcc and aaBBCc d aabbCC and AaBbCC 30 Assume that the model in question 29 applies in the real world and that there is very little environmental variation A breeder selects only those individuals with scores of 11 or 12 for breeding Which alleles would you expect to decline in frequency a a and A b b and B c