Answer:
Transcription and translation occur simultaneously in prokaryotes.
Explanