You selfpollinate a pea plant that is heterozygous for round
Solution
You self-pollinate a pea plant that is heterozygous for round/wrinkled alleles at the r locus (Rr). You want to figure out the probability of finding 2 round and 4 wrinkled seeds in a 6 seed pod, Round is dominant to wrinkled, so Rr plants are round. Ar in an Fa monohybrid cross like this,
what proportion of the progeny will be round (p) and what proportion of the progeny will be wrinkled (q)? Note: p q B. Using Pascal\'s triangle to identify the coefficients for each possible outcome (hint: p means 6 round peas in a pod),
R---------round --------dominant allele
Recessive allele ---r--------wrinkled
Rr x Rr
R
r
R
round Rr
round Rr
r
round Rr
wrinkled rr
Probability of round seed= ¾=0.75---------------p
Probability of wrinkled seed= 1/4=0.25-----------q
of finding 2 round and 4 wrinkled seeds in a 6 seed pod,
n=6
(p+q)^n=
(p+q)^6=p^6+ 6p^5q^1+15p^4q^2+20p^3q^3+15p^2q^4+6p^1q^5+q^6=
=15p^2q^4 = 2 round and 4 wrinkled
=15*0.75^2*0.25^4 =0.0329—answer
Coefficient
p^6
1
Six round
6p^5q^1
6
5 round and 1 wrinkled
15p^4q^2
15
4 round and 2 wrinkled
20p^3q^3
20
3 round and 3 wrinkled
15p^2q^4
15
2 round and 4 wrinkled
6p^1q^5
6
1 round and 5 wrinkled
q^6
1
6 wrinkled
| R | r | |
| R | round Rr | round Rr |
| r | round Rr | wrinkled rr |

