Answer:
D
Explanation:
because it shows the features of a bird
Identifying the genetic sex of a child is based on finding intracellular Barr bodies that consist of inactive chromatin material.
Inactive chromatin material is the one where no transcription takes place. It is also known by the name heterochromatin. It appears as a dark condensed form in the chromatin.
Barr bodies are the inactive X chromosomes. These are mad inactive by a process termed as lyonization. It is essential to make the chromosome inactive in organisms with XY type of sex determination. They are present at the periphery of the nucleus. Inactivation of X chromosome makes the amount of X chromosomes equal in both, males and females.
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Decomposers break down the remains of producers and consumers.
Answer:
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
Explanation:
Let the green color of the seed be depicted by "G" and the yellow color of the seed be depicted by "g"
Let the smooth the seed be depicted by "R" and wrinkled seed be depicted by "r"
F1 cross -
true breeding smooth green plant ( RRGG) and true breeding wrinkled yellow (rrgg)
F1 gamete will be RG, RG, rg, rg
F1 offspring will be RrGg , Thus all F1 offspring will be heterozygous smooth and yellow.
Thus, R is dominant over r and g is dominant over G
F2 Generation –
RrGg x RrGg
Gametes will be RG, Rg, rG, rg
RG Rg rG rg
RG RRGG RRGg RrGG RrGg
Rg RRGg RRgg RrGg Rrgg
rG RrGG RrGg rrGG rrGg
rg RrGg Rrgg rrGg rrgg
R is dominant over r and g is dominant over G
Genotypes are –
RRGG - 1 (Smooth Green)
RRGg-2 (Smooth yellow)
RrGG-2 (Smooth Green)
RrGg-4 (Smooth yellow)
RRgg- 1 (smooth yellow)
Rrgg – 2 (Smooth yellow)
rrGG – 1 (wrinkled Green)
rrGg – 2 (Wrinkled yellow)
rrgg – 1 (wrinkled yellow)
Yellow smooth - 9
Yellow wrinkle - 3
Green smooth- 3
Green Wrinkle - 1
