Let the normal gene be represented by P, and the two mutations be represented by p# and p*.
Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the
gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes
obtained will be:
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1)
In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and
p*p* appear only once (1).
Since, both the mutations are recessive, three phenotypes would be obtained in the ratio:
mutant 1 : normal : mutant 2 :: 1:2:1
Three genotypes will be obtained in the ratio:
p#p# : p#p* : p*p* :: 1:2:1
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1)
Solution
Let the normal gene be represented by P, and the two mutations be represented by p# and p*.
Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the
gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes
obtained will be:
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1)
In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and
p*p* appear only once (1).
Since, both the mutations are recessive, three phenotypes would be obtained in the ratio:
mutant 1 : normal : mutant 2 :: 1:2:1
Three genotypes will be obtained in the ratio:
p#p# : p#p* : p*p* :: 1:2:1
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1).
Let the normal gene be represented by P, and the two mutations be re.pdf
1. Let the normal gene be represented by P, and the two mutations be represented by p# and p*.
Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the
gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes
obtained will be:
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1)
In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and
p*p* appear only once (1).
Since, both the mutations are recessive, three phenotypes would be obtained in the ratio:
mutant 1 : normal : mutant 2 :: 1:2:1
Three genotypes will be obtained in the ratio:
p#p# : p#p* : p*p* :: 1:2:1
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
2. (1)
Solution
Let the normal gene be represented by P, and the two mutations be represented by p# and p*.
Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the
gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes
obtained will be:
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
p#p*
(1)
p*p*
(1)
In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and
p*p* appear only once (1).
Since, both the mutations are recessive, three phenotypes would be obtained in the ratio:
mutant 1 : normal : mutant 2 :: 1:2:1
Three genotypes will be obtained in the ratio:
p#p# : p#p* : p*p* :: 1:2:1
p#
p*
p#
p#p#
(1)
p#p*
(1)
p*
