Question

In: Biology

3. A farmer finds that some of his pot plants have spindly leaves and others have...

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.

Solutions

Expert Solution

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.

gladiator answered 1 year ago

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