Answer:
The disadvantages of the given instance are mentioned below.
Explanation:
This analysis seems to be a tool device used to evaluate the proportion of the population.
<u>Disadvantages:</u>
- The effectiveness is dependent on a significant portion of the number of people is being recorded.
- Animal marks may have a drastic impact on either the animals.
- This approach does indeed have a certain drawback because that's not a very suitable technique.
just ist five inanimate objects
Alright:
If you cross color stripes (A) and yellow stripes (Y) you use punned squares to figure out the rest.
In genetics, you can have 3 combinations. Aa, AA, and aa (or Yy, YY, and yy). If you cross each of these in a punned square with each combination, you have 9 different possibilities- each that have 4 possibilities inside of them, making 36 total possible outcomes.
Although, most of the 36 are repeating, so you can get one of two answers: dominant (if it has a capital letter) and recessive (no capital letters)
To answer your question, you will have the following:
-a grasshopper with both colors.
-a grasshopper with yellow.
-a grasshopper with neither.
-a grasshopper with both.
Answer:
The map distance between sn and ct is = 25 m.u or map unit.
Explanation:
Given, sn ct+ × sn+ ct
In F1 generation the progenies were interbred.
In F2 generation,
sn ct = 13 (recombinant)
sn ct+ = 36 (parental)
sn+ ct = 39 (parental)
sn+ ct+ = 12 (recombinant)
<em>Linkage Map distance = (no. of Recombinant progeny/total progeny) </em>
<em> × 100 </em>
<em>Recombination frequency = (no. of Recombinant progeny/total progeny) </em>
<em> × 100</em>
= ( 13+ 12) /100 × 100 m.u
= 25/100 × 100
= 25 %
∴ The distance between sn and ct = 25 m.u or map unit
∴ Linkage happened between sn and ct as the map distance is much less than 50
