Answer:
1. 2.8%
2. 97.1%
Explanation:
- If the birth rate of this disorder is 1 in every 5000 births it means that the frequency of the recessive homozygous genotype is 1/5000=0.0002
If we name the gene for the disorder with A, than the genotypes are aa (homozugous recessive), Aa (heterozygous, carrier), AA (dominant homozygous).
According to the Hardy-Weinberg equation:
frequency of AA is p2 (P)
frequency of Aa is 2pq
frequency of aa is q2 (Q) and
p2+2pq+q2=1 (P+Q=1)
Q=0.0002 P=1-0.0002=0.9998
q=
0.0002=0.014
p=
0.9998=0.999
2pq=0.028=2.8%
2. According to the Hardy-Weinberg equation:
frequency of AA is p2 (P)
frequency of Aa is 2pq
frequency of aa is q2 (Q) and
p2+2pq+q2=1 (P+Q=1)
The frequency of the recessive homozygous genotype (Q) is 1/5000=0.0002
Q=0.0002
q2=0.0002 2pq=0.028 p2=?
p2= 1-0.0002-0.028 =0.9718=97.1%
Answer:
If the tasmanian devil has XY, then it is a boy. If the tasmanian devil has XX, then it will be a female
Explanation:
Just like other mammals, the tasmanian devil is a marsupial, that possess 14 chromosomes including the sex chromosomes with the male being heterogametic (XY) and the female producing two similar chromosome (XX). Thus, if the tasmanian devil has XY, then it is a boy. If the tasmanian devil has XX, then it will be a female, just like in the human normal chromosomes.
This cell division is called mitosis, and it takes place during the last four phases of the cell life cycle: Prophase,Metaphase, Anaphase, and Telophase. The purpose of mitosis is to copy the genetic material so that whenTelophase has ended, the result is two genetically identical cells.
The correct answer is A. R and S
Hope this helps