Answer:
The genotype for each of the parents must be
parent 1 : Gg
parent 2 : Gg
Explanation:
Please note that a dominant trait is a trait that is expressed phenotypically in a heterozygous state, while a recessive trait is a trait that can only be expressed in a homozygous state.
Now, since gray face (G) for Oompa Loompas is dominant, and orange face (g) is recessive, for an offspring to be orange faced, it means that the genotype of the offspring must be 'gg'. Also, since both parent contribute an allele in the pair of alleles in the offspring, both parents must have the recessive (g) in their genotype. Moreover, we are told that both parents are gray-faced, meaning that their genotypes were 'Gg' and 'Gg'. To confirm, let me do the cross
G g
G GG Gg
g Gg gg
from the cross above, we find out that out of 4 offspring, 3 were gray face (GG, Gg ) while one was orange face (gg).
Prophase. Mitosis begins with prophase, during which chromosomes recruit condensin and begin to undergo a condensation process that will continue until metaphase
Prometaphase
Metaphase
Anaphase
Telophase and Cytokinesis.
<h2>Koch's Postulates.</h2>
Explanation:
The next step should be isolate the microorganism from the mice and check if it is identical to the original microbe.
According to Koch's postulates, a microbe can be considered as a causative agent of a disease only after all the following are established:
- The microorganism should be present in all the organism suffering from the disease.
- The isolated microorganism should be grown in pure culture.
- The organism grown in the culture should be reintroduced in the susceptible but healthy host.
- Finally the microorganism must be re-isolated from the experimental host an cultured to see whether it is identical to the original specific microorganism.
Answer:
The order must be K2→K1, since the permanently active K1 allele (K1a) is able to propagate the signal onward even when its upstream activator K2 is inactive (K2i). The reverse order would have resulted in a failure to signal (K1a→K2i), since the permanently active K1a kinase would be attempting to activate a dead K2i kinase.
Explanation:
- You characterize a double mutant cell that contains K2 with type I mutation and K1 with type II
mutation.
- You observe that the response is seen even when no extracellular signal is provided.
- In the normal pathway, i f K1 activat es K2, we expect t his combinat ion of two m utants to show no response with or without ext racell ular signal. This is because no matt er how active K1 i s, it would be unable to act ivate a mutant K2 that i s an activit y defi cient. If we reverse the order, K2 activating K1, the above observati on is valid. Therefore, in the normal signaling pathway, K2 activates K1.
Answer:
Polytene chromosomes may be defined as the giant chromosomes that contain alternate dark and light bands when view under the microscope. These chromosomes are found in the salivary gland of Drosophila.
These glands are functionally highly active and contain thousand of DNA strands. These chromosomes are aligned parallel with each other as the replication is normal but the cells are unable to separate and fails to undergo the process of cytokinesis.
