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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Answer:
option B is correct
Explanation:
the effect of external physical factor on cell division is clearly seen on in density dependent inhibition( a phenomenon in which crowded cell stop dividing)
A) Deoxyribose + phosphate group + thymine
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If this is the picture you are talking about, the right answer is non-disjunction
Non-disjunction is the non-separation of homologous chromosomes at the time of cell division which results in the formation of ova or spermatozoa leading to an abnormality in the number of chromosomes of the egg thus fertilized by the spermatozoa.
The fertilized egg consists of either a single chromosome, what is called monosomy, or, conversely, three chromosomes, this is called trisomy. While naturally, the fertilized egg has a single pair of chromosomes.
