Water is selectively permeable to an egg's cell membrane.
A mutation. Adaption and resistance - evolution, essentially. Those who survived the antibiotic developed a mutation which was able to resist the antibiotic and this became passed down to offspring.
Hope this helps!
<span>Osmosis, and Movement Across a Membrane ... of the diffusing particle, the more permeable the membrane </span>will<span> be; All </span>else<span> being equal, smaller particles </span>will diffuse<span> more ... How </span>Will Water<span> Move Across Semi-Permeable Membrane?</span>
Answer:
3.5%
Explanation:
The genes crossveinless and Stubble are linked and 7mu apart. That means that the frequency of recombination between them during meiosis will be 7%.
The alleles for crossveinless are:
- cv-c+ wildtype, dominant
- cv-c mutation, recessive
The alleles for Stubble are:
- Sb mutant, dominant
- Sb+ wildtype, recessive
A dihybrid female Drosophila with genotype cv-c Sb+/cv-c+Sb is testcrossed (crossed with a homozygous recessive male):
cv-c Sb+/cv-c+Sb X cv-c Sb+/ cv-c Sb+
<u>The male can only produce one type of gametes:</u> cv-c Sb+
<u>The female can produce 4 types of gametes:</u>
- cv-c Sb+ Parental, 46.5%
- cv-c+Sb Parental, 46.5%
- cv-c Sb Recombinant, 3.5%
- cv-c+Sb+ Recombinant, 3.5%
The frequency of recombination between cv-c and Sb is 7%, and 2 recombinant gametes are formed, so each of them will appear 3.5% of the times. The parental gametes will have a frequency of 100%-7%=93%, and there are 2 of them so each will have a frequency of 46.5%.
Only when the recombinant gamete cv-c+Sb+ joins the gamete generated by the male parent will the offspring be wild-type for both genes, so the proportion of phenotypically wild-type individuals in the progeny will be 3.5%.
These include the chemical. cellular. tissue organ. organ system and the organism level
