Why do cells undergoing mitosis require one set of divisions, whereas those undergoing meiosis require two sets of divisions? wh
at is the end result of meiosis?
1 answer:
Mitosis, you see, is somatic (body) cell division, and the DNA in the nucleus during Interphase can replicate by doubling the chromosomes before the actual mitosis or cell division process of splitting into TWO identical cells (and they function the same with the same number of identocal chromosomes too! So they are called Diploid cells with 46 chromosomes!)
Meiosis is sperm/egg division and must divide twice to produce FOUR sperm or egg cells. Sperm goes through I- PMAT just like mitosis, but goes through PMAT again after that! They create four Haploid gametes (sex cells) with 23 chromosomes so that when an egg and a sperm unite in sexual reproduction, the zygote (fertilized egg) will have exactly 46 chromosomes like the normal number of chromosomes in a somatic cell. ^^
Meiosis is sperm/egg division and must divide twice to produce FOUR sperm or egg cells. Sperm goes through I- PMAT just like mitosis, but goes through PMAT again after that! They create four Haploid gametes (sex cells) with 23 chromosomes so that when an egg and a sperm unite in sexual reproduction, the zygote (fertilized egg) will have exactly 46 chromosomes like the normal number of chromosomes in a somatic cell. ^^
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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)
