Learning Genetics Can Be Fun
1. For Labrador retrievers, black fur is dominant to yellow. yellow. a) Explain how two black dogs can have different genotypes. b) ould a black black dog have the same same genotype as a yellow dog! c) "ow could two black dogs have a yellow pup! #. $ pea plant with round seeds is crossed with a pea plant that has wrinkled seeds. For the cross, indicate each of the following% a) the genotype of each of the parents if the round seed plant is hetero&ygous. b) the gametes gametes produced by the parents parents c) the genotypes and phenotypes of the F1 generation d) the F# generation if two round plants from the F 1 generation were crossed '. $ pea plant with a tall phenotype is pollinated by a short plant, and the seeds of the first generation hybrid produce '#( tall plants and '#1 short plants. ive the genotypes of all the plants. *. +n a certain species of plant, one purebred variety has hairy leaves and another pure variety has smooth leaves. $ cross of the # varieties produces offspring that all have smooth leaves. redict the ratio of phenotypes in the F # generation. -. +n summer suash, white fruit color /yep, it0s a fruit) is dominant and yellow is recessive. $ suash plant that is homo&ygous for white is crossed with a homo&ygous yellow one. redict the appearance of a) the F1 generation b) the F# c) the offspring of a cross between an F 1 individual and a homo&ygous white individual. . For 2almatian dogs, the spotted condition is dominant to non3spotted. a) 4sing a unnett suare, show the cross between two hetero&ygous parents. b) $ spotted female 2almatian dog is mated to an unknown male. From the appearance appearance of the pups, pups, the owner concludes concludes that the unknown male was a 2almatian. 5he owner notes that the female had six pups, three spotted and three non3spotted. 6hat are the genotype and phenotype of the unknown male! (. +n horses, the trotter characteristic is dominant to the pacer characteristic. $ male trotter mates with three different females, and each female produces a foal. 5he first female, a pacer, gives birth to a foal that is a pacer. 5he 5he second female, also a pacer, gives birth to a foal that is a trotter. 5he third female, a trotter, gives birth to a foal that is a pacer. 2etermine the genotypes of the male, all three females, and the three foals sired. 7. +magine for hair color that 8 gives brown hair and b gives blonde hair. 4se a unnett suare to determine the following in a cross of two hetero&ygous parents. a) 6hat are the chances of the offspring being homo&ygous brown haired! b) 6hat are are the chances of the the offspring offspring having blonde hair! hair! c) 6hat are the chances of the offspring being hetero&ygous brown haired! d) 6hat is the genotypic ratio! e) 6hat is the phenotypic ratio! f) +s there a hetero&ygous blonde haired offspring! 6hy! 6hy! g) +f curly hair is dominant to straight hair, what letters will we use to show these genes! h) $ hetero&ygous hetero&ygous curly haired male marries a straight haired female. 6hat are their genotypes! i) 6hat would be the gametes for the male parent! 9) 6hat would would be the gametes gametes for the female female parent! k) 6hat are the chances of the offspring being homo&ygous curly haired! l) 6hat are the chances of the offspring having straight hair! m) 6hat are the chances of the offspring being hetero&ygous curly haired! n) 6hat is the genotypic ratio! o) 6hat is the phenotypic ratio! p) +s there a hetero&ygous straight straight haired offspring! offspring! 6hy! 6hy!
:. 5halassemia is a serious human genetic disorder that causes severe anemia. 5he homo&ygous condition /5 m5m) leads to severe anemia. eople with thalassemia die before sexual maturity. 5he hetero&ygous condition /5 m5n) causes a less serious form of anemia. 5he genotype 5n5n causes no symptoms of the disease. +ndicate the possible genotypes and phenotypes of the offspring if a male with the genotype 5 m5n marries a female of the same genotype. 1;. +n pea plants, tall is dominant to short and yellow is dominant to green.
5all, t > short 8 > brown hair, b > blonde hair ross a homo&ygous tall, hetero&ygous brown haired male with a hetero&ygous tall, blonde haired female. 4se a unnett suare to show parents and gametes. a) 6hat is the phenotypic ratio! b) "ow many offspring are homo&ygous for both characteristics! c) "ow many offspring are hetero&ygous for both characteristics! 1-. $ssume that curly hair /) is dominant to straight hair /c). $lbinism is a condition in which cells which normally produce pigment do not do so. 5he allele for skin albinism /a) is recessive to the normal allele /$). $ woman with curly hair and albinism and a man with straight hair and normal pigmentation have a child that has straight hair and is an albino. 6hat are the genotypes of the parents! 1. ?4 is a recessive disorder.
17. +n a disputed paternity case, a woman with blood type 8 has a child with type @, and she claimed that it had been fathered by a man with t ype $. 6hat can be proved from these facts! 1:. Bultiple alleles control the intensity of pigment in mice. 5he gene 2 1 designates full color, 2 # designates dilute color and 2' is deadly when homo&ygous. 5he order of dominance is 2 1 C 2# C 2'. 6hen a full color male is mated to a dilute color female, the offspring are produced in the following ratio% two full color to one dilute to one dead. +ndicate the genotypes of the parents. #;. Bultiple alleles control the coat color of rabbits. $ gray color is produced by the dominant allele . 5he ch allele produces a silver3gray color when present in the homo&ygous condition, chch, called chinchilla. 6hen ch is present with a recessive gene, a light silver3gray color is produced. 5he allele h is recessive to both the full color allele and the chinchilla allele. 5he h allele produces a white color with black extremities. 5his coloration pattern is called "imalayan. $n allele a is recessive to all the other alleles. 5he a allele results in a lack of pigment, called albino. 5he dominance hierarchy is C ch C h C a. 5he table below provides the possible genotypes and phenotypes for coat color in rabbits. Dotice that four genotypes are possible for full color but only one for albino. henotypes Full color hinchilla Light gray "imalaya $lbino
enotypes , ch, h, a chch chh, cha hh, ha aa
a) +ndicate the genotypes and phenotypes of the F1 generation from the mating of a hetero&ygous "imalayan coat rabbit with an albino coat rabbit. b) 5he mating of a full color rabbit with a light gray rabbit produces two full color offspring, one light gray offspring, and one albino offspring. +ndicate the genotypes of the parents. c) $ chinchilla color rabbit is mated with a light gray rabbit. 5he breeder knows that the light grey rabbit had an albino mother. +ndicate the genotypes and phenotypes of the F1 generation from this mating. d) $ test cross is performed with light gray rabbit, and the following offspring are noted% five "imalayan color rabbits and five light gray rabbits. +ndicate the genotype of the light3gray rabbit. #1. $ geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. +f oval radishes are crossed with oval radishes, the following phenotypes are noted in the F1 generation% 1;; long, #;; oval, and 1;; round radishes. 4se symbols explain the results obtained for the F1 and F# generations. ##. alomino horses are known to be caused by the interaction of two different genes. 5he allele r in the homo&ygous condition produces a chestnut, or reddish color, horse. 5he allele m produces a very pale cream color, called cremello, in the homo&ygous condition. 5he palomino color is caused by the interaction of both the chestnut and cremello alleles. +ndicate the expected ratios in the F 1 generation from mating a palomino with a cremello. #' +n mice, the gene causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. +ndividuals without pigment are albino. $nother gene, 8, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. 5he recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light3brown, color is produced. 5he genes that produce black or tan coat color rely on the gene , which produces pigment, but are independent of it. +ndicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F1 generation from the following crosses% a) 88 x cbb /b) cc88 x c8b /c) c8b x ccbb /d) c8b x c8b #*. 5he mating of a tan mouse and a black mouse produces many different offspring. 5he geneticist notices that one of the offspring is albino. +ndicate the genotype of the tan parent. "ow would you determine the genotype of the black parent! #-. 5he gene A produces a rose comb in chickens. $n independent gene, , which is located on a different chromosome, produces a pea comb. 5he absence of the dominant rose comb gene and pea comb allele /rrpp) produces birds with single combs. "owever, when the rose and pea comb alleles are present together, they interact to produce a walnut comb /A).
+ndicate the phenotypes of the parents and give the genotypic and phenotypic ratios of the F1 generation from the following crosses% a) rr x AApp /b) Arp x AA /c) Ar x rr /d) Arp x Arp #. For shorthorn cattle, the mating of a red bull and a white cow produces a calf that is described as roan. Aoan results from intermingled red and white hair. Bany matings between roan bulls and roan cows produce cattle in the following ratio% 1 red, # roan, 1 white. +s this a problem of codominance or multiple alleles! Explain your answer. #(. For $8@ blood groups, the $ and 8 genes are codominant but both $ and 8 are dominant over type @. +ndicate the blood types possible from the mating of a male who is blood type @ with a female of blood type $8. ould a female with blood type $8 ever produce a child with blood type $8! ould she ever have a child with blood type @! #7. For mice, the allele produces color. 5he allele c is an albino. $nother allele, 8, causes the activation of the pigment and produces black color. 5he recessive allele, b, causes the incomplete activation of pigment and produces brown color. 5he alleles and 8 are located on separate chromosomes and segregate independently. 2etermine the F 1 generation phenotypes from the cross c8b x c8b. #:. 5he genotype of F1 individuals in a tetrahybrid cross is $a8b2dEe. $ssuming independent assortment of these four genes, what are the probabilities that F# offspring would have the following genotypes! a. aabbddee b. $a8b2dEe c. $a8bddee d. $$8bddEE ';. ooh had a colony of tiggers whose stripes went across the body. "is $merican pen3pal, ogi, sent him a tigger whose stripes ran lengthwise. 6hen ooh crossed it with one of his own animals, he obtained plaid tiggers. +nterbreeding among the plaid tiggers produced litters with a ma9ority of plaid members, but some crosswise3 and lengthwise3striped animals were also produced. 2iagram the crosses that ooh made, showing the genotypes of the tiggers which account for the coat patterns observed. '1. 6hat are the possible blood types of the children in the following families! a. 5ype $ mother, 5ype @ father b. 5ype 8 mother, 5ype 8 father c. 5ype $ mother, 5ype $8 father d. 5ype @ mother, 5ype @ father e. 5ype 8 mother, 5ype $ father '#. olydactyly /extra fingers and toes) is due to a dominant gene. $ father is polydactyl, the mother has the normal phenotype, and they have had one normal child. 6hat is the probability that a second child will have the normal number of digits!
