Question Details till 22/10/2022
Question | |
Chapter Name | Principles Of Inheritance And Variatio |
Subject | Biology (more Questions) |
Class | 12th |
Type of Answer | Video |
Question Language | In Video - English In Text - English |
Students Watched | 5.1 K + |
Students Liked | 0 + |
Question Video Duration | 5m34s |
Inheritance is the transfer of genes from the parents to the offspring. The inheritance occurs via sexual reproduction, and the offspring generated from the parents are known as F1 offspring.
F2 offspring are obtained by crossing the F1 hybrid.
In general, the probability of occurrence of a homozygous trait “XX or xx” is ¼, and the probability of occurrence of a heterozygous trait “Xx” is ½.
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a)aabbccdd b)AaBbCcDd c)AABBCCDD d)AaBBCcDd
e)AaBBCCdd
Please help me on tackling problems like this in a systematic way.
© BrainMass Inc. brainmass.com October 24, 2022, 7:53 pm ad1c9bdddf//brainmass.com/biology/dna-chromosomes-and-genomes/tetrahybrid-cross-probabilities-genotypes-94734
Please see attached file where formatting is conserved.
If you do not have to list the alleles in your answers, this would be a better and easier way to calculate the probabilities.
The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring will have the following genotypes? a)aabbccdd b)AaBbCcDd c)AABBCCDD d)AaBBccDd
e)AaBBCCdd
We have four groups
A Parent 1 Parent 2 A A
a a
Possible combinations A A A a a A
a a
Thus there is a total of 4 combinations
B Parent 1 Parent 2 B B
b b
Possible combinations B B B b b B
b b
Thus there is a total of 4 combinations
C Parent 1 Parent 2 C C
c c
Possible combinations C C C c c C
c c
Thus there is a total of 4 combinations
D Parent 1 Parent 2 D D
d d
Possible combinations D D D d d D
d d
Thus there is a total of 4 combinations
A genotype has the 4 groups together
Therefore possible number of ways of forming the genotype = 4 x 4 x 4 x 4= 256
Probability is the number of ways in which the desired ...