Answer: The estimated population is 1250 mice
Explanation: The method use was marked and recapture, in which individuals are marked in the first capture and after some time biologist trap a new group of individuals that can be or not marked
Whit this data is possible to estimate the size of a population applying the Peterson method but is important to make some considerations such as:
1. All indivuals have the same probability to be capture
2, The population remain constant in terms of birth and death rate.
Taking this into account, the formula that allows to determine the size of the population is:
N=CM / R
where N is the size of the population, C is number of indivuals trapped in recapture, M is number of individuals marked in the first capture and R is the number of marked animals trapped in recapture
In this case:
N = 250 * 200 / 40 = 1250
Answer:
b. The enzyme and substrate would be stuck together.
Explanation:
Enzymes are proteins whose active site binds to specific chemical reactants (i.e., substrates), thereby forming a complex that is similar to the interaction between a lock and its key. This active complex lowers the energy of the reaction and promotes a conformational change in the substrate to break down it into multiple products. When the enzyme contains mutations in its active site, the ability to bind the substrate is altered. In this case, the enzymatic reaction can't occur because the interaction enzyme-substrate doesn't produce an active complex.
Answer:
B. Traditional view
Explanation:
the traditional view is that nothing positive results from a conflict
Answer:
36
Explanation:
A two-point test-cross is a cross between an individual with a double heterozygote genotype and a homo-zygous recessive individual in order to determine the recombination frequency between two linked genes. In genetics, one map unit (m.u.) can be defined as the measure of the distance (i.e., genetic distance instead of physical distance) between genes for which one (1) product of meiosis in one hundred (100) is recombinant. In this case, 36 of the offspring have the recombinant phenotype, while the remaining 64 offspring are not recombinant, and therefore both genes are separated by 36 mu (64 + 36 = 100 >> 36 mu).
