Answer:
Personally, I believe the information helps.
Explanation:
Without this information the virus would not be contained, we would not have done research, taken samples, tested out the virus, or even attempted to contain it. With this information we can attempt to prevent a break out of the disease again. This is my personal opinion though.
Answer:
Probability that both cats described will have an offspring who is heterozygous for both traits is 50%, according to the scenario given.
Explanation:
When a cat amazing and bashful is crossed with a cat average and boring, it is a crossing where two traits are taken into account that will be segregated independently by the parents, and the result can be obtained by analyzing their genotypes and making the corresponding crossing in a Punnett square.
<u>Genotypes</u>:
- <u>Male cat</u> AABb
- <u>Female cat
</u> aabb
<u>Crossing</u>
:
Punnett's square (simplified)
<em>Alelles AB Ab</em>
<em>ab AaBb Aabb</em>
<em>ab AaBb Aabb</em>
<u>Offspring</u>:
- <em>AaBb amazing and bashful (heterozygous for two traits) 50%</em>
- <em>Aabb amazing and boring (heterozygous for trait A ans homozygous recessive for B) 50%</em>
<u><em>Probability that they will have an offspring who is heterozygous for both traits is 50%.</em></u>
B is the answer because it the ER helps with protein transport.
