To be honest im glad its there cause it is helpful
Answer:
1 . The stage on the first meiotic division when the homologous chromosomes move to opposite poles but the sister chromatids remain together
: b. Anaphase I
2 . The stage in the second meiotic division where sister chromatids migrate to opposite poles
: c. Anaphase II
3 . A structure on the chromosome that holds a pair of chromatids together during replication
: f. centromere
4 . A double-stranded chromosome following replication attached by a centromere
: d. chromatid
5 . A condition where non-sister chromatid of homologous chromosomes exchange genes
: e. crossing over
6 . The stage in the first meiotic division where the homologous chromosomes line up as a pair
: a. Metaphase I
7 . The stage in the second meiotic division where the chromatid pair lines up at the equator of the cell: g. Metaphase II
Explanation:
DNA replication occurs during the S phase of the interphase of the cell cycle. The replicated DNA molecules are accommodated in two sister chromatids of a chromosome that are held together by a centromere.
During prophase I, the chromatids of a homologous chromosome pair exchange a genetic segment. This process is called crossing over. It generates recombinant chromatids with new combinations of genes.
Metaphase I of meiosis I includes the alignment of homologous pairs of chromosomes at the cell's equator. This is followed by separation and movement of homologous chromosomes to the opposite poles of the cell during anaphase I.
Metaphase II of meiosis II includes the alignment of individual chromosomes, each with two sister chromatids, on the cell's equator. During anaphase II, splitting centromere separates the sister chromatids which then move to the opposite poles of the cell.
Answer:
Explanation:
Bacterial count in stock- 1.85x10^6 cfu/ml
Dilution methods
Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.
Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.
Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.
Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml
So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.
If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.
Answer:
Land Insecurity
Very Poor Living Conditions
Unemployment
Crime.
To name a few
Answer:
All the organisms of one kind in an ecosystem are called a population. For example, a pond ecosystem might have populations of frogs, waterlilies, insects, duckweed, and protists. Populations living and interacting with each other form a community.
Explanation:
