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Name-l-?------+-~~ ----jf-r-- _ Section Studies in Genetics Introduction °lll/cG I~ which these plants are growing were produced from a cross of hybrid parents (Gg x Gg). Genetic studies show that characteristics are passed on from parent to offspring in reproduction. Questions Objectives I. How many plants are growing? Having completed the lab on genetics, the student will be able to: 2. How many plants are green? 1. Define and apply terms such as genotype, phenotype, homozygous, heterozygous, dominant, recessive, codominant, and sex linkage. _ 3. How many plants are albino? 4. Using a Punnett Square, determine the expected ratio in a cross (Gg x Gg) producing green and albino plants: G 2. Solve genetic problems involving monohydrid cross with dominant and recessive traits, codominant traits, and sex linked traits. 3. Solve dihybrid problems. Monohybrid Cross in Corn Introduction When one pair of characteristics in an individual is crossed, the individuals possessing mixed genes are called monohybrids. The presence of chlorophyll in plants is due to a dominant gene; the lack of chlorophyll (albinism) is due to a recessive gene. Materials Tray of corn seedlings showing a hybrid cross for green and albino plants. Procedure Examine a tray of corn seedlings showing both green and albino plants. The seeds from Green Albino~ 5. Was there a difference between the expected ratio and the ratio of plants growing in the tray? Why? or why not? --j) -/ CoUIt//- /1T LE~5/ :;OOc::PfIt/Vf5 6. Why would albinism in plants be considered a "lethal gene factor"? ~ ....51 CI/'!OkQJ//j/t AIu Stt~IfO ~~ --7' tf?7"r $;li7f{te; Monohybrid Genetic Problems 2. A man with normal pigmentation, whose father was an albino, marries an albino. What chance is there that their first child will be an albino? 1. In humans, normal pigmentation (A) is dominant to albinism (a). Set up a Punnett Square to determine the phenotype and genotype of the offspring for each of the following: a. AA x aa: ~(}J6 Phenotype:,_If--:.....-_ _ Genotype: 3. A man with normal pigment marries a woman with normal pigment. They have an albino child. What are the genotypes of all (three individuals? /J' /I . !l-Q.,; rr _ ~ 11 etlf b. Aa x aa: c. Aa x Aa: InA nO(,., "~~ Woman At{.. Child aa Cl~ - 11 : i Phenotypic ratio:J Genotypic ratio: 1.M: 11«..-. C(. 4. Some people are able to detect the bitter taste of phenylthiocarbamide (P.T.c.) while other people are unable to taste it. The following diagram representing a family, some of whom are "tasters" and some of whom are "non-tasters". The squares represent males; the circles represent females. The shaded squares and circles represent "tasters". Phenotypic ratio.:_~1-f/ic...:......:.:~1~a..,.,-~_ Genotypic ratio: t Aa.,: L &<.-Q..... I/-ev /1_____ nty§ tt 11't Man )C &l...--' ,.A«-: t It.t:t/ 8 5 d. aa x aa: qq Phenotype:,_t(,, __ Genotype: a. Under the symbols for each person write their phenotype: "taster" or "non" q q.., T =T;;5 I EI(, 2 ~ -;.. #0111- M$I€/Z-' c. Give the phenotype (Normal=N and disorder=n) and genotype for each individual in the following sets of parents: b. Which is the recessive trait? A/oN- rr17TI3[L c. Above the symbol for each person, write their genotype (IT, Tt, or tt) A Couple 1: Phenotype Genotype A y Aa X A C(. < Couple 2: Phenotype CA:-- 4 Genotype aa. y4~ 5. In chickens rose comb (R) is dominant over single comb (r). A rose-combed male is mated to a rose-combed female. Eighteen chicks are produced, ten of which are rose combed and eight of which are single-combed. What are the genotypes of the parents? Couple 3: Phenotype ,4/".4 Genotype /lcrr X dO\ 7. The diagram below represents a family history for a particular genetic disorder. The open squares and circles represent "nafflicted individuals, the shaded squares and circles represent afflicted individual . __----'" 19((, ,qct,. Male_R--'LL _ Female & rfll1 fC-. 