Answer:
I would say B.
Explanation:
A would only fix biodiversity temporarily, until humans come and wipe them out again. C is just kind of impossible, and containing them could also be problematic. D is also unreasonable, and some animals would not be able to thrive in new ecosystems. B would help save the environment, which would save the animals, which would help biodiversity.
Answer:
1/1024 is the proportion of the F2 genotypes will be recessive for all five loci
Explanation:
When crossing between both parents, all genotypes will give us 100% AaBcCcDdEe. When a self-fertilization is performed this means that it can have a cross for example of two flowers of the same plant with the genotype AaBcCcDdEe In this way, given the law of independent segregation which states that the alleles of two or more different genes are distributed in the gametes independently of each other. The proportion that at this junction the alleles are aa (1/4) bb (1/4) cc (1/4) dd (1/4) and ee (1/4). The proportions (1/4)* (1/4)*(1/4)*(1/4)*(1/4) are multiplied, obtaining a value of 1/1024
1. four, bond
2. Cohesion
3. Adhesion is correct ... looked in my bio book sorry
Answer:
The correct answer is peroxisome
Explanation:
Answer:
Each mutant would be mated to wild type and to every other mutant to create diploid strains. The diploids would be assayed for growth at permissive and restrictive temperature. Diploids formed by mating a mutant to a wild type that can grow at restrictive temperatures identify the mutation as recessive. Only recessive mutations can be studied using complementation analysis. Diploids formed by mating two recessive mutants identify mutations in the same gene if the diploid cannot grow at restrictive temperature (non-complementation), and they identify mutations in different genes if the diploids can grow at restrictive temperature (complementation).
Explanation:
Recessive mutations are those whose phenotypic effects are only visible in homo-zygous individuals. Moreover, a complementation test is a genetic technique used to determine if two different mutations associated with a phenotype colocalize in the same <em>locus</em> (i.e., they are alleles of the same gene) or affect two different <em>loci</em>. In diploid (2n) organisms, this test is performed by crossing two homo-zygous recessive mutants and then observing whether offspring have the wild-type phenotype. When two different recessive mutations localize in different <em>loci</em>, they can be considered as 'complementary' since the heterozygote condition may rescue the function lost in homo-zygous recessive mutants. In consequence, when two recessive mutations are combined in the same genetic background (i.e., in the same individual) and they produce the same phenotype, it is possible to determine that both mutations are alleles of the same gene/<em>locus</em>.
