CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In maize, coloured endosperm (C) is dominant over colourless (c) and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F1 generation was test crossed, it produced four phenotype in the following percentage
Coloured and Full = 45%
Coloured - Shrunken = 5%
Colourless - Full = 4%
Colourless - Shrunken = 46%
From these data what would be distance between the two allelic genes

A
48 units
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
9 units
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4 units
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12 units
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 9 units
Distance between the two allelic genes
= Percentage of recombinant offsprings
= Percentage of Coloured – Shrunken offspring + Percentage of Colourless – Full offspring
= 5 + 4 Map units
= 9 Units
(Here recombinant offspring means different than F1 and F2.)

flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Chromosome Theory of Inheritance
BIOLOGY
Watch in App
Join BYJU'S Learning Program
CrossIcon