In the F<span>1 </span>generation of a Mendelian cross,
only the dominant trait is visible.only the recessive trait is visible.neither the dominant nor recessive trait is visible.<span>both the dominant and recessive traits are visible.
-I believe the correct answer is "ONLY THE DOMINANT TRAIT IS VISIBLE" in F1 generation, it is when the two true breeds, both homo (same genes) cross for example, HH and hh, since H will always be present in a punnet square, the answer is ONLY THE DOMINANT TRAIT IS VISIBLE. key word VISIBLE, the dominant trait is H</span>
Specifically, urease catalyzes the hydrolysis of urea to produce ammonia and carbamate, the carbamate produced is subsequently degraded by means of spontaneous hydrolysis to produce another molecule of ammonia and carbonic acid. [1] Urease activity tends to increase the pH of the medium in which it is due to the production of ammonia. It is produced by bacteria, fungi and several higher plants. Urease, functionally, belongs to the superfamily of amidohydrolases and phosphotriesterases. [2]
Answer:
Let the "barred" allele be caller B and the "non-white" allele b. Since chickens use chromosomes 
and 
to determine sex, hens would have chromosomes 
, and roosters would have chromosomes 
. A Z-linked gene is represented as a superscript on the chromosome, 
for the dominant allele and 
for the ressesive allele.
A barred hen would have a copy of B on its Z chromosome, a non-barred rooster would have both copies of b on both Z chromosomes. Using Punnet squares to represent the crosses we get the following cases:
That is a ratio of two barred heterozygote roosters to two non-barred hens. Crossing them we get:
That is a ratio of one barred heterozygote rooster to one barred hen to one non-barred rooster to one non barred hen.
Answer:
take the shape of the container it is in.
Explanation:
