Learning Genetics Can Be Fun

Learning Genetics Can Be Fun

1. For Labrador retrievers, black fur is dominant to yellow. Explain how two black dogs can have different genotypes. Could a black dog have the same genotype as a yellow dog?

2. Cystic fibrosis is regulated by a recessive allele, c. Explain how two normal parents can produce a child with this disorder.

3. A 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 heterozygous.

b) the gametes produced by the parents

c) the genotypes and phenotypes of the F₁generation

d) the F₂generation if two round plants from the F₁generation were crossed

4. Long stems are dominant over short stems for pea plants. Determine the phenotypic and genotypic ratios of the F₁ offspring from the cross pollination of a heterozygous long stem plant with a short stem plant.

5. A pea plant with a tall phenotype is pollinated by a short plant, and the seeds of the first generation hybrid produce 327 tall plants and 321 short plants. Give the genotypes of all the plants.

6. In a certain species of plant, one purebred variety has hairy leaves and another pure variety has smooth leaves. A cross of the 2 varieties produces offspring that all have smooth leaves. Predict the ratio of phenotypes in the F₂ generation.

(3:1, smooth to hairy)

7. In summer squash, white fruit color is dominant. Yellow is recessive. A squash plant that is homozygous for white is crossed with a homozygous yellow one. Predict the appearance of

a) the F₁ generation (F₁ = all white)

b) the F₂ (F₂ = 3/4 white and 1/4 yellow)

c) the offspring of a cross between an F₁ individual and a homozygous white individual.

8. For Dalmatian dogs, the spotted condition is dominant to non-spotted.

a) Using a Punnett square, show the cross between two heterozygous parents.

b) A spotted female Dalmatian dog is mated to an unknown male. From the appearance of the pups, the owner concludes that the unknown male was a Dalmatian. The owner notes that the female had six pups, three spotted and three non-spotted. What are the genotype and phenotype of the unknown male?

9. For Mexican hairless dogs, the hairless condition is dominant to hairy. A litter of eight pups is found; six are hairless and two are hairy. What are the genotypes of the parents?

10. In horses, the trotter characteristic is dominant to the pacer characteristic. A male trotter mates with three different females, and each female produces a foal. The first female, a pacer, gives birth to a foal that is a pacer. The second female, also a pacer, gives birth to a foal that is a trotter. The third female, a trotter, gives birth to a foal that is a pacer. Determine the genotypes of the male, all three females, and the three foals sired.

11. Imagine for hair color that B gives brown hair and b gives blonde hair. Use a Punnett square to determine the following in a cross of two heterozygous parents.

a) What are the chances of the offspring being homozygous brown haired?

b) What are the chances of the offspring having blonde hair?

c) What are the chances of the offspring being heterozygous brown haired?

d) What is the genotypic ratio?

e) What is the phenotypic ratio?

f) Is there a heterozygous blonde haired offspring? Why?

g) If curly hair is dominant to straight hair, what letters will we use to show these genes?

h) A heterozygous curly haired male marries a straight haired female. What are their genotypes?

i) What would be the gametes for the male parent?

j) What would be the gametes for the female parent?

k) What are the chances of the offspring being homozygous curly haired?

l) What are the chances of the offspring having straight hair?

m) What are the chances of the offspring being heterozygous curly haired?

n) What is the genotypic ratio?

o) What is the phenotypic ratio?

p) Is there a heterozygous straight haired offspring? Why?

12. Thalassemia is a serious human genetic disorder that causes severe anemia. The homozygous condition (T^mT^m) leads to severe anemia. People with thalassemia die before sexual maturity. The heterozygous condition (T^mTⁿ) causes a less serious form of anemia. The genotype TⁿTⁿ causes no symptoms of the disease. Indicate the possible genotypes and phenotypes of the offspring if a male with the genotype T^mTⁿ marries a female of the same genotype.

13. In pea plants, tall is dominant to short and yellow is dominant to green. Show the F₁ and F₂ results of a cross between a homozygous tall pea plant that produces yellow seeds, and a short plant that produces green seeds.

14. In guinea pigs, black coat color (B) is dominant to white (b), and short hair length (S) is dominant to long (s). Indicate the genotypes and phenotypes from the following crosses:

a) Homozygous for black, heterozygous for short hair guinea pig crossed with a white, long hair guinea pig.

b) Heterozygous for black and short hair guinea pig crossed with a white, long hair guinea pig.

c) Homozygous for black and long hair crossed with a heterozygous black and short hair guinea pig.

d) For each of these crosses, give the probability that an offspring will have:

(i) black coat, long hair

(ii) black coat, short hair

(iii) white coat, long hair

15. For guinea pigs, black fur is dominant to white fur color. Short hair is dominant to long hair. A guinea pig that is homozygous for white and homozygous for short hair is mated with a guinea pig that is homozygous for black and homozygous for long hair. Indicate the phenotype(s) of the F₁ generation. If two hybrids from the F₁, generation are mated, determine the phenotypic ratio of the F₂ generation.

