The answer is 0 percent.
If we represent dominant trait (wrinkled seeds) with A and recessive trait (non-wrinkled seeds) with a, then genotypes are AA (dominant homozygous), Aa (heterozygous), and aa (recessive homozygous). The phenotype of wrinkled seed is determined by genotypes either Aa or AA. Since the cross of a plant with wrinkled seed with a plant with non-wrinkled sees gives only wrinkled-seeded offspring, the plant with wrinkled seeds must be dominant heterozygous.
Let's look at both examples.
If wrinked-seeded parent contains one non-wrinkled allele, it must be heterozygous (Aa):
Parents: Aa x aa
Offspring: Aa Aa aa aa
So, the 50% of offspring has wrinkled seeds and 50% has <span>non-wrinkled seeds.
If </span><span>the wrinkled-seeded parent does not contain any non-wrinkled allele:
</span>Parents: AA x aa
Offspring: Aa<span> Aa Aa Aa
</span><span>
Thus, al</span><span>l of the offspring are heterozygous and all of them have wrinkled seeds.</span>
Answer:
anaphase ll
Explanation:
anaphase ll
<em><u>BRAINILIEST</u></em><em><u> </u></em><em><u>PLEASE</u></em>
Answer:
C
Water moves out or in to balance cell concentration
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