4. Lab 2 Answers - LAB 2 WORKSHEET NAME STUDENT # Group...

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Unformatted text preview: LAB 2 WORKSHEET NAME STUDENT # Group Members , -—_,. 1. (2 marks) Proteins are responsible for the diversity of function within an individual, within a population and within the biosphere. This is because proteins can be made from 20 different amino acids. Using what you know about probability, calculate how many unique proteins could there be that are 100 amino acids in length? Eur/{A mum Mal w w (pro-Wm tau/id km W m dbiw DOari--L.;ngpac.€al$ (are) Ll: w piolcm HCLS '39 “hashibé’m (Lack one. comp-t be Maj aw 0i“ JO'9MSI'lgl-r a .a'; l ; his O ’00 3-H 9t VET-Y). ‘01 .* € 2. (2.5 marks) Draw an acrocentric chromosome, labelling the ca rome'e, the p arm, and the q arm. Ob arm 1!; f"! WI. l5 Shark,er CEl’li’W‘G‘i’tlei'fi, . “We. 1 arms 3. (4 marks) In a particular species, the amount of DNA is measured in cells found at the various stages of meiosis. If there is 7.4 picograms (pg) in a cell at metaphase ll, indicate how much DNA is present at the following stages. (31 4o 4 29 Prophase | 1408 p g (32 Mg pg Following telophase II and cytokinesis i } Pg 4. (5 marks total) A cell has two pairs of acrocentric chromsosomes, 1a, 1b, 2a, 2b. (The a & b of each chromosome are the homologues.) Allele M is located on the long arm of chromosome 1a, and allele m is located at the same position on chromosome 1b. Allele P is located on the short arm of chromosome 1a, and allele p is located at the same position on chromosome 1b. Allele R is located on the long arm of chromosome 2a, and allele ris located at the same position on chromosome 2b. a. (3 marks) Draw these chromosomes, labelling genes M, m, P, p, R and r, as they might appear in metaphase l of meiosis. b. (2 marks) What process(es) generates different combinations of the alleles of genes M and P? What processes generate variation among the alleles of genes M and R? GIQMQS M «tr P are Gin fl’té’ Saméi CHVOMOSOME’ “hit/LS, re comm/i441???) (vied Clogging} om.» g€w€wéks allelic Cflmioitmgi‘ifl’l’ifi {3i f’hftzfv’ €93 . Hflwe-v’tri beam, M + l2 axe. local—ital. On d-l‘iltcrw aiii’vmoao-mfis “MW at {affirm/cl when c WWW.) in whom of W Cid/WM are generated be) Mdepenciewi" . ,. ¥ :4. 5. (2 marks) For many pedigree questions you are often told an allele is rare in the population. When analyzing ago fl “6 pedigrees what implications does this have for a dominant trait? What implications does it have for a recessive trait? Why a dominant trait “3 if rare in populayhfo n flee/la most Mimic»; ' .2’ indiviqu it. We. halo-Cm E1403 o-—u..§. new ha: a! luc It "I". ' t‘flamtgfg‘ I JI C it . Riv r r (639% “it aii ' m) cm twice-Cb; ’iwtsi‘ , imis indium-i 1*)" W “W “' J3 t . . . Eta-r: my ,mg J. "recesji m. , allfllfi q 6. (3 marks total) A horse has 64 chromosomes and a donkey has 62 chromosomes. A cross between a female horse and a male donkey produces a mule, which is usually sterile. A) (1 mark) How many chromosomes would a mule have in its somatic cells? B) (2 marks) Explain, in a few sentences, why most mules would be sterile. #0 TM Wm)? hag C05 ohmmwsomes in its swim/hr sails... Ci“?r “M91053 33- imfi m artist/r“ t ‘53 {at} W» flat? 0C CAM E” ‘i/Mr'g’ but? big 01 Ohm mama Moot has "FLO 'ytmflrwy’tw : ,b‘ilfbtet-xQ-j MSW’VJL‘ "iii/kw.) CM www Wm}; 3" “Li M In- Pm/ ("yum/he h (Vi/1V (is ""i’i-ka5tn by Q, 9.; 6-312.566 “PM \1 High MC, Yl-xr (thatg‘figtfi-a=ia.€s {Err “grow. 0’1 (1&ch SW“? WW.) mm— fg’fi’dm b P - t . ‘ --',x ii'rtJ-rm'ec'fi‘ . hummer 5 EH @2qu aw i irruuwwy W mpmper for w. 333+ ...r ewssrb \@ 10mm 7. ( marks) In tomatoes, red fruit colour (R) is dominant to yellow fruit colour (r). A tomato plant true breeding for g _ J. 'TQ .. ‘wh red fruits is crossed with a true-breeding yellow-fruited plant. The F1 are intercrossed to produce the F2. o-F (unfitng a. (6 marks) Diagram the above crosses and give the genotypes and phenotypes (& their proportions, where "’ 7‘93. applicable) of the parents, the F1, and the F2. I ‘P my ' a O R genemh‘on (14”) RP X r: l~ 307‘ b. (3 marks) Diagram a backcross between the F1 and the yellow parent and provide the genotypes and phenotypes and their proportions of the offspring. Rr V r‘ i” red) WWW \1/ if! Rr {Err Viredr 'lgrYelLOt/U 8. (5 marks) in cats, blood type A results from an allele (IA) that is dominant to an allele (:8) that produces blood type B. There is no 0 blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of each litter next to their blood type. Are there any matings where the offspring don’t provide enough information to determine the precise genotype of either parent? If so, describe why in the notes column Blood type of male arent . mt \L "7 F12. fiWWflh‘Qn 1/4 bRT Eli} h" Pro Fl 1 ! LEW 00' (Mfl damn/me Precise Wehee 0F either Pawns . I ’ ,1 i ' J , .2 *3 .. «har- l'b‘ Or that). we parent one permi- is h6n1021.-r(Em-rfi «:11: V E"? 45w Maw- is hemmgws Ear I"a L3 l Alkaptonuria results from an allele (3) that produces a phenotype recessive to that of a normal phenotype produced by the A allele. Sally’s father has alkaptonuria, and her mother has normal metabolism. Sally, who is normal, has a child with a man with alkaptonuria. a) Using pedigree symbols (Ch. 2), and noting genotypes diagram this cross, including the genotypes of Sally’s parents, Sally, and her husband. % 9. (6 marks total) In humans, alkaptonuria is a metabolic disorder in which affected persons produce black urine. For Sdlho‘s calm} we. the {MW «aghast; Clair! fan out») warnbule am ‘a‘ aileiffpmlmbiw 0tr U “‘19: the. Pmkfiabils‘i‘lg first-r" is“; claim Edam-9% iii. M‘anwria is 0.3 (“W13 Cha-hmfitihj {1m -: .3 flit/mi. W “ a.” ancie- . 10.(7 marks) ln jimsonweed, purple flowers (P) are domlnantt white flowers (p), and spiny pods (N) are dominant to smooth pods (n). The progeny from one mating consists of 610 purple-flowered, spiny-pod plants; 656 purple- flowered, smooth-pod plants; 220 white-flowered, spiny-pod plants; and 194 white-flowered, smooth-pod plants. andfiufld Whatwere the genotypes oft‘he parents in this cross? Tofu! Planing:- H730 low-{.r'rmf; P: pwqfig N: Siflt‘lfi W5 , . i NW K” .12. '3» WW“ h ‘-' Smooth pods Tia/H0 are oqspnns CM ‘04? used *0 Find parent-S! genotvipfi. 5 Ha” ha COMM” Gad/1 eriiv indwemo‘t’xmtfio} PM?” Rowers um ’09: swab/Mao 3‘ i raho ‘3 33']. Purple‘ whit? l W5th 9i.ew£..-r .3. 33 Q 4 mile 4 M 5:56;) .510, Wes. grates w «m prim; 7' pg 7 .3 r ifgi’rlififf'; r and: m was f - “was g a (N ) ’3“. “film! “kwggwfl Cm PhO‘OhiPic ratio) when hemaflfjolf n is named +0 bx homo'ivifitgwg reCESSive (an) '. ‘ Palm“: genohfipes PP rt X i"! V”? . -—-_-vwv-w—u——_...—— . ...___-_. .—_....... - . . . . _ ————-H m __- -.....-_--_. . _ . 2 WW?‘WW bra/huh MWOd deem t tel-nge W iv I'm/MW bW'C/hm “"95. [3 :H" \S'. or we,.mgct} a Wand O; momma “if ..Pmblewts. invotvrm _ mw'lhéf ~ $646“ i do 65103 e “t Wan/u neat i‘nstcmotem then m ‘h‘ a? gen“)? 7 11. (13.5 marks) onsider 5 gene pairs. each represented by a differ ’t letter. In each 53% the mrcase a Eele 5 5‘17? codes for a phenotype dominant to that of the lowercase allele. These 5 genotypes assort independently of each 7"“ “Wit-Mo other. Calculate the probability of obtaining the following: The following 2 genotypes are crossed: AaBchddEe x AabchDdEe. What will the proportion of the following genotypes be among the progeny of this cross? Solve parts a, b, and c using a branch diagram. In addition, solve b using a Punnett square as well. Show all your work. a. (3.5 marks) AaBchDdEe paw/ink A“ 8}) C¢ old Ee )1 Act l9!) Cc Dd tie- ‘Wé’ l“:- - d/fiwmwgt: Pm, 50m Amxl‘ta a V; 01w f2)” Memo 0 an em ewes rV—L 98w C ~ chance 0? Cc from (c. ICC». 