Answer:
The correct answer is - 3/16 or 18.75% portion will be resistant to both Bt and DDT.
Explanation:
It is given that bollworms that are susceptible to Bt are represented by either RR or Rr and resistance one are shown by rr genotype and bollworms that are resistant to DDT are represent TT or Tt genotype and susceptible to DDT have tt genotype.
Cross: RR tt with rr TT
Gametes: Rt and rT
F1 cross:
RT
rT RrTt ( susceptible Bt and resistant DDT)
F1 Cross with F1
Gametes: RT, Rt, rT and rt
cross:
RT Rt rT rt
RT RRTT RRTt RrTT RrTt
Rt RRTt RRtt RrTt RrTt
rT RrTT RrTt rrTt rrTt
rt RrTt Rrtt rrTt rrtt
The bold genotypes are the three out of 16 that are resistant to both DDT and Bt.
Choice C. Educated guess or a prediction that can be tested, based on limited evidence.
The bacteria was not killed off. In those three days symptoms had stop because the bacteria was dormant. They must have been activated because Jim provided a suitable environment for their growth
Answer:
This question lacks options, options are:
a) indirect
b) mosaic
c) determinative
d) regulative
e) direct.
The correct answer is d.
Explanation:
The egg (ovum) like the embryo during the first embryonic divisions, do not possess any sign of polarity. This development mechanism can be called regulative development(in contrast to the mosaic model) since the fate of the cells that originate is not fixed and can be modified during development. In regulative embryos, part of the embryo can be removed and the remaining cells can compensate for the loss and give a complete individual as the final product.The strongest evidence that continues to support this regulatory model is based on the plasticity or potential that mammalian cells possess before implantation. It is known that the blastomeres produced by the first divisions can be replaced with each other or even can be eliminated without apparently altering the embryonic development and therefore they are equivalent (without polarity).
Answer:
The blank is:
C1 through T1 and L1 through S4
Hope this Helps