16 Black coat color (B) in cocker spaniels is dominant to white coat color (b). Solid coat pattern (S) is dominant to spotted pattern (s). The pattern arrangement is located on a different chromosome than the one for color, and its gene segregates independently of the color gene. A male that is black with a solid pattern mates with three females. The mating with female A, which is white, solid, produces four pups: two black, solid, and two white, solid. The mating with female B, which is black, solid, produces a single pup, which is white, spotted. The mating with female C, which is white, spotted, produces four pups: one white, solid; one white spotted; one black, solid; one black, spotted. Indicate the genotypes of the parents.

17. T = Tall, t = short B = brown hair, b = blonde hair

Cross a homozygous tall, heterozygous brown haired male with a heterozygous tall, blonde haired female. Use a Punnett square to show parents and gametes.

a) What is the phenotypic ratio?

b) How many offspring are homozygous for both characteristics?

c) How many offspring are heterozygous for both characteristics?

18. Assume that curly hair (C) is dominant to straight hair (c). Albinism is a condition in which cells which normally produce pigment do not do so. The allele for skin albinism (a) is recessive to the normal allele (A). A 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. What are the genotypes of the parents?

19. PKU is a recessive disorder. Suppose two people who were heterozygous for PKU married and had a child. What is the probability that the child will have PKU? (1/4)

20. For chickens, the gene for rose comb (R) is dominant to that for single comb (r). The gene for feather legged (F) is dominant to that for clean legged (f). Four feather legged, rose combed birds mate. Rooster A and hen C produce offspring that are all feather legged and mostly rose combed. Rooster A and hen D produce offspring that are feathered and clean, but all have rose combs. Rooster B and hen C produce birds that are feathered and clean. Most of the offspring have rose combs, but some have single combs. Determine the genotypes of the parents.

21. For human blood type, the alleles for types A and B are codominant, but both are dominant over the type O allele. The Rh factor is separate from the ABO blood group and is located on a separate chromosome. The Rh+ allele is dominant to Rh-. Indicate the possible phenotypes from the mating of a woman, type,O, Rh-, with a man, type A, Rh+.

22. In a disputed paternity case, a woman with blood type B has a child with type O, and she claimed that it had been fathered by a man with type A. What can be proved from these facts?

23. Multiple alleles control the intensity of pigment in mice. The gene D¹ designates full color, D² designates dilute color and D³ is deadly when homozygous. The order of dominance is D¹ > D² > D³. When 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. Indicate the genotypes of the parents.

24. A geneticist notes that crossing a round shaped radish with a long shaped radish produces oval shape radishes. If oval radishes are crossed with oval radishes, the following phenotypes are noted in the F₁ generation: 100 long, 200 oval, and 100 round radishes. Use symbols explain the results obtained for the F₁ and F₂ generations.

25. In mice, the gene C causes pigment to be produced, while the recessive gene c makes it impossible to produce pigment. Individuals without pigment are albino. Another gene, B, located on a different chromosome, causes a chemical reaction with the pigment and produces a black coat color. The recessive gene, b, causes an incomplete breakdown of the pigment, and a tan, or light-brown, color is produced. The genes that produce black or tan coat color rely on the gene C, which produces pigment, but are independent of it. Indicate the phenotypes of the parents and provide the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) CCBB x Ccbb (b) ccBB x CcBb (c) CcBb x ccbb (d) CcBb x CcBb

26. The mating of a tan mouse and a black mouse produces many different offspring. The geneticist notices that one of the offspring is albino. Indicate the genotype of the tan parent. How would you determine the genotype of the black parent?

27. The gene R produces a rose comb in chickens. An independent gene, P, which is located on a different chromosome, produces a pea comb. The absence of the dominant rose comb gene and pea comb allele (rrpp) produces birds with single combs. However, when the rose and pea comb alleles are present together, they interact to produce a walnut comb (R_P_).

Indicate the phenotypes of the parents and give the genotypic and phenotypic ratios of the F₁ generation from the following crosses:

a) rrPP x RRpp (b) RrPp x RRPP (c) RrPP x rrPP (d) RrPp x RrPp

28. For shorthorn cattle, the mating of a red bull and a white cow produces a calf that is described as roan. Roan results from intermingled red and white hair. Many matings between roan bulls and roan cows produce cattle in the following ratio: 1 red, 2 roan, 1 white. Is this a problem of codominance or multiple alleles? Explain your answer.

