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%
The mass of one atom of carbon in the correct number of significant figures is 1.99×10-²³. There are four significant figures each in both the atomic mass of carbon (12.01 g) and Avogadro's number, so the answer must also have 4 significant figures. The operation is shown below:
<span>12.01 / (6.022x10²³) = 1.994x10</span>₋²³<u />
<span>I is dominant, i is recessive. The A's and B's are just show which allele I is. When there is just one dominant allele, it masks the recessive in blood typing. Remember IA and IB are codominant.
O is always ii
A is IAi (heterozygous) or IAIA (homozygous)
B is IBi (heterozygous) or IBIB (homozygous)
AB is always IAIB
Remember: You get one allele from each parent!
1. Father must be ii, mother must be ii, so all children must be ii.
2. Father is IAIA (the homozygous one), the mother is IBIB, so the only possibility for the children is IAIB, because you get one allele from the father and one from the mother.
3. Father is IAi, mother is IBi, so the children can be any of the blood types, because they can have all the combinations of genotypes.
4. Father is ii, mother is IAIB. Children can only be IAi or IBi.
5. Father is IAIB, mother is IAIB. Children can be IAIA, IBIB, or IAIB.
Example of Punnett square:
3. Father is type A, heterozygous, mother is type B, heterozygous
Father must be IAi (heterozygous)
Mother must be IBi (heterozygous)
_______IA ____ i
IB____ IBIA____IBi
i _____ IAi______ii
Sorry, that was difficult on here, hope it's understandable.
The father's alleles run across the top, the mother's are on the side, you follow to where they meet to find the possibilities for the children. IBIA (AB blood type), IBi (B), IAi (A), and ii (O) are the possibilities in this case.
Hope that helps!</span>
Increasing salinity<span> also increases the </span>density<span> of sea </span>water<span>. Less dense </span>water<span> floats on top of more dense </span>water<span>. Given two layers of </span>water<span> with the same </span>salinity<span>, the warmer </span>water<span> will float on top of the colder </span>water<span>. ... </span>Temperature<span> has a greater </span>effect<span> on the </span>density<span> of </span>water<span> than </span>salinity<span> does</span>
