B) short because it has both t alleles and no T alleles
The right answer is metaphase II.
The process is performed in two nuclear and cytoplasmic divisions, called first and second meiotic division or simply meiosis I and meiosis II. Both include prophase, metaphase, anaphase, and telophase. First division prophase is long and consists of 5 stages: leptotene, zygotene, pachytene, diplotene, and diakinesis. It is at this point that genetic recombination takes place at the level of chiasmus.
During meiosis I, the members of each homologous pair of chromosomes are paired during prophase, forming bivalents. During this phase, a protein structure, called synaptonemal complex form, allows recombination between homologous chromosomes. Subsequently, a large condensation of the bivalent chromosomes occurs and go to the metaphase plate during the first metaphase, resulting in the migration of n chromosomes to each of the poles during the first anaphase. This reduction division is responsible for maintaining the number of chromosomes characteristic of each species.
In meiosis II, as in mitosis, the sister chromatids comprising each chromosome are separated and distributed between the nuclei of the daughter cells. Between these two successive steps, there is no DNA replication. The maturation of the daughter cells will result in the gametes.
Half .
I think that’s right .
Answer:
ice /freezing fosilization results when the whole animal is preserved
1.This is a point mutation where thymine(T) has been replaced with adenine(A) in the DNA.
2.In the messenger RNA the codon will be changed Adenine(A) is replaced with uracil(U)
3.The changed codon will cause a different amino acid to be placed. Glutamic acid is replaced with valine. This replacement will cause the protein to be deformed and have an altered function.
