3. A farmer finds that some of his pot plants have spindly leaves and others hav
ID: 187234 • Letter: 3
Question
3. A farmer finds that some of his pot plants have spindly leaves and others have normal leaves. From among his plants, he chooses 6 pairs and mates them, with the following results:
1) spindly X normal: 56 spindly & 61 normal
2) spindly X spindly: 63 spindly & 0 normal
3 ) normal X normal : 0 spindly & 44 normal
4 ) spindly X normal: 59 spinldy & 0 normal
5) spindly X spindly: 122 spindly & 41 normal
Assuming that the spindly/normal phenotypes result from a single well-behaved Mendelian gene locus,
a) Which of the two phenotypes is dominant to the other?
b) Using the symbols sp and sp+ for the two traits, what are the genotypes of the parents of each cross?
c) For each of crosses 2, 4 and 5, how many of the spindly progeny would you expect to produce some normal progeny when self-fertilized? Briefly explain your answer.
Explanation / Answer
It is given that
1) spindly X normal: 56 spindly & 61 normal
2) spindly X spindly: 63 spindly & 0 normal
3 ) normal X normal : 0 spindly & 44 normal
4 ) spindly X normal: 59 spinldy & 0 normal
5) spindly X spindly: 122 spindly & 41 normal
a. Case shows the ratio of : that is standard mandelian ratio for law of independant assortment.
It shows that the spindly is the dominant(because it is a cros of two heterozygous spindle which is possible when atleast one allele is of spindle type.
Aa X Aa= AA, Aa, Aa, aa(Aa, AA are spindle type)
b. taking the sp and sp+
cross for 1. spindly X normal: 56 spindly & 61 normal (approx ratio is 1:1)
let say sp is dominant for spindly and sp+ is recessive for normal
spsp+ X sp+sp+
offspring: spsp+, spsp+, sp+sp+, sp+sp+
2. spindly X spindly: 63 spindly & 0 normal
spsp xspsp+
spsp+,spsp+, spsp,spsp
3. normal X normal : 0 spindly & 44 normal
sp+sap+ x sp+sp+
all resultant will be sp+sp+
4. spindly X normal: 59 spinldy & 0 normal
spsp xsp+sp+
All spsp+ which are spindly
5. spindly X spindly: 122 spindly & 41 normal
spsp+spsp+
spsp, spsp+, spsp+, sp+sp+
c. if normal progeny self fertilized
in case of 2
spsp+xspsp+
in this case 1/4th of the progeny will be normal
in case4,
spsp X spsp
no progeny will be normal
in case of 5
it is already self crossed so result will be same as above in case 5.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.