1) making haploid cell for sexual reproduction
2)it's long story u can read miosis in text books im sure u will undrestand it if u want I can suggest u some books
3)befor meiosis in interphase
in bacterial cells dividing cell into two control the number of chromosomes in cell in if binary fission takes long time the chromosome of bacteria will replicate again and cell will contain 3 chromosomes from main chromosome but in eukaryotic cell there is inhibition after dna replication that avoid cell to do that
4)2 times
5)I couldn't understand your question
Answer:
3/4 axial flowers and 1/4 terminal flower
Explanation:
A cross between two true breeding with the axial flowers been dominant to the terminal flowers will produce first generation offspring with axial flowers that are not true breeding (heterozygous in genotype but still axial flowers in phenotype). In the F2 generation, which might be a cross between the the F2 generations will produce
3/4 axial flowers and 1/4 terminal flower
Astronomy is everything outside the earth and astrology is the positioning of the stars and planets that affects the way of events occur on earth
Answer: yes it did he figured that depending on which island the finches were on they would adapt to that island depending on resources for example on finch had a very large beak in order to crack nuts on the island because it was abundant on said island
Explanation:
Answer:
<h2>
Allele frequencies for B and b.
</h2><h2>
"b"(q) allele frequency = 0.60
</h2><h2>
</h2><h2>
"B"(p) allele frequency = 0.40
</h2><h2>
</h2><h2>
Genotype frequencies;
</h2><h2>
BB = 0.16
</h2><h2>
Bb = 0.48
</h2><h2>
Bb = 0.36
</h2>
Explanation:
Given
Non-baldness (B) is dominant on baldness(b), so B is dominant over b.
Homozygous pattern baldness male (bb) = 360,
Heterozygous non- baldness (Bb)= 480,
Homozygous non-baldness (BB)= 160.
So, we can also denote then by genotypes only,
BB= 160;
Bb= 480;
bb= 360;
Total= 1000
Allele frequency q² (bb) = 360/1000=0.36
allele fequency for q( b)= √o.36=0.60.
Allele frequency for p²(BB) = 160/1000=0.16
allele frequency p(B)= √0.16 = 0.4
Expected genotype frequencies;
BB = 160/1000 = 0.16
Bb = 480/1000= 0.48
Bb = 360/1000= 0.36