6. Using the family history below, answer these questions: (Shaded symbols represent a genetic disorder) generation? Couple -I iJftf£),.AI~ r Using the pedigree, answer the questions: a. Is this disorder expressed in every j.. III L b. Can two people phenotypically normal produce a child that is afflicted with this disorder? )C tta... J ) ~ /11() a-ev c. In the mating of two afflicted individuals are there any children that are not afflicted with this disorder? 11/.1 /I a..-- . /la." a. Is this disorder carried as a dominant or & recessi ve characteristic? ? f) r' . . .tA.GV/' d. Is this affliction a dominant or recessive trait? f CtS5,'y/c b. What evidence supports your answer to question a? CDu.fLt& W,IL- ,#T Q,£V y II /lp/EV.5? 0 It i acv L!k ~ ~1t£i1/o yfll / 8. Andalusian chickens may be either black, white, or gray. The gene for black is not dominant over the gene for white, nor is the gene for white dominant over the gene for black. When a black rooster is mated to a white hen, all gray chicks are produced. Materials Metric ruler, Procedure Determine your phenotype and possible genotype for the following characteristics by reading the descriptions of the traits in the following paragraphs: a. What term is used to describe a condition when neither gene shows dominance over its homologous allele? Co -J)O!l1 ,11/111/r Example: Albinism (a) is caused by a recessive gene. If you ARE albino your phenotype is .-1L and your genotype is aa. b. Using a Punnett Square, demonstrate the inheritance of feather color in Andalusian chickens by crossing a black rooster with a white hen. If you are NOT albino your phenotype is A and your genotype could be AA or Aa. Since you do not know use A-- for your genotype. Trait iv the col r(5) of the ChiCkS:~) c. Why would it be impossible to establish! true breeding flock of gray Andalusian chickens? T3LJo BW B W /3/3 8W GbV Phenotype 1. Attached ear lobes 2. Widow's Peak 3. Tongue Roller 4. Hitchhiker's Thumb 5. Bent little finger 6. Long palmar muscle 7. Pigmented iris of eye 8. Mid-Digital hair 9. Freckles I BtAck ~ Gtl1.Rt( 1 wl/;r£ I _w--, Human Inheritance Introduction Many inherited human characteristics result from the interaction of a number of genes, but frequently the variation is due to a dominant or recessive gene trait. Where possible, determine your phenotype (expression) and your genotype (genetic make up) for each trait listed below. When you express a dominant trait, it is not possible to determine whether you carry two genes for this trait (homozygous) or whether you carry one dominant and one recessive (heterozygous). 10. Hair form 11. Long eyelashes 4 Genotype 3. What is the ratio of appearance for these four phenotypes derived above. (?PS:?Ps:?pS:?ps) (Hint: To detennine a ratio divide all 4 numbers by the lowest number.) Dihybrid Cross of Corn Introduction Two of the many genetic characteristics observed in corn kernels are color (purple or yellow) and texture (smooth or wrinkled). Purple (P) is dominant to yellow (p) and shows a 3: 1 ratio in a monohybrid cross. Smooth (S) is dominant to wrinkled (s) and shows a 3: 1 ratio in a monohybrid cross. The quality of endosperm (food in the seed) determines the texture of the seed. Wrinkled kernels have a sweet endosperm; smooth kernels have a starchy endosperm. _ _PS:_ _PS:_--IpS:--p s 4. Using a Punnett Square, set up a dihybrid problem to detennine the expected ratio from a cross between ppSs x ppSs following these steps: a. For the F2 generation, parents with the genotype ppSs are crossed. Each plant would produce the following gametes: Materials One cob of com produced from parents with a ppSs genotype. Parent Parent (PPSs):~'S) Procedure 1. Before working with the cobs of com, consider this problem: A plant with genes homozygous for the purple and smooth seeds (PPSS) is crossed with a plant with yellow and wrinkled seeds (ppss). Each plant would produce the following gametes: Parent (frs,S):~,.;""r_'S l"~ .. ~, (~s): " ?fs.j &~/ TQS'J>S r ff- 1t) fS)f.;w.S- b. The gametes from one parent are placed on one side of the Purmett Square; gametes from the other parent are placed on the opposite side of the square. Complete the cross to give the F2 generation. _ .. Parent (ppss):,_~(f-5+-=_ _~_ _ The Fl generation of seeds produced from this cross would have thl~ following: genotype: "GPp S$ T_S phenotype:_..... _ 2. For the F2 generation, parents with the genotype ppSs are crossed, producing four different phenotypes. Using the traits listed in the table below, classify and tally the kinds of kernels in fOUT rows of the com cob. Phenotypes Purple-smooth CPS) Purple-wrinkled CPs) Yellow-smooth (pS) Yellow-wrinkled 5. Give the expected phenotypic ratio from the Punnett Square: Number Tallied 6. Compare the expected phenotypic ratio with the actual ratio produced in answer #3. (ps) 6 V))@ d. BBRr x BbRr: (;,ffl''1'; Genetic Problems: Dihybrid and Sex-linked (g:J 11 br abR~ 8r - b~ B73 Kr ~Br" ,, i /")' e. BbRr x BbRr: ..E--,--:!1.r--_ _ .... BU!!~LJtr~~ Phenotype Genotype· __ I")t b. BBRR x bbRR: Ut'A"" Bit. Bil RPt - r;, /ltt!! 'I s is,\() I I 9: 3;3: i Phenotype Genotype f3,-y---;;;~=-=- Phenotypic Ratio ~'IJR. 3: 13r 3: b R. ;. !b r Genotypic Ratio ,.~ L ~ ----=/:-:-~-Z=-=-:--=Z=-",:-.. '""",.=-----/- Z: ,,'--.-. _ -----,13£ RR c. ~r x ~: r;,/Ifl~i£' J}.r ~'r ~} ~ OJ c::::;y I3r bR Bbftr r I 2. A black, rough male guinea pig is mated to a white, rough female. Out of several litters totaling 27 individuals, 9 are black, rough; 7 are black, smooth; 8 are white, rough; 3 are white, smooth. What are the genotypes of the parents? (Hint: List what is given for the 0 parents and offspring first.) br Phenotypic Ratio t BJr: 1. Pr : t b~; Genotypic Ratio_----.-- br -- - a. BBRR x bbrr: '{3/ T~ f} o..,e...L.l-...-_~8.-".._ _~~...-----=--." 1. In guinea pigs, black coat color (B) is dominant to white (b), and a rough coat (R) is dominant to smooth (r). Using a Punnett Square, determine the expected genotypic and phenotypic results of the following crosses. .~ ~k -,---,--_ ~ 86Kr ~:J. brr: i b'bR~ ~ t 6~,..r vJ 8 g ~ ~b - Male: ab RrFemale: hl;.R. Y' / J3- R_ "1 75 _ r-r e b6R- 3 hb rr 5. A roan bull which is heterozygous for the polled condition is mated to several cows of identical genotype to his. Using a Punnett Square, determine how many roan, polled animals should be produced out of 16. 3. In rabbits, black fur (B) is dominant to brown (b), and long hair (L) is dominant to short hair (I). A male is mated to several brown, short-haired females. These matings result in the following offspring: 11 brown, long-haired; 16 black, long-haired; 12 brown, short-haired; 15 black, short-haired rabbits. Express the genotype and phenotype of the male. (Hint: List what is given about the ~1 female and the offspring first.) g :: ;SLf.JCk 1- -; Lv N6 ____ ~ bb II I') B- " /1, Male: b I Phenotype:_O_h I ':. S Iff) 66 " 73ffC/C ~l'l t, H~ - J..R 11 HfllAj IfhX'f Hhll~J HII'i-W HltlJ~ 11 h Jl. 'tI IIh w'W HhK/L (111 ,-11/ f//l1-/t hAR.W flhRVJ /fhw,J tAil "tJ Ah wi\! HI(ilfL b = g~uJfV J1 !bL :1(, Tk_L Hlt I /VI issi";~ All,fl e > J Genotype:& LI b_ Number of Roan, polled calves_ _ 4. In cattle, the polled condition (H) is dominant to the horned condition (h). A cross between an individual with red coat (R) and white coat (W) results in roan (RW). A polled, red bull is mated to three cows. With cow A, which is horned and white, a polled roan calf is produced. With cow B, which is horned and roan, a horned, red calf is produced. With cow C, which is polled and red, a horned calf is produced. Give the genotypes of all individuals. Bull JI.h- "RJZ COW Allh 11/ Calf A JIb ".j Cow B AARWI Calf B h" RB Cow C HA "jl J!