<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>
Pythons are affecting everglades very negatively they are causing chaos in the food pyramid attacking all levels and not having predators prey on them pythons have become a huge problem that must be dealt with quickly because their affects are being felt quickly and very largely.
<span>pernicious form of anemia</span>
Answer:
I've responded to each scenario individually, please see below:
1. mutation that prevents ribosome binding to the mRNA
No transcription occurs
2. 5' UTR mutation that changes region 1 tryptophan codons into alanine codons
Transcription when alanine is low
3. mutation that creates a stop codon in region 1 of mRNA 5' UTR
Transcription occurs, a truncated protein will be made during translation
4. deletions in region 2 of the mRNA 5' UTR
Transcription occurs, region 1 and 2 might get paired together
5. deletions in region 3 of the mRNA 5' UTR
Transcription occurs, region 3 and 4 could combine to form a hairpin loop structure
6. deletions in region 4 of the mRNA 5' UTR
Transcription occurs, region 3 and 4 could combine to form a hairpin loop structure
7. deletion of the string of adenine after region 4 of the mRNA 5' UTR
Transcription occurs, however, it will not be as stable a molecule because of losing its polyA tail
Hope that answers the question, have a great day!
