Question

A pure breeding tall plant with smooth seeds was crossed with a pure breeding short plant with wrinkled seeds. All the F1 plants were tall with smooth seeds. Two of these F1 plants were crossed and four different phenotypes were obtained in the 320 plants produced. How many tall plants with wrinkled seeds should you expect to find.  

Answer 1

As we can see when tall plant with smooth seeds were crossed with short plant with wrinkled seeds, all the F1 progeny were tall with smooth seeds which tells us that the tall and smooth seed characters are the dominant phenotypes. Hence let us consider allele of tall plant as 'T' and short plant allele as 't' and smooth seeds as 'S' and wrinkled seeds as 's'. Hence the genotype of the F1 progeny is TtSs.

A cross between two F1 progeny will result in the ratio as given below:

9/16 - Tall, smooth

3/16 - Tall, wrinkled

3/16 - short, smooth

1/16 - short, wrinkled

A punnett square cross is given in the table below:

	TS	Ts	tS	ts
TS	TTSS tall, smooth	TTSs tall, smooth	TtSS tall, smooth	TtSs tall, smooth
Ts	TTSs Tall, smooth	TTss tall, wrinkled	TtSs tall, smooth	Ttss tall, wrinkled
tS	TtSS tall, smooth	TtSs tall, smooth	ttSS short, smooth	ttSs short, smooth
ts	TtSs tall, smooth	Ttss tall, wrinkled	ttSs short, smooth	ttss short, wrinkled

So if the total number of population is 320 then the tall plants with wrinkled seeds will be in the 3/16 proportion.

Total tall plants with wrinkled seeds = 320*3/16 = 60 plants

Hence the expected number of plants with tall, wrinkled seeds phenotype will be 60 plants out of 320 plants.

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