In a breeding program a cross was made between two true breeding cow-pea types: one with round seeds and the other with wrinkled seed. If roundness is dominant why wrinkleness is recessive, in the first filial generation all the seeds produced will be
25% wrinkled, 75% round
100% wrinkled
50% wrinkled, 50% round
75% wrinkled, 25% round
100% round
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Discussions (34)
D questn says d@ d cross was made btwn 2 TRUE BREEDING COWPEA,whr R is dominant nd r is recessive..i.e RR * rr = Rr Rr Rr Rr.which means d ans is E(100%)
Please guys.. listen to justice.. This question clearly states the cow peas are true breeding, not carriers.. So it RR and rr. Therefore first filial g. will be 100% rounded
Lets asume R stands for dominant roundnes and r 4 recesive wrinklnes. Rr + Rr=RR Rr Rr and rr(d 3 wit capital letter R shows dominance while d smal letter shws recesvnes. Hence in %,75% are dominant/round while 25% is recesive/wrinkld
when we have RR as d dominant and rr as d ressessive den after whc d allili will form as a result of meiosis having R R * r r den we cross after crossing d f1 phenotype are all heterozygous round Rr. that z option (E) is correct 100%
The answer taken is very wrong. The key word here is (True). The question says 2 true breeding; round(R) and wrinkle(r). True breeding of round(RR) whle true breeding of wrinkle(rr). meanwhle, crossing it gives heterozygous all through Tt Tt Tt Tt. 100%%%%
the questn said "true breeding"which means Homozygous and the round cowpea is dorminant over wrinkled recessive cowpea ryt?lets assume dat 'R' stand 4 round and 'r' 4 wrinkled.since round is dorminant, it has 2 b written wit capital letter.therefore parents generatn=(RR) x (rr),parent gametes=(R)(R) x (r)(r),first filial generatn=(Rr) x (Rr).xo since R representn round is dorminant,al f1 generatn should b round thus 100%.Note,d cross was made btw 2 true breedn cowpeas which stands 4 homozygous.Note again dat d questn requires f1 nd nt f2 xo plz dnt confuz ursef b/coz of d ansa given coz it's wrong.tnkz.
The answer is E,R is dominant and r recesive if we cross breed RR and rr ,we get Rr,Rr,Rr,Rr.The answer is therfore 100% cause R is dominant over r
it's clearly stated in the question . True breeding means it's homozygous. since round is dominant while wrinkled is recessive. The alleles will be RR and rr.
| R | R |
| | |
r| Rr | Rr |
| | |
r | Rr | Rr |
| | |
100% round
Pls I have never understood this crossbreeding someone should pls explain in details tanks
the supposed answer is incorrect because if the first filial generation Rr and Rr cross, you are meant to have 75% round and 25% wrinkled. please check and confirm it
Again a stock of round-seed pea plants dat always produces round-seed offspring when crossed is said 2 b "true breedn"it has two genes 4 roundness.these round-seed pea plant is said 2 form a pure stock.xo d ansa is wrong.
the problem here is. How do we know if the dominant round cowpea is heterozygous or homozygous
