The best answer to go with is c you’re welcome help it help
The probability of the offspring of a heterozygous father and homzygous mother having five fingers is 50%.
<h3>How to calculate genotype of a cross?</h3>
According to this question, a gene coding for the number of fingers in humans is involved. The allele for six fingers (F) is the dominant trait while the allele for five fingers (f) is the recessive trait.
If a cross between a heterozygous father that posseses a genotype of Ff and a homzygous mother that posseses a genotype of ff, the following offsprings will be produced:
This shows that the probability of the offspring of a heterozygous father and homzygous mother having five fingers is ½ (50%).
Learn more about genotype at: brainly.com/question/12116830
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You are likely to see a connotation used in both advertisement and also some university lectures. If only one answer is permitted, B is the best answer.
The answer is hypnotic do u have answer choices?
Answer:
The correct answer would be C) 75% large-toothed and 25% small-toothed.
It can be explained with the help of the law of dominance which states that the dominant allele expresses itself completely over the recessive allele in the heterozygous condition.
So, the genotype TT, as well as Tt, will result in the production of dominant trait or character in the offspring.
Thus, three (1 TT and 2 Tt) out of four offspring would have large teeth and only one offspring (tt) would have small teeth.
Hence, we can conclude that 75% of the offspring will have large teeth and only 25% will have small teeth.
