Question

the leaves of pineapples may have three phenotypes: spiny, spiny tip, and piping (non spiny). Consider the following cross: p1) piping x spiny-tip f1) piping f2) 12/16 piping 3/16 spiny-tip 1/16 spiny

a. assign genotypes to the p1 and f1 generations (use letters A, B)

b. give two possible genotypes for f2 plants with piping leaves.

Answer 1

Let the spiney be S, spiney tip be ST and piping be P

we have a cross of spiney tip X piping that is ST x P

16 progenies has been obtained with the following phenotypoes

P-12 ST-3 S-1 from this we can infer that the spiney is a recessive character and piping is dominant.

also the ratio is 12:3:1 which is a modified version of 9:3:3:1 which suggests that the number of genes involved is 2 that is this ia a dihybrid. and there is a link between genes otherwise the two genes should have been independently assorted giving two hybrids with the dihybrid ratio 9:3:3:1

so consider two genes A and B with A,B notation as dominant allele and a,b as recessive allele

without doubt we can write spiney as a/a,b/b as it is the recessive

the other combinations would have been

A/A,B/B, A/A,B/b, A/a,B/B  and A/a,B/b usually dominant here piping

A/a,b/b ,, A/A,b/b and a/a,B,b a/a,B/B - former or later can be ST hybrids

Since nothing has been mentioned and no gene interactions has to be considered im goning with a/a,B,b a/a,B/B for ST so the other will be piping. we can go with the other also the end result will be same

so the genotypes corresponding to the phenotypes of F1 will be as follows

P- pining will be all with A dominat that is A/A,B/B, A/A,B/b, A/a,B/B A/a,B/b A/a,b/b ,, A/A,b/b

ST- will be a recessive B dominat that is  a/a,B,b a/a,B/B

and S recessive a/a,b/b

A)

So the parents were Px ST so we can assign any genotypes from above for our cause . I am going with a/a,B/b (ST) x A/a,B/b (P) so that all the combinations are possible

(P1 should have ideally A/a,B/b x A/a,B/b) for all combinations)

B)

For the cross two (F2) to be pipeting from the above data any genotype combination of P will be possibe. since the parents are not given im going with  A/a,B/b and A/a,B/B. a cross of spiney and ST will nor generate a piping. keep that in mind also.