Genetics Problems ans

Genetics Problems Name ANSWER KEY Problems 1-6: In tomato fruit, red flesh color is dominant over yellow flesh color, ...

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Genetics Problems

Name ANSWER KEY

Problems 1-6: In tomato fruit, red flesh color is dominant over yellow flesh color, Use R for the Red allele and r for the yellow allele. 1. What would be the genotype for a tomato which is homozygous for red flesh color? RR 2. What would be the genotype for a tomato which has a yellow-flesh color? rr 3. What are the possible gametes produced from a homozygous red-fleshed tomato? R 4. What type of allele(s) will the yellow plant pass to its offspring? r 5. Draw a punnett square showing a cross between a homozygous red and a yellow tomato. R r Rr 6. a. What is the phenotype of the F1 generation from the previous cross? Red b. What are the possible gametes from an F1 plant? R & r c. Draw a punnett square for the F1 cross which will represent the F2 generation. R R genotypes RR 1 RR : 2 Rr : 1 rr r r Rr Rr phenotypes rr 3 red : 1 yellow d. What are the ratios of phenotypes and genotypes for the F2 generation? above e. If 500 tomatoes were picked from the vines of the F2 generation, approximately how many would be red? Yellow? 350 Red : 125 yellow Problems 7-11: In tomatoes, tall (T) is dominant over short (t) and red flesh is dominant over yellow flesh. A homozygous tall, red-fleshed tomato was crossed with a homozygous short yellow tomato. 7. a. What is the genotype for each parent? TTRR & ttrr b. What are the possible gametes for each? TR & tr

8. Draw a punnett square for this cross TR tr TtRr

9. What are the phenotypes and genotypes of the F1 generation? Tall-red tomato 10. The F1 generation is self-pollinated, what are the possible gametes for an F1 plant? TR - Tr - tR - tr 11. a. Draw a punnett square for this cross. TR TR genotype phenotype TTRR 1 – TTRR 9 – tall red Tr Tr 2 – TTRr 3 – tall yellow TTRr TTRr 2 – TtRR 3 – short red tR tR 1 – TTrr 1 – short yellow TtRR TTrr TtRR 4 - TtRr tr tr 1 - ttRR TtRr TtRr TtRr TtRr 2 - Ttrr 2 - ttRr Ttrr ttRR Ttrr 1 - ttrr ttRr

ttRr ttrr

b. What are the genotypes and phenotypes of the F2 generation? Give the ratios above

Incomplete dominance: A hairy plant allele is dominant over a hairless allele. A heterozygous plant will show a plant with few hairs. This is an example of incomplete dominance. 12. A very hairy plant is crossed with a hairless plant. If the pollinated, what will be the appearance of the F2 generation? phenotype ratios of the F2. Alleles: H – hairy, h – hairless H Parent: HH (hairy) x hh (hairless) F1: Hh (few hairs) h Hh F2: genotype: 1 HH : 2 Hh : 1 hh phenotype: 1 hairy: 2 few hairs : 1 hairless

F1 generation is selfGive both genotype and H HH h Hh hh

Sex Determination: A male will contain an X and a Y chromosome where a female will carry two X chromosomes. 13. Does a male child inherit his X chromosome from his mother of father? Explain. Sex chromosomes Father: XY X X Offspring Mother: XX XX XX (females) 1/2 Y X XY (males) 1/2 XY XX The male child must inherit his X XY chromosome from his mother. Sex-linked: The X and Y chromosomes carry specific traits which can be linked to the gender of a child. Due to the difference between the X and Y chromosomes some traits can be expressed from an X without the Y being able to mask it. 14. A woman who is a carrier for red-green color blindness marries a man with normal vision. Each time they have a child what is the probability that the child will be a. color blind male 25% b. a color blind female 0% b. normal male 25% d. normal female 25% alleles Parents X – normal X X Xc & XY Xc – color-blind gene X X X Y – normal Y XX Genotypes Y Xc XX – normal female XY X Xc X Xc – carrier female X Xc – color-blind female XY – normal male Xc Y Xc Y – color-blind male Genetic Problems: 15. In meadow mice, dark coat color is dominant over cream color. If a homozygous dark individual is crossed with a homozygous cream one. What are the genotype and phenotype ratios of the F1 offspring? Parents DD & dd D d Dd F1 Dd - dark

16. In rabbits, brown fur color is dominant over albino. Short hair is dominant over angora (long) hair. a. A homozygous brown short-haired rabbit is crossed with an albino long-haired rabbit. Parents: BBSS & bbss Give the phenotype(s) and genotype (s) of the F1 generation F1 - genotype: BbSs phenotype: brown-short hair b. If two rabbits from the F1 generation are mated, what are the expected genotypes and phenotypes of the F2 generation and in what ratio? Dihybrid heterozygous cross: (look at 11a. for an example) genotypes phenotypes BBSS – 1 BbSS – 1 brown-short hair – 9 BBSs – 2 Bbss - 2 brown-long – 3 BbSS – 2 bbSs – 2 albino-short – 3 BBss - 1 bbss - 1 albino-long – 1 BbSs – 4 17. Two short tailed (Manx) cats are bred together. They produce three kittens with long tails, five with short tails and two without any tails. From these results, how would you think that tail length in cats is inherited? Show genotypes to support your answer. 3 long: 5 short: 1 no tail Short tail would appear to be a mix of long and no L L LL - long tail alleles. The ratio is also close to the Mendelian l LL l Ll - Manx ratio of 1:2:1, so it would appear that the Manx is Ll Ll ll- no tail heterozygous (Ll) with L probably being the allele ll for long and l the allele for the short. This is an example of incomplete dominance. 18. In the tiger plant, striped leaves (S) are dominant over non-striped leaves (s). A heterozygous striped leaf plant was crossed with a plant having non-striped leaves. Predict the genotypes and phenotypes of the offspring. Parent – Ss & ss S s F1 genotypes F1 phenotypes s Ss s Ss – ½ striped – ½ ss Ss ss - ½ non-striped – ½ ss

19. a. In squirrels, gray (G) fur is dominant over black fur (g). By employing a testcross, explain how a squirrel breeder could determine if a particular gray squirrel is homozygous or heterozygous for gray fur color. Test crosses involve breeding as organism possessing G g a dominant trait with another having the recessive ? Gg g trait. If all the offspring show dominance, then the Gg or Gg one parent was homozygous dominant. If half of the gg Gg offspring show the recessive phenotype, then the parent or gg was heterozygous. Phenotypes: Gg (heterozygous = gray) gg (homorecessive = black) b. Explain why it is important that the squirrel breeder carry out a large number of testcross matings before coming up with a sure conclusion. Small samples don’t necessarily follow to the predicted probabilities. Larger number of offspring will better fit the correct ratios. 20. A woman with normal vision marries a man with normal vision and they have a colorblind son. Her husband dies and she marries a color-blind man. Show the types of children that might be expected from this marriage and the proportion of each. The woman must be a carrier for color-blindness. By marrying a color blind man: Xc Xc Children from #2 X Xc c c c c c Y XX X Y XX X X X – color-blind female c c c XY XY XY XX XXc – carrier female XY XY XcY – color blind male XY – normal male The proportions for the offspring are all ¼ or 25% 21. Suppose a woman comes to you for advice as a genetic counselor. She tells you that her brother is a hemophiliac, but both parents are normal. She wishes to marry a man who has no history of hemophilia. She wants to know the probability of her children having the disease. What will you tell her, explain your answer? X X If her brother is a hemophilic (XhY) and Chromosomes X – normal X Y XX Xh both parents are normal, then her mother is XY XXh a carrier for hemophilia. As a female from Xh – hemophilic X Y – normal Y this, she has a 50% (1/2) chance of being a h XY carrier (XXh vs. XX). The only risk to her children is if she is a carrier (which would look like the above punnett square). If she is a carrier and her husband is normal, then the only child to be a hemophilic is a son (XhY). This probability is equal to her probability times the sons probability [ ½ x ¼ = 1/8] . She has a 12. 5% chance of having a hemophiliac child.

22. A man files for divorce based on the grounds of infidelity. Both man and wife have normal vision, but they have a baby girl who is color blind. If you were the man’s lawyer could you use this as evidence? How would you explain it? To get a color blind daughter (XcXc), the father has to be color blind (XcY) and the mother is either a carrier (XXc) or color-blind herself(XcXc). Since he is not colorblind, he cannot have a color blind daughter. Look at the punnett squares on problem 20. 23. In the radish plant, the long and round traits are incompletely dominant and result in an oval shape. The red and white colors are incompletely dominant also. If two plants that are purple and oval were crossed, what would the phenotypes ratios of the offspring be? alleles LR LR Parents: LlRr & LlRr long – L F1 genotypes F1 phenotypes round – l Lr LLRR Lr LLRR – 1 long red - 1 red – R LLRr – 2 long purple - 2 white – r lR LLRr LLRr lR LlRR – 2 oval red - 2 LLrr – 1 long white - 1 lr LlRR LLrr LlRR lr LlRr – 4 oval purple - 4 llRR – 1 round red - 1 LlRr LlRr LlRr LlRr Llrr – 2 oval white - 2 llRr – 2 round purple - 2 Llrr llRR Llrr llrr – 1 round white - 1 llRr

llRr llrr

The phenotype and genotype ratios are the same due to the incomplete dominance. The dominant allele doesn’t mask the recessive, instead they blend.

24. In certain breeds of dogs, black color is dominant and red color is recessive; solid color is dominant to white spotting. A homozygous black and white spotted male is crossed with a red, white spotted female. What is the probability of them producing a solid black puppy? parents: BBss & bbss All the F1 generation will be Bbss Alleles black – B Bs bs which is black-white spotted. None red – b Bbss of the offspring can be solid because solid – S neither parent had the dominant white spotting – s solid allele (S).

25. The black and yellow pigments in the coats of cats seem to be controlled by a sixlinked pair of alleles in such a way that the heterozygote is the familiar calico. The following genotypes and phenotypes are possible XBXb calico female XbXb yellow female XBXB black female XB Y black male Xb Y yellow male a. Explain why the next calico cat will be female and not male. Calico cats can only be female (XBXb) and males cannot. b. What types of kittens will be produced from a black female and a yellow male? Parents: XBXB & XbY Xb XB F1 offspring B b B b X X - calico female ( ½ ) X X B Y X XBY - black male ( ½ ) XBXb XBY XBY

c. If one of the female kittens is crossed with a yellow male, what kinds of kitten are produced? Parents Xb XB F1 offspring B b B b b Y X X X XBXb – calico female ( ½ ) X X XbY XB Y XBXb XBY - black male ( ¼ ) XbY - yellow male ( ¼ ) XbY