Question

1. A researcher studied six independently assorting genes in a plant. Each gene has a dominant and a recessive allele: R black stem, r red stem; D tall plant, d dwarf plant; C full pods, c constricted pods; O round fruit, o oval fruit; H hairless leaves, h hairy leaves; W purple flower, w white flower. The following cross was setup:
2. 1. Parent 1 Rr Dd cc Oo Hh Ww      X    Parent 2 Rr dd Cc oo Hh ww
3. 2. How many different types of gametes Parent 1 can form?
4. 3. What is the probability of obtaining a black, constricted, oval, hairy, purple phenotype in the progeny?

Answer 1

Answer:

Parent 1 Rr Dd cc Oo Hh Ww X Parent 2 Rr dd Cc oo Hh ww ---Parents

b)

Rr produces two types of gametes, R & = 2

Dd = 2

cc = 1

Oo = 2

Hh = 2

Ww = 2

Number of different types of gametes = 2*2*1*2*2*2 = 32

c).

Rr x Rr = R_ (black-3/4) & rr (red-1/4)

Dd x dd = Dd (tall-1/2) & dd (dwarf-1/2)

cc x Cc = Cc (full-1/2) & cc (constricted-1/2)

Oo x oo = Oo (round-1/2) & oo (oval-1/2)

Hh x Hh = H_ (hairless-3/4) & hh (hairy-1/4)

Ww x ww = Ww (purple=1/2) & ww (white-1/4)

The probability of obtaining a black, constricted, oval, hairy, purple phenotype = R_cc oo hh Ww = ¾ * ½ * ½ * ¼ * ½ = 3/128