A. The trait for the disease is recessive, and each offspring had a 50% chance of receiving the recessive allele and developing
the disease.
B. The trait for the disease is dominant, and each offspring had a 50% chance of receiving the dominant allele and developing the disease.
C. The trait for the disease is dominant, and all offspring had a 100% chance of receiving the dominant allele and developing the disease.
D. The trait for the disease is recessive, and all offspring had a 100% chance of receiving the recessive allele and developing the disease.
B. The trait for the disease is dominant, and each offspring had a 50% chance of receiving the dominant allele and developing the disease.
C. The trait for the disease is dominant, and all offspring had a 100% chance of receiving the dominant allele and developing the disease.
D. The trait for the disease is recessive, and all offspring had a 100% chance of receiving the recessive allele and developing the disease.
1 answer:
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Solution :
1. They can be reproduced on a human timescale --- renewable resources
2. They take millions of years to form -- non renewable resources
3. They are produced naturally -- renewable resources
4. They are extracted by mining -- non renewable resources
5. Their use as an energy source is increasing because they reduce pollution -- renewable resources.
a) Evolutionary fitness can be characterized as a person's commitment to the cutting edge's quality pool,which is chiefly founded on the aggregate or the genotype of the individual.This principally identifies with the conceptive achievement of an organic entity.
It is fundamentally determined by the offspring that can endure and pass on the characteristics/traits.
b) Graph is already uploaded in the below attached document.
c) The population 3 will have the most elevated frequency of heterozygotes,45% of the complete population.
d) Since population 2 is as of now having a low number of heterozygotes, the occasion would totally eliminate them from the population or decreases their number
Explanation:
Frequency of animals in population 1 having both alleles 1 and 3
frequency of allele 1 =0.60
frequency of allele 3= 0.25
Frequency of both allele 1 and 3 would be allele 1 + allele 3 frequency x 100
= 0.60+0.25 X 100= 85%
