Answer:
<h2>E. normal roosters & hens and bald roosters & hens.</h2>
Explanation:
The sex chromosomes of the chickens are represented by ZZ for male chicken that is rooster and ZW for the female chicken that is hen.
In the given question, the genotype of the bald rooster is zz and the genotype of the normal hen is ZW.
So the F1 progeny are, Zz that is normal male and the zW that is bald female.
So F1 × F1
Zz × zW
Thus, F2 progeny are Zz ( normal rooster), zz ( bald rooster), ZW ( normal hen), zW ( bald hen).
Answer:
sorry I don't know understand
Answer:
Explanation:
Answer: Cornfield
Well in a cornfield, its only corn and if your a hebirvore, its heaven there, but for canviroe's it doesnt work out to well, also corn field's are normally in a farmer's land, so meat eating animals dont really liketo go there, or they migth get shot
Answer:
25% of the offsprings will be BBCC
Explanation:
This is a typical dihybrid cross involving two distinct genes. One coding for fur colour and the other for claw sharpness. The allele for brown fur (B) is dominant over the allele for black fur (b) in the first gene while the allele for sharp claws (C) is dominant over the allele for dull claws (c) in the second gene.
In a cross between parents with genotypes: BbCc x BBCC , each parent will produce four possible allelic combinations of gametes as follows:
BbCc: BC, Bc, bC, bc
BBCC: BC, BC, BC, BC
Using these gametes in a punnet square (see attached image), 16 possible offsprings will be produced with four distinct genotypes:
BBCC (4)
BBCc (4)
BbCC (4)
BbCc (4)
According to the question, an offspring that is homozygous dominant for both traits will possess a genotype: BBCC
N.B: Homozygous dominant means contains same alleles for the dominant trait.
Hence, offsprings with genotype, BBCC, from this cross are expected to be 4 out of the 16 possible offsprings. Hence, the percentage is 4/16 × 100 = 25%.
False. Agribusiness is agriculture conducted on a large, commercial scale.