Answer:
Alleles for feather colour exhibit incomplete dominance or co-dominance.
50% gray offspring + 50% black offspring
Explanation:
<em>It means that the alleles for feather colour in the hen exhibit incomplete dominance or co-dominance over one another.</em>
Assuming the allele for white colour is B, white colour will be b while the heterozygote Bb gives the gray phenotype.
Gray rooster + gray hen = 15 gray chicks, 6 black chicks and 8 white chicks.
15:6:8 is roughly 2:1:1 which is phenotypic ratio obtainable from crossing two heterozygous individuals as pointed out by Mendel.
Bb x Bb = 1BB, 2Bb, and 1bb
Crossing the gray rooster (Bb) with a black hen (bb):
Bb x bb = Bb, Bb, bb, and bb
= 2Bb (gray):2bb (black)
50% of the offspring will be gray while the remaining 50% will be black.
The correct term is histological section.
<h3>What are histological sections?</h3>
Histology has to do with the study of tissues. Sectioning has to do with cutting very thin slices of objects.
Thus, histological sectioning is an act of cutting very thin slices of tissues for the purpose of mounting such on microscope slides. To enhance views and differentiate different parts of cells, sections of tissues are often stained.
More on histological sections can be found here: brainly.com/question/15603731
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Answer:
Neither the Wolffian ducts nor the Mullerian ducts will develop
Explanation:
SRY gene is the gene is located on the Y chromosome and it is responsible for the development of testes.
MIS is Anti-Müllerian hormone which has the role to prevent the development of Müllerian ducts (they develop into female internal genitalia).
Androgen insensitivity syndrome is a disorder characterized by unfunctional androgen receptors. If androgen receptors are not working properly, the Wolffian ducts (they develop male internal organs) will not develop.
So, with MIS and withandrogen insensitivity syndrome, this fetus will not develop neither the Wolffian ducts nor the Mullerian ducts.
C) to lessen his/her personal introduction of atomspheric CO2