29. For ABO blood groups, the A and B genes are codominant but both A and B are dominant over type O. Indicate the blood types possible from the mating of a male who is blood type O with a female of blood type AB. Could a female with blood type AB ever produce a child with blood type AB? Could she ever have a child with blood type O?

30. For mice, the allele C produces color. The allele c is an albino. Another allele, B, causes the activation of the pigment and produces black color. The recessive allele, b, causes the incomplete activation of pigment and produces brown color. The alleles C and B are located on separate chromosomes and segregate independently. Determine the F₁ generation from the cross CcBb x CcBb.

31. Baldness (H^B) is dominant in males but recessive in females. The normal gene (Hⁿ) is dominant in females, but recessive in males. Explain how a bald offspring can he produced from the mating of a normal female with a normal male. Could these parents ever produce a bald girl? Explain your answer.

Learning Genetics Can Be Fun - Solutions

1. Two black dogs could be homozygous black (BB) or heterozygous black (Bb). Yellow must be homozygous, therefore cannot be the same genotype as black.

2. P Cc x Cc

F₁ CC, Cc, Cc, cc both parents are normal but “carry” the allele for CF. One in four children will inherit it.

3. a) P Rr x rr

b) R, r and r, r

F₁Rr, Rr, rr, rr 1 round:1 wrinkled

F₂ RR, Rr, Rr, rr 3 round:1 wrinkled

4. P Ll x ll

F₁ Ll, Ll, ll, ll 1 long:1 short

5. P L_ x ll

F₁ 327 tall: 321 short - almost 1:1 therefore unknown parent must be heterozygous. Note: homozygous (LL) would give ALL tall plants in F₁.

6. The presence of all smooth in the offspring means smooth is dominant.

P SS x ss

F₁ Ss

F₂ 3:1

7. P WW x ww

a) F₁ Ww all white

b) F₂ WW, Ww, Ww, ww 3 white:1 yellow

c) P Ww x WW

F₁ Ww, WW all white

8. a) P Ss x Ss

F₁ SS, Ss, Ss, ss

b) P S_ x _ _

The female must be heterozygous as she produced non-spotted pups! The unknown male must be homozygous recessive (ss). If he were homozygous dominant, all pups would be spotted. If he were heterozygous, you would expect a 3:1 ratio in pups.

9. A ratio of 3:1 observed, so we conclude the parents are both heterozygous.

10. (i) P T_ x tt

F₁tt

(ii) P T_ x tt

F₁ Tt

(iii) P T_ x Tt

F₁ tt

The male must be heterozygous (Tt) to be able to produce both trotters and pacers. If he were homozygous dominant her would produce only trotters.

11. P Bb x Bb

F₁ BB, Bb, Bb, bb

a) 1/4

b) 1/4

c) 1/2

d) 1 homozygous brown:2 heterozygous brown:1 homozygous blonde

e) 3 brown:1 blonde

f) not possible because blonde (b) is recessive

g) C = curly; c = straight

h) P Cc x cc

F₁ Cc, Cc, cc, cc

i) C, c

j) c, c

k) 0

l) 1/2

m) 1/2

n) 1 heterozygous:1 homozygous recessive

o) 1 curly:1 straight

p) No. Straight hair is recessive so individual MUST be homozygous (cc).

12. P T^mTⁿ x T^mTⁿ

F₁ T^mT^m, 2T^mTⁿ, TⁿTⁿ (1 severe : 2 mild : 1 normal)

13. The trick here is that the parents produce all yellow or all green seeds. That means they muts be homozygous, otherwise they would produce a mixture.

P TTYY x ttyy

F₁ TtYy

F₂ 9:3:3:1

14. B - black; b - white; S - short; s - long

a) P BBSs x bbss

F₁ BbSs, Bbss 1 black, short:1 black, long

b) P BbSs x bbss

F₁ BbSs, Bbss, bbSs, bbss 1 black, short:1 black, long: 1 white, short:1 white, long

c) P BBss x BbSs

F₁ BBSs, BBss, BbSs, Bbss 1 black, short:1 black, long

d) i) (a) 1/2 (b) 1/4 (c) 1/2

ii) (a) 1/2 (b) 1/4 (c) 1/2

iii) (a) 0 (b) 1/4 (c) 0

15. B - black; b - white

S - short; s - long

P bbSS x BBss

F₁ BbSs

F₂ 1 BBSS, 2 BBSs, 2 BbSS, 2 Bbss, 4 BbSs, BBss, bbSS, 2 bbSs, bbss (typical heterozygous dihybrid cross phenotypic ratio of 9:3:3:1 )

16. B - black; b - white; S - solid; s- spotted

male female

a) P B_S_ bbS_

F₁ 2 BbS_ , 2 bbS_

Some white pups so the male must be Bb. The absence of any non-spotted pups suggests that female A is SS but we can’t say for sure.

b) P BbSs B_S_

F₁ bbss

the presence of white, non-spotted pups means that female B must be BbSs

c) P BbSs bbss

F₁ bbSs , bbss , BbSs , Bbss

The genotype of female C can be determined from her phenotype.

