A researcher is studying two populations of mice-one population where all genotype d/d and
another population where all mice have the genotype D/d. In this scenario, the d/d mice would be
considered and the DVd mice are O heterozygous; homozygous dominant O monohybrids;
heterozygous homozygous recessive; hybrids homozygous dominant; homozygous recessive
heterozygous; homozygous recessive
Solution
Hi,
Because mitosis is occurred in somatic cells of eukaryotes.
3. Answer of third question is(b) a variation in the neuclieotide sequence that may or may not
result in detectable phenotype.
Because for example in case of heterozygous dominant alles result may not same as homozygous
alleles.
4.answer of this question is(a)viruse.
Because virus is known as connective link between living and non living things.
I hope my answer is satisfactory.
Thanks!!.
A researcher is studying two populations of mice-one population where.pdf
1. A researcher is studying two populations of mice-one population where all genotype d/d and
another population where all mice have the genotype D/d. In this scenario, the d/d mice would be
considered and the DVd mice are O heterozygous; homozygous dominant O monohybrids;
heterozygous homozygous recessive; hybrids homozygous dominant; homozygous recessive
heterozygous; homozygous recessive
Solution
Hi,
Because mitosis is occurred in somatic cells of eukaryotes.
3. Answer of third question is(b) a variation in the neuclieotide sequence that may or may not
result in detectable phenotype.
Because for example in case of heterozygous dominant alles result may not same as homozygous
alleles.
4.answer of this question is(a)viruse.
Because virus is known as connective link between living and non living things.
I hope my answer is satisfactory.
Thanks!!
