Answer:
C.
Explanation: It lasts until all the codons are read,so it encounters a stop codon.
Complete question:
Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited.
A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time.
Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).
Answer:
The probability that the second child will also have the disease is 1/16.
Explanation:
<u>Available data:</u>
- Two genes independently inherited: one for blood type, the other for disease
- Man with type A blood x Woman with type B blood
- Both parents are heterozygous for the gene that causes the disease; Dd
If the man has A blood, and the woman has B blood, and they already have an affected child, this means that they must be heterozygous for blood type too.
Cross:
Parentals) AiDd x BiDd
Gametes) AD Ad iD id BD Bd iD id
Punnett square) AD Ad iD id
BD ABDD ABDd BiDD BiDd
Bd ABDd ABdd BiDd Bidd
iD AiDD AiDd iiDD iiDd
id AiDd Aidd iiDd iidd
F1) <u>Genotype</u>:
1/16 ABDD
2/16 ABDd
1/16 ABdd
1/16 AiDD
1/16 BiDD
2/16 AiDd
2/16 BiDd
1/16 Aidd
1/16 Bidd
1/16 iiDD
2/16 iiDd
1/16 iidd
<u>Phenotype:</u>
3/16 A/B normal
4/16 A normal
4/16 B normal
3/16 0 normal
1/16 0 affected by the disease.
I would say D because mechanical weathering is the breaking up of rocks by physical forces
<span>In the duct system the atop testis where sperm mature from is known as the epididymis</span>