3 92.. Ojfihe D " Change mt bot Mm dd Y DA 1': V1 aft/Chant? ogbfifiégrmg '6. fiemofijyc f E “* chem Lt at: E7? EPKIBWi ELF Egg! .1." U2 ’ AC! Eric? Cic‘ ('3- ‘ If? fl 1 ,5 gm '5 I/a.’ A; /_1 If; (2. r /;7_ "' Vast b. (6.5 marks) Aabchddee i‘m not direct.”er out t-“trte Dunne/h Sq) new "the: point was h; show You how Mfiw-C‘dtj iw, Dunnetl'sqfvta/e is m-tmh‘w, h) We brand/x method :— '/ W“? A ' Mama 01c A“ Mm Fm“ pm I 2“ Cinema: ob bgfgpw‘ufl "E. fiemohjpe ‘jwc B—chamz 0;? tab van 35.5: xbb = ’2 Aabbadfliw . (33mg Y) *- chamu (‘5’P alo’t 4mm 0M! Dal :: V2 .-,- 7L4— c. (3.5 marks) An abode phenotype (i.e., the recessive phenotype for all of the genes involved) This W{sh'0n, in an incl/t'fed" WM, pit/ornate? qom with We amol'gpe ofthe: Offspring: 1:“ “"f wearer“ “t HAL to? of "fine "if Gates than??? awhile? ' is dflwdnam'i +0 ‘b'} .536 Mg} pkewOMW ‘73 J“-"‘~€ . M3 ageimtww a? as: lab M (M as. Para/1H 9““ E"O Cc dbl Ea x Act blo Cc Dal Ea 530w Pr v' chm/tot recesst ?h€fio"l/)Pf «pi/om A—fik x A01 2: [/47 W B- mound: ‘YICMH (me'WO'I‘VIrfl my“ 35 xbb : yz (Temp C -" L? M 514.4%? tripe...“ aha Ftp .pw M (H c X C C : V4 W1) —~ GNU/nu? “cud/(H fimm tidy/bat: t/Dr W E " chat/hut mew/14W. pwwt’qpfi 'fi/ow. Ex Ma c M; V.“ cm“ E Pkwfiqpe abdd-t. 4 Is Vi'Vz'VAr'Vza : V2.59 / 0h ditfi’xem‘r clamme (wk'i‘i 4317.) 12.(12 marks total) In watermelons, bitter fruit (8) is dominant ove weet fruit (b), and yellow spots (E) are dominant over no spots (9). The genes for these two characteristics as ort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. 3. (3 marks) Diagram the crosses described above, indicating genotypes of all individuals, with the exception of the F2 individuals. P ‘ 1515 BE 10 '0 c e‘ Wfl’fim X 5M9} MIL", . \Ieiwwspofi \L VW 5‘70": a 3,9 E, x 13L, Ee UNWOW‘EQA) kiwi/Nit, \QNWw Spot; ‘2 b. (6 marks) Using a Punnett square & then a branch diagram, determine the proportion of F2 offspring that show the dominant phenotype for both traits. 8b 98 x 9239 E e r want atomimm phemo‘rum)? Ger em), lmii. Bib X $‘0 (monohvsievfak W953 " 3hr dominant phenompe EC X 5?. (monvhijbn’cl CNS) ” 3/4 40Wth WWWVWF @491 boil/x 6W5 fr 0. (3 marks) A F1 plant is mated with a bitter, non-spotted individual. Determine, using a branch diagram, the probability of bitter, non-spotted offspring in the next generation. wofigvnhgb 9mm ‘ WC, this inlViCJLl/tai [40$ 5‘ domnah’r 3’96 “’> WV. n WWW” Puma ()6 PW Wt +2291” “4. Wyam. ‘ r ‘ S I i Eb x 8b X ,. Kai/“E We 1M1? “Cog/(Er ‘ “09% ,1, Win03 E73366 0K 6’1 (:8 bimkg M” - 3/4 ‘¥°Ef$prm5 1/ 5 "ail w'H km 01* W“ W ' 4' /8 C [as i B) (as a i aw.) )3, (WM {ti/1L6 nonswmfie X 55 Ge xse g 4/ I” : A?” “Our-)Pfikd #- 4 + fl Vaxl .: 3/4x'/.;: /8 fl /9 8 all MW 9901"}— (L 13.(9 marks) In cucumbers, dull fruit (D) is dominant to glossy fruit (d), orange fruit (R) is domiylzit to cream fruit (r). and bitter cotyledons (B) are dominant to nonbitter cotyledons (b). Genes located on differ nt pairs of chromosomes encode the three characters. A plant homozygous for dull, orange fruit and bitter cotyledons is crossed with a plant that has glossy, cream fruit and nonbitter cotyledons. The F1 are intercrossed to produce the F2. What proportion of the cucumbers in the F2 will produce only dull, orange fruit with bitter cotyledons if mated to a glossy, cream-fruited plant with nonbitter cotyledons. Diagram the crosses, including unknowns and solve using a method other than a Punnett square. D :d/ull 5R: Orange 8" hwy WWW“? "HAM" Parent" muglr homomgoug, dam-Emmi" in molar For Au.- oltspnnglo mire? dom’nmrt need +0 ole/Warring Phcnvhjlm pmbaloi'lih/J 0? Fl genclrabon pdeu'ng 6m F10CFsgyrinfl F'— 0 ...
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4. Lab 2 Answers - LAB 2 WORKSHEET NAME STUDENT # Group...

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