- Calf C 6. In human, the determination of sex depends upon whether the male sperm carries an "X" chromosome (resulting in a daughter) or a "y" chromosome (resulting in a son). In other words, body cells of females carry two X chromosomes, and those of the males carry one X and one Y. During meiotic division, the egg of the female must of necessity carry one X, whereas segregation of the X and Y in spermatogenesis results in half of the sperm that are Y bearing, and half that are X bearing. One human abnormality, called red-green color blindness, is the result of a recessive gene carried on the X chromosome. It has no allele on the Y. Consequently, the genotypes CC, Cc and cc are possible in females, but a male must be either CY or cY. "h n~ 9 Answer the following questions: What proportion of their daughters may be a. Is it possible for a female to be color-blind? expected to y.l.5 XeX C If one of their sons whose vision is normal marries a woman of the genotype CC, can they have any color-blind children? Dem~trate with a ~ett Square. b. Can two persons with normal vision produce a color-blind daughter? a color-blind son? Demonstrate using a Punnett Square. xC--;. tTIovY't1 {,y be;:'~dIlG4'M; >< e.~ f Jot. (J l, it!iJ X X L c 'I 'I' DaUghter:----'&b~~_ _ Son: I 7. In Drosophila frui flies, sex determination is the same as in the human. Normal flies have bright red eyes, a certain recessive sex linked gene is responsible for white eyes. From a pure-breeding strain of red-eyed flies, a female is selected and bred to a male from a pure-breeding strain of white-eyed flies. Set up a Punnett Square to answer the following .... c~V' questions. 'fR.4- c. Can two color-blind parents produce a child with normal vision? Demonstrate using a Punnett Square. X X -:. X Y -;. F£/lIIU /hllile X\ 'ACJ) -=- R Xr-=- /,JlfiTt.--... r r d. Suppose that a woman with normal vision, whose father was color-blind, marries a color blind man. Using a Punnett Square, determine the fOll0c..ng: f( ~r / It 111 1 y R Rr ~:L I f1V a. Will all their offspring have eyes of the same color? c y Xy 1€5 b. What will be the color of the male flies? What proportion of their sons may be expected to be color-blind? hoj{5 Jv5 c. What will be the color of the female flies? 0 10 /'j,J..,£", ~ -l-/~\I 1/ Gametes produced by man: 8. The males and females produced in the above problem mate at random to produce the F2 generation. Exactly 100 of their offspring are collected (50 males and 50 females). Set up a Punnett Square for this problem and answer the following questions: ~y Rn 'Rv- rYJ II/I) Gametes produced by woman: a HI htA./ Punnett Square: KY Y RyJ R , 1 . a.h 11// a./t O/If /f trtt /1A It' a. How many of the 50 females should have 0 white eyes? % b. How many of the "50 males should have white eyes? J () c. How many of the 50 females should have red eyes? d. How many of the 50 males should have red eyes? ~t>~ a. Assuming that four sons are born to this union, what different combinations of traits might they exhibit in regard to aniridia and hemophilia? 1.. lAMII Ii/1/i{1. i0,'4 f1/£11d? !lit/I} i 1 ..... 9. In humans, there is a type of blindness known as aniridia, which is due to a dominant gene. Another gene, hemophilia is a disease in which the blood does not clot properly, and results from a sex-linked recessive gene. A non-hemophilic man who is blind from aniridia, but whose mother was not blind, marries a non-hemophilic woman who is not blind, but whose father was afflicted with hemophilia. Set up a Punnett Square (dihybrid) after determining these answers: Genotype of man:__ Genotype of woman:_ (A) !/AJitZ,'Di/f is 1'9/t/ AI(705(}/hIf~ ~11/1 I1I11'JJ is .D(jP1"~I9,.vr;- A?Eisol1/' S II 8J.../~O W /711 /12/t,''' tJr ,II~ AI1, II f:.fVI0P!t,·UI/ is Se;c -tlitl/[~p X: -1 w,Tt-I lIU11d9~/~/'tt0j w,71-1 /'Jl/lifl,p,ir 1/ Itf is ~CU1t?L b. Assuming that four daughters are born to this union, what different combinations of traits might they exhibit in regard to aniridia and hemophilia? c. What percentages or ratios of the children would exhibit the following combinations of traits: Aniridia Only_---,~,L~_yLL----- Hemophilia only_ I Both aniridia and hemophilia,_YfL-Neither aniridia or hemophilia _ ~ 12