''. 5he
+n the pedigree above, a. is the trait being expressed characteristic of a dominant or recessive allele! Explain your reasoning. b. +f individuals : and # were to have children together, what chance do they have of having a child with the trait! Explain your answer.
'*. 5he pedigree above represents the family history of the Gohnson family. 4sing the symbols F for a dominant trait and f for a recessive trait, answer the following uestions. a. +s the trait a dominant or recessive trait! b. 6hat is the genotype of individual ! c. 6hat is the genotype of individual 1#! d. +f individual 1- and #; were to have a child, what are the chances that child would be homo&ygous recessive! e. +f individual 1- and individual 1( were to have a child, what are the chances they would have a child with the trait!
Learning Genetics Can Be Fun - Solutions
1. 5wo black dogs could be homo&ygous black /88) or hetero&ygous black /8b). ellow must be homo&ygous, therefore cannot be the same genotype as black. #. a) Ar x rr b) A, r and r, r c) F1 Ar, Ar, rr, rr 1 round%1 wrinkled d) F# AA, Ar, Ar, rr ' round%1 wrinkled '. L x ll F1 '#( tall% '#1 short 3 almost 1%1 therefore unknown parent must be hetero&ygous. Dote% homo&ygous /LL) would give $LL tall plants in F 1. *. 5he presence of all smooth in the offspring means smooth is dominant. << x ss F1 curly= c > straight h) c x cc F1 c, c, cc, cc i) , c 9) c, c k) ; l) 1H# m) 1H# n) 1 hetero&ygous%1 homo&ygous recessive
o) 1 curly%1 straight p) Do. 'H* x 'H* x 'H* > #(H* b) 1 3 #(H* > '(H* c) /all disease) > /pp) x /pp) x /pp) > 1H* x 1H* x 1H* > 1H* d) 1 3 1H* > 'H*
1(. A 3 rose= r 3 single F 3 feather= f 3 clean $x F1 all feather, mostly rose /FA and Frr) $x2 F1 feathered and clean but all rose /FA and ffA) 8x F1 feathered and clean, most rose but some single /FA, Frr,ffA, ffrr ) $ FfAr 8 FfAr FfAr 2 FfAA /!) 17. 8 x $ F1 ii the man could be father but this is not proof 1:. 212' x 2#2' F1 #21 , 1 2#2', 1 2'2' the presence of two recessive alleles in one offspring means each parent must have donated one #;. a) ha x aa F1 ha, aa 1 himalayan%1 albino b) a x cha F1 # , ch , aa c) chch x cha F1 chch, cha 1 chinchilla%1 light gray d) chh x aa test cross F1 - ha, - cha #1.
F1 Ar /walnut), rr /pea) d) Arp /walnut) x Arp /walnut) F1 AA /walnut), #AAp /walnut), AApp /rose), #Ar /walnut), *Arp /walnut), # Arpp /rose), rr /pea), # rrp /pea), rrpp /single) #. odominance gives a typical 1%#%1 ratio. Bultiple alleles would give ratios other than 1%#%1 because of any dominance hierarchy. #(. ii x + $+8 F1 +$i, +8i $8 could produce $8 offspring if were type $, 8, or $8= she could never produce type @ in F 1 because she always donates either $ or 8. #7. 3 color= c 3 albino 8 3 black= b 3 brown c8b x c8b F1 88, 8b, c88, c8b, bb, cbb, cc88, cc8b, ccbb 3 the expected ratio of :%'%'%1 becomes :%'%* due to epistasis. #:. "8 3 baldness 3 dominant in males= recessive in females "n 3 normal 3 dominant in females, recessive in males "n"n x "8"n F1 "n"8 3 a bald girl is impossible because the father cannot donate " 8 ';. a) * /24" J ee, challenging) b) 2 3 normal= d 3 recessive 2d x 2d c) /F) dd x /) dd F1 /B) dd, /D) dd
