In rats, black color is dominant and white color is recessive.

1. In rats, black color is dominant and white color is recessive. • Cross heterozygous black rat with a white rat

2. B b b b B=blackb=whiteBb x bb P: 2/4 black, 2/4 white 1:1 G: 2/4 Bb 2/4 bb 1:1

3. Can two white rats have a black offspring? • Show the Punnett Square to confirm your answer

4. b b b b B=blackb=whitebb x bb  2 white rats P: 4/4 white 1:0 G: 4/4 bb 1:0 NO, 2 white rats cannot have a black offspring.

5. In humans the ability to taste a certain chemical is dominant. • Cross a homozygous taster with a heterozygous taster.

6. T T T t T=tastert=non-tasterTT x Tt G: 2/4 TT 2/4 Tt 1:1 P: 4/4 Tasters 1:0

7. Albinism – lack of pigment – is caused by a recessive allele. • Cross an albino with a heterozygous, normally pigmented person (remember to choose your letters using the letter of the dominant allele)

8. n n N n N=normal pigmentn=albinonn x Nn G: 2/4 Nn, 2/4 nn 1:1 P: 2/4 Normal 2/4 albino 1:1

9. Cross a homozygous roller (tongue, of course) with a heterozygous roller

10. R R R r R=rollerr=non-rollerRR x Rr G: 2/4 RR 2/4 Rr 1:1 P: 4/4 Rollers 1:0

11. Write the genotype of a heterozygous curly haired, homozygous albino, heterozygous tongue roller, homozygous taster. • Remember, you should have two alleles for each trait

12. C=curly, c=straight • N=normal pigment, n=albino • R=roller, r=non-roller • T=taster, t=non-taster • CcnnRrTT

13. Write the phenotype for the following person using these letters for your alleles • C = curly hair, c = straight hair • N = normal pigment, n = albino • R = roller, r = non-roller • T = taster, t = non-taster ccNnrrTT

14. ccNnrrTT •  straight hair, normal-pigment, non-tongue rolling, taster

15. A spotted rabbit when crossed with a solid colored rabbit produced all spotted offspring. When these F1 rabbits were crossed within themselves, they produced thirty-two spotted rabbits and ten solid-colored rabbits. Which characteristic is dependent on the dominant allele?

16. 32 spotted, 10 solid-colored •  ~ 3:1 ratio of spots:solids • Therefore, the spotted allele is dominant

17. How many of the spotted rabbits in the F2 generation in the preceding problem would be expected to be homozygous? • How many of the solid-colored F2 rabbits would be homozygous?

18. S s S s S=spotteds=solidSs x Ss G: ¼ SS, ¼ ss, 2/4 Ss 1:1:2P: ¾ spotted ¼ solid 3:1 ¼ Homozygous spotted, ¼ homozygous solid

19. In guinea pigs, the coat may be rough or smooth. • Certain rough coated guinea pigs when crossed with smooth coated guinea pigs produce all rough offspring. • Other rough coated guinea pigs when crossed with smooth coated guinea pigs produce equal numbers of rough and smooth coated guinea pigs

20. Smooth coated guinea pigs crossed together always produce smooth coated offspring • Explain these results • Write the genotypes for all animals involved in the crosses

21. R R r R r r r r r r r r Rough x smooth  All rough - RR x rrRough x smooth  ½ rough, ½ smooth - Rr x rrSmooth x smooth  All smooth - rr x rr R=roughr=smooth

22. Yellow guinea pigs crossed with white ones always produce cream colored offspring. • Two cream colored guinea pigs, when crossed together, produce yellow, cream and white offspring in a 1:2:1 ratio. • How are these colors inherited?

23. G=guinea pig, GY=yellow, GW=white • GYGW = cream-colored • 1:2:1 ratio of yellow homozygous, cream heterozygous, white homozygous •  Shows incomplete dominance • There is “blending”

24. In tomatoes the texture of the skin may be smooth or peach (hairy) • The Ponderosa variety has fruit with smooth fruit • The Red variety has fruit with peach (hairy) fruit • A cross between the two varieties produces all smooth fruit

25. Crosses between the smooth fruited F1 plants produced 174 peach textured fruits and 520 smooth textured fruits • How are these skin textures inherited?

26. 174 hairy, 520 smooth • What ratio does this look close to?? • 3:1 • Where have you seen this before? • When you cross two heterozygotes • Smooth is the dominant allele, hairy (peach) allele is recessive

27. In horses black is due to a dominant allele and chestnut is due to a recessive allele. • The trotting gait is due to a dominant allele and the pacing gait is due to a recessive allele

28. B = black, b = chestnutT = trotter, t = pacer • If a homozygous black pacer is mated to a chestnut homozygous trotter, what will the resulting F1 generation look like (this is a two – factor cross)

29. BBtt x bbTT G: 16/16 BbTt, 1:0P: 16/16 black trotters, 1:0

30. In cats black fur color and yellow fur color are codominant and tortoise-shell is produced in the heterozygote. • This fur color is also X - linked

31. XB = black, XY = yellow, XB XY = tortoise-shell • Remember XX = female and XY = male • Cross a black male and a yellow female • Cross a yellow male with a tortoise-shell female

32. XBY x XYXY XYY x XBXY G: ¼ XBXY, ¼ XYXY, ¼ XBY, ¼ XYY. 1:1:1:1 P: ¼ tortoise female, ¼ yellow female, ¼ black male, ¼ yellow male G: 2/4 XBXY, 2/4 XYY P: 2/4 tortoise female, 2/4 yellow males

33. XN = normal vision, Xn = color blind,male = XY, female = XX IA = Type A blood, IB = Type B blood, i= Type O blood, • A colorblind man with type O blood marries a heterozygous normal vision woman with type AB blood. • What will their sons be in regard to vision and blood type?