Answer:
The flowchart is not seen in your question. The labeling cannot be done without seeing the flowchart.
Here are the processes of bacterial transformation:
Explanation:
Bacterial transformation is defined as the change in the properties of bacteria which is caused by the introduction of foreign and naked DNA.
DNA is an hereditary material in organisms that contains their genetic information.
Here are the processes of bacterial transformation:
Step 1: Donor cell forms a Donor cell lyses
Step 2: Donor cell homologous binds to a receptor site on the recipient cell.
Step 3: One strand of donor cell DNA is degraded, and transformed DNA Pairs with homologous region on recipient cell.
Step 4: Finally, recombines with recipient cell chromosome
Regular physical activity can help prevent or manage many health conditions including high blood pressure, diabetes, depression, Osteoporosis, and Obesity. Im not sure if its your text book answer but i am a personal trainer and i gave you more than one benefit to choose from. Physical activity should be a way of life for everyone
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.
