Phalaenopsis orchids are grown in a facility in Taiwan. The flower spikes are on
ID: 48207 • Letter: P
Question
Phalaenopsis orchids are grown in a facility in Taiwan. The flower spikes are on average 20cm long and contain nine flowers. The variance is 2cm and 2 flowers. The 22cm spikes with 11 flowers are crossed and the resulting offspring have an average spike length of 21cm with 10 flowers. A group of clonal plants to those described above are grown in a greenhouse in New England. These have an average flower spike length of 15cm with 5 flowers and the variance is 3cm and 2 flowers. When the 17cm spike with 7 flowers plants are crossed the resulting offspring have 16cm spikes with on average 6 flowers.
1-Calculate the narrow sense heritability for both traits under both growth conditions. Are these different results consistent?
2-What types of traits are these? How is the environment impacting the phenotype?
3-Assume this selection was done for a few generations. How would the broad sense heritability be useful for the breeder?
Explanation / Answer
1. narrow sense heritability (h2 is calculated using the following equation:
h2 = R/S ----- (1)
where,
R- Selection response
S- Selection differential
a. calculating narrow sense heritability for both traits for first growth conditions as:
(i) Spike length:
S= 22-20= 2
R= 21-20 = 1
Substituting values in (1) we get,
h2 = R/S
= 1/2 = .5
=50%
(ii) Number of flowers:
S= 11-9= 2
R= 10-9 = 1
Substituting values in (1) we get,
h2 = R/S
= 1/2 = .5
=50%
b. Calculating narrow sense heritability for both traits for second growth conditions as:
(i) spike length:
S= 17-15= 2
R= 16-15 = 1
Substituting values in (1) we get,
h2 = R/S
= 1/2 = .5
=50%
(ii) Number of flowers:
S= 7-5= 2
R= 6-5 = 1
Substituting values in (1) we get,
h2 = R/S
= 1/2 = .5
=50%