17. P TTBb x Ttbb

F₁TTBb, TtBb, TTbb, Ttbb

a) 1 tall, brown:1 tall, blonde

b) 1/4

c) 1/4

18. P C_aa x ccA_

F₁ ccaa this means the parents must both be heterozygous (Ccaa and ccAa)

19. P Pp x Pp

F₁ PP, Pp, Pp, pp chance of PKU is 1/4

20. R - rose; r - single

F - feather; f - clean

P A x C

F₁ all feather, mostly rose (F_R_ and F_rr)

P A x D

F₁ feathered and clean but all rose (F_R_ and ffR_)

P B x C

F₁ feathered and clean, most rose but some single (F_R_, F_rr,ffR_, ffrr )

A FfRr

B FfRr

C FfRr

D FfRR (?)

21. Female OORh-Rh- Male AORh+Rh-

AORh+Rh-, AORh-Rh-, OORh+Rh-, OORh-Rh-

AORh+Rh+ AORh+Rh-, OORh+Rh-

AARh+Rh- AORh+Rh-, AORh-Rh-

AARh+Rh+ AORh+Rh-

possible phenotypes are A+, A-, O+, O-

22. P B_ x A_

F₁ ii the man could be father but this is not proof

23. P D¹D³ x D²D³

F₁ 2D¹_, 1 D²D³, 1 D³D³ the presence of two recessive alleles in one offspring means each parent must have donated one

24. S^R - round; S^L - long

P S^RS^L x S^RS^L

F₁ S^RS^R, S^RS^L, S^RS^L, S^LS^L (incomplete dominance)

25. a) P CCBB (black) x Ccbb (brown)

F₁ CCBb (black), CcBb (black)

b) P ccBB (albino) x CcBb (black)

F₁ CcBB (black), CcBb (black), ccBB (albino), ccBb (albino)

c) P CcBb (black) x ccbb (albino)

F₁ CcBb (black), Ccbb (brown), ccBb (albino), ccbb (albino)

d) P CcBb (black) x CcBb (black)

F₁ CCBB (black), CCBb (black), CcBB (black), CCbb (brown), CcBb (black), Ccbb (brown), ccBB (albino), ccBb (albino), ccbb (albino)

note: you would get the normal 9:3:3:1 as in any heterozygous dihybrid cross but ccBB, ccBb, and ccbb all combine to give 4 albino (a bit tricky, eh?)

26. P Ccbb x CcB_

A test cross would be necessary to determine the genotype of the black parent. Cross it with a CCbb mouse. If any albino offspring are present in F₁, the black mouse is heterozygous. If all black, the black mouse is homozygous.

27. a) P rrPP (pea) x RRpp (rose)

F₁ RrPp (walnut)

b) P RrPp (walnut) x RRPP (walnut)

F₁ RRPP (walnut), RRPp (walnut), RrPP (walnut), RrPp (walnut)

c) P RrPP (walnut) x rrPP (pea)

F₁ RrPP (walnut), rrPP (pea)

d) P RrPp (walnut) x RrPp (walnut)

F₁ RRPP (walnut), 2RRPp (walnut), RRpp (rose), 2RrPP (walnut), 4RrPp (walnut),

2 Rrpp (rose), rrPP (pea), 2 rrPp (pea), rrpp (single)

28. Codominance gives a typical 1:2:1 ratio. Multiple alleles would give ratios other than 1:2:1 because of any dominance hierarchy.

29. P ♂ ii x ♀ I^AI^B

F₁ I^Ai, I^Bi

AB ♀ could produce AB offspring if ♂ were type A, B, or AB; she could never produce type O in F₁ because she always donates either A or B.

30. C - color; c - albino

B - black; b - brown

P CcBb x CcBb

F₁ CCBB, CCBb, CcBB, CcBb, CCbb, Ccbb, ccBB, ccBb, ccbb

- the expected ratio of 9:3:3:1 becomes 9:3:4 due to epistasis.

31. H^B - baldness - dominant in ♂; recessive in ♀

Hⁿ - normal - dominant in ♀, recessive in ♂

P ♂ HⁿHⁿ x ♀ H^BHⁿ

F₁ HⁿH^B

- a bald girl is impossible because the father cannot donate H^B