Scientific Law, A rule of nature.
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>
A larger overall leave size is the answer.
NOTE: Marking this answer as brainliest would be much appreciated! :)
Answer:
Anaphase is the right answer
Explanation: