1.A well organized lab report should begin with the experimental procedure.

Biology

1 answer:

Yakvenalex [24]3 years ago

8 0

False
D)
true
A)
A)
false
false
C)
B)
D)
I'm pretty sure my answers are correct. Hope this helps! :)

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two pure breeding lines of pea, one tall and hairy, the other short and smooth, are crossed. the f1 hybrid is tall and hairy. th

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The genotypes A B_, A bb, aaB_, and aabb are matched to the four phenotypic groups.

<h3>How do phenotypic classifications work?</h3>

The genotypes A B_, A bb, aaB_, and aabb are matched to the four phenotypic groups. If any of the single homozygous recessive genotypes (i.e., A bb or aaB_) exhibits the same phenotype as the double homozygous recessive (aabb), then a 9:3:4 phenotypic ratio will result.

<h3>What does phenotypic classification entail?</h3>

phenotype, all of an organism's observable traits that emerge from the interaction of its genotype (total genetic inheritance) and environment. Observable features include things like behavior, biological characteristics, colour, shape, and size.

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Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to tas

dexar [7]

Complete question:

Suppose "A" is a dominant gene for the ability to taste phenylthiocarbamide and "a" is a recessive gene for the inability to taste it. Which couples could possibly have both a child who tastes it and a child who does not?

a. father AA, mother aa

b. father Aa, mother AA

c. father Aa, mother Aa

d. father AA, mother AA

Answer:

c. father Aa, mother Aa

Explanation:

According to the given information, the ability to taste phenylthiocarbamide is a dominant trait and is imparted by the allele "A". This phenotype would be expressed in both homozygous and heterozygous conditions. The non-taster phenotype would be expressed in the homozygous recessive genotypes only.

To have both taster and non-taster children, both the parents should have at least one copy of the recessive allele. Among the given options, the father with genotype Aa and the mother with genotype Aa have the possibility to have both taster and non-taster children.

Aa x Aa= 3/4 taster (1/4 AA and 1/2 Aa): 1/4 non-taster (1/4 aa)