wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A cross is made between true breeding lines of pea plants with axillary and terminal flowers. If 1240 progeny are obtained, how many plants are homozygous with axillary flowers and heterozygous with axillary flowers in the F2 generation?


A

310 & 620

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

620 & 620

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

310 & 310

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

620 & 310

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

310 & 620


The F2 genotypic ratio of Mendel’s monohybrid cross is 1:2:1. Out of the F2 progeny 1/4th are homozygous plants with axillary flowers. Homozygous plants with axillary flowers =1240×14=310 . Heterozygous plants with axillary flowers =1240×12=620. Thus option A is correct.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mendel's Experiments
BIOLOGY
Watch in App
Join BYJU'S Learning Program
CrossIcon