GENETIC PROBLEMS Question 1 How many different kinds

GENETIC PROBLEMS

Question 1 • How many different kinds of gametes could the following individuals produce? 1. 2. 3. 4. 5. aa. Bb CCDdee AABb. Cc. DD Mm. Nn. Oo. Pp. Qq UUVVWWXXYYZz

Answer • Remember the formula 2 n • Where n = # of heterozygous 1. 2. 3. 4. 5. aa. Bb CCDdee AABb. Cc. DD Mm. Nn. Oo. Pp. Qq UUVVWWXXYYZz =2 =2 =4 = 32 =2

Question 2 • In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w) • WW, Ww = wire haired • ww = smooth haired

Answer If a homozygous wire-haired dog is mated with a smooth-haired dog, what type of offspring could be produced W w w Ww Ww Ww F 1 generation all heterozygous

Question 3 • What type(s) of offspring could be produced in the F 2 generation? • Must breed the F 1 generation to get the F 2. • Ww x Ww

Answer W w W WW Ww ww F 2 generation genotype: 1: 2: 1 ratio phenotype: 3: 1 ratio

Question 4 • Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. • If these, two wire-haired dogs mate again, what are the chances that they will produce another smooth-haired pup? • What are the chances that the pup will wirehaired?

Answer W w W WW Ww ww F 2 generation - 1/4 or 25% chance for smooth-haired - 3/4 or 75% chance for wire-haired

Question 5 • A wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smoothhaired. • What are the phenotypes and genotypes of the pups they could produce? • Breed: Ww x ww

Answer W w w Ww ww phenotypes: 2: 2 ratio genotypes: 2: 2 ratio

Question 6 • In snapdragons, red flower (R) color is incompletely dominant over white flower (r) color. • The heterozygous (Rr) plants have pink flowers. RR Rr rr - red flowers - pink flowers - white flowers

Question 6 • If a red-flowered plant is crossed with a whiteflowered plant, what are the genotypes and phenotypes of the plants F 1 generation? • RR x rr

Answer R R r Rr Rr phenotypes: genotypes: F 1 generation 100% pink 100% heterozygous

Question 7 • What genotypes and phenotypes will be produced in the F 2 generation? • Rr x Rr

Answer R RR Rr rr F 2 generation phenotypes: 1: 2: 1 ratio genotypes: 1: 2: 1 ratio

Question 8 • What kinds of offspring can be produced if a red-flowered plant is crossed with a pinkflowered plant? • RR x Rr

Answer R RR RR r Rr Rr 50%: red flowered 50%: pink flowered

Question 9 • What kind of offspring is/are produced if a pink -flowered plant is crossed with a whiteflowered plant? • Rr x rr

Answer R r r Rr rr 50%: white flowered 50%: pink flowered

Question 10 • In humans, colorblindness (cc) is a recessive sex-linked trait. • Remember: XX - female XY - male

Question 10 • Two normal people have a colorblind son. • What are the genotypes of the parents? • XCX_? x XCY • What are the genotypes and phenotypes possible among their other children?

Answer XC Y XC X CX C X CY Xc X CX c X c. Y parents 50%: female (one normal, one a carrier) 50%: male (one normal, one colorblind)

Question 11 • A couple has a colorblind daughter. • What are the possible genotypes and phenotypes of the parents and the daughter?

Answer Xc Y XC X CX c X CY Xc X c. Y parents: Xc. Y and XCXc or Xc. Xc father colorblind mother carrier or colorblind daughter: Xc. Xc - colorblind

Question 12 • In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). • Dimpled cheeks (D) are dominant to nondimpled cheeks (d).

Question 12 • Two persons with freckles and dimpled cheeks have two children: one has freckles but no dimples and one has dimples but no freckles. • What are the genotypes of the parents? Parents: d x Ff_Dd_ F_f D_ Children: F_dd x ff. D_

Question 12 b • What are the possible phenotypes and genotypes of the children that they could produce? • Breed: Ff. Dd x Ff. Dd • This is a dihybrid cross

Answer • Possible gametes for both: FD Fd f. D fd FD FFDd Ff. DD Ff. Dd Fd FFDd FFdd Ff. Dd Ffdd f. D Ff. Dd ff. DD ff. Dd fd Ff. Dd Ffdd ff. Dd ffdd

Answer 12 b Phenotype Freckles/Dimples: Freckles/no dimples: no freckles/Dimples: no freckles/no dimples: 9 3 3 1 Phenotypic ratio will always be 9: 3: 3: 1 for dihybrid crosses.

Answer 12 b Genotypic ratio: FFDD FFDd- 2 FFdd - 1 Ff. DD - 2 Ff. Dd - 4 Ffdd - 2 ff. DD - 1 ff. Dd - 2 ffdd - 1 -1

Question 13 • What are the chances that they would have a child whom lacks both freckles and dimples? • This child will have a genotype of ffdd • Answer: 1/16

Question 14 • A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but not dimples whose father did not have freckles or dimples. • Breed: Ff. Dd x Ffdd • Possible gametes: FD Fd f. D fd x Fd fd

Question 15 • In dogs, the inheritance of hair color involves a gene B for black hair and gene b for brown hair b. • A dominant C is also involved. It must be present for the color to be synthesized. • If this gene is not present, a blond condition results. BB, Bb - black hair bb - brown hair CC, Cc - color cc - blond

Question 16 • A brown haired male, whose father was a blond, is mated with a black haired female, whose mother was brown haired and her father was blond. Male: bb. Cc (gametes: b. C bc) Female: Bb. Cc (gametes: BC Bc b. C bc) • What is the expected ratios of their

Answer 16 BC Bc b. C bc b. C Bb. Cc bb. CC bb. Cc bc Bb. Cc Bbcc bb. Cc bbcc Offspring ratios: Black: Brown: Blond: 3/8 2/8 or 1/4

Question 17 • Charlie Chaplin, a film star, was involved in a paternity case. The woman bringing suit had two children, on whose blood type was A and the other whose blood type was B. • Her blood type was O, the same as Charlie ’s! • The judge in the case awarded damages to the woman, saying that Charlie had to be the father of at least one of the children.

Answer 17 • Obviously, the judge should be sentenced to Biology. For Charlie to have been the father of both children, his blood type would have had to be what? IA IB Answer i I Ai I Bi