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Macalester CHEM 111 - LS107 Spring 2020 Practice final exam

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1. Which'of'the'following'statements'is'an'example'of'independent'assortment?''a. An'individual'with'red'hair'is'no'more'or'less'likely'to'be'taller'than'anyone'else.'b. More'men'than'women'are'color-blind.'c. Some'traits'are'more'likely'to'be'inherited'than'others.'d. A'diploid'individual'passes'on'only'one'copy'of'a'gene'to'her'offspring.'e. Possessing'a'Mendelian'trait'is'independent'of'genetic'background.''2. Achondroplasia'is'a'fully'penetrant,'autosomal'dominant'trait.'An'unaffected'couple'has'a'first'child'born'with'achondroplasia.'What'is'the'mo st'lik ely 'ex p la na tio n ? ''a. autosomal'recessive'inheritance'b. A'new'mutation'c. variable'expression'd. incomplete'p en e tra n ce 'e. X-linked'rec e ss ive 'in h erit an c e ''3. For'a'population'of'synthetic'RNAs'made'with'a'5:3:2'ratio'of'C:G:A,'what'fraction'of'the'codon s'w ill'enco d e'arginine?''a. .03'b. .18'c. .21'd. .28'e. Unable'to'tell''4. Trisomy'21'can'result'from'nondisjunction'at'meiosis'I'or'II'in'either'the'mother'or'the'father.'For'the'chromosomes'of'a'trisomic'child,'which'of'the'following'is'TRUE?''a. Nondisjunction'at'meiosis'I'will'result'in'chromosomes'21'with'three'different'haplotypes'b. Nondisjunction'at'meiosis'I'will'result'in'chromosomes'21'with'two'different'haplotypes'c. Nondisjunction'at'meiosis'II'will'result'in'chromosomes'21'with'three'different'haplotypes'd. Nondisjunctio n 'at 'm e io sis 'II'co u ld 're s ult 'in 'ch ro mosom e s '21 'th a t'a re 'h o m o z yg o u s'a c ro ss 'all't h ree 'chromosomes'e. Three'different'haplotypes'could'result'from'nondisjunction'at'either'meiosis'I'or'II.''5. For'a'recessive'disorder'with'a'carrier'frequency'of'1/20,'what'is'the'ch a n ce 'th a t't h e'fir st'c h ild 'o f'an 'unaffected'couple'with'no'family'history'of'the'disease'will'be'affected?''a. 1/4'b. 1/40'c. 1/400'd. 1/800'e. 1/1600''6. For'a'recessive'disorder'with'an'incidence'of'1/2500,'what'is'the'carrier'frequency?''a. 1/10'b. 1/25'c. 1/50'd. 1/100'e. 1/1250''7. To'make'a'knockout'mouse,'you'design'a'CRISPR'guide'RNA'and'inject'it'into'the'male'pronucleus'of'a'fertilized'egg.'The 'exp erim e nt'is'su cce ssfu l'an d'on e'o f'the'a lleles'o f'the're sultin g'fou n de r'm ou se 'has 'a'sm a ll,'frameshifting 'de letion .'All'of'th e'following'are'true'EXCEPT:''a. The'founder'mouse'is'heterozygous'for'the'mutation.'b. The'founder'mouse'is'homozygous'for'the'mutation.'c. Breeding'the'mouse'with'an'unaffected'mouse'will'provide'F1'animals,'half'of'which'will'have'the'mutation.'d. The'knockout'will'be'observed'in'the'F2'generation'after'breeding'two'heterozygous'F1'animals.'e. One'quarter'of'the'F2'animals'will'be'homozygous'for'the'mutation.''8. Inherite d 'fo rms'of'ca n ce r'r es u ltin g 'fr o m'either'p r o to-oncogene'or'tumor'suppressor'mutations'exhibit'all'of'the'following'EXCEPT:'''a. Inherite d 'ca n c er s'd u e 'to 'oncoge n e 'mutatio n s'c a n 'b e'domin a n tly 'in h e rite d .'b. Inherited'can ce rs 'd u e 'to 'tu mor'su p p re ss o r'mutatio n s 'ca n 'b e 'd o minant ly'in h e rit ed .'c. Tumors'resulting'from'oncogene'mutations'have'a'second'mutation'on'the'other'allele.'d. Tumors'resulting'from'tumor'suppressor'mutations'have'a'second'mutation'on'the'other'allele.'e. Tumors'resulting'from'oncogene'mutations'typically'have'activated'the'mutated'gene'while'mutations'affecting'tumor'suppressors'typically'have'inactivated'the'mutated'gene.'''10.'A'bacterial'reference'sequence'encoding'a'small'peptide'is'5’-AUG'GUG'UAC'GAC'AAG'AGA'UAA-3’.'A.'If'nucleotide'9'is'changed'from'C'to'G,'what'are'the'consequences'for'the'protein?'B.'A'suppressor'screen'identifies 'b ot h 'in tra ge n ic 'an d 'extragenic'suppressors.'Among'the'suppressors,'what'amino'acids'could'be'inserted 'in 'pla c e'o f't h e's to p 'co d o n ? 'C .'W h e n 'th e 'sc re e n'was'ana lyz e d,'t h ere 'were'2'of't h e'p o te n tia l'substituting'amino'acids'that'were'never'found ,'no'matter'ho w'm any'sup presso rs'were'iden tifie d .'In 'n o 'more'than'two'sentences,'how'can'this'result'be'explained?'Please'label'your'answers'A,'B'and'C.''''9. The'pedigree'is'of'a'family'with'a'Mendelian'disease.''A.'' What'is'the'inheritance'patte rn'(autosomal'dominant,'autosomal'recessive,'X-linke d'd o minant'o r'X-linked 'recessive)?''''''''!B.'' For'each'of'the'following'individuals,'indicate'the'most%likely'genotype'(AA,'Aa'or'aa)?'I-2:'' 'II-3:'' 'II-6:'' 'II-7:' 'III-10:' !'A" B" C" D"Which"of"the"above"shows"the"chromosomes"in"a"cell"that"has"just"completed"mitosis?"11. Two pairs of chromosomes are shown in the germline cell below. The location of the A and B genes are shown on the chromosomes. A and a are two different alleles of the A gene, and B and b represent two different alleles of the B gene. If this cell divides normally to produce sperm, what are the possible sperm genotypes? a) A, a, B, b b) Aa, Aa, Bb, Bb c) AB, ab, Ab, aB d) Aa, Bb, AB, ab, Ab, aB 12. All of the four cells shown above are found in the same individual. Which of the above shows the chromosomes of a cell that has just completed mitosis? A B C D none of above 13. Recombination frequencies near 50% suggest which of the following? a) two genes are on different chromosomes, or are far apart on the same chromosome b) two genes are on different chromosomes c) two genes lie very close together on the same chromosome d) two genes are on the same chromosome but lie very far apart -A a- -b B-14. Genes A, B, and C are linked (in this order) on a chromosome. The distance between A and B and between B and C is 30 mu and 16 mu, respectively. Assume no interference. What proportion of the progeny from a selfing of aBC // Abc will be abc // abc? a) 2.30 x 10-3 b) 1.59 x 10-2 c) 5.76 x 10-4 d) 6.35 x 10-2 e) 2.52 x 10-1 15. Drosophila females heterozygous for three recessive mutations a , b , and c were crossed to males homozygous for all three mutations. The cross yielded the following results: Which gene is in the middle? a) a b) b c) c d) Not enough information to decide 1. (6 points) A woman with type AB blood has a baby with type B blood. Two different men, Kobe and LeBron, claim to be the father. Kobe has type A blood, while LeBron has type B blood. Which statement is correct? (a) Only Kobe can be the father. (b) Only LeBron can be the father. (c) Neither man can be the father. (d) Either man can be the father. 2. (6 points) Drosophila females

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