Answer:
25% of the heterozygous cross are short, and the offspring of a homozygous dominant and homozygous recessive pea plant will always display the dominant trait (phenotype), because they are heterozygous.
Explanation:
In this explanation, I'm assuming that the allele "T" for tall plants is dominant to the allele "t" for short plants, like in Gregor Mendel's pea plant experiment.
A homozygous tall pea plant will have the genotype "TT" and a homozygous short plant will have the genotype "tt" because homozygous means that both alleles are identical. Since "T" is dominant over "t", any plant with at least one "T" allele will be tall (the dominant trait), regardless of what the other allele is. Let's look at a Punnett square for this cross:
Explanation:
Answer:
The fraction of the offspring that will show the phenotype produced by the recessive c allele is 1024/4096 = 0.25 A-BbccDdE-Ff = 25%
Explanation:
You can calculate the fraction of the offspring that will show the phenotype produced by the recessive c allele by making the punnet square for each gene and then multipling the phenotypic proportions, like this:
Cross) AAbbCcDDEeff x AaBBCcddEEFF
Cross For each gene by separately:
Gametes) A A A a
F1) 2/4 AA
2/4 Aa
Gametes) b b B b
F1) 4/4 Bb
Gametes) C c C c
F1) 1/4 CC
2/4 Cc
1/4 cc
Gametes) D D d d
F1) 4/4 Dd
Gametes) E e E E
F1) 2/4 Ee
2/4 EE
Gametes) f f F F
F1) 4/4 Ff
So, fraction of the offspring that will show the phenotype produced by the recessive c allele is:
4/4 A- x 4/4 Bb x 1/4 cc x 4/4 Dd x 4/4 E- x 4/4 Ff =
1024/4096 = 0.25 A-BbccDdE-Ff =25%
Among the following solid fuels given in the question, coke has the highest heating value. The correct option among all the options that are given in the question is the last option or option "D". The heating value of coke is calculated to be between 28000 kJ/Kg to 31000 kJ/kg. I hope the answer helps you.
It breaks down lactose into the galactose and glucose. So the answer here is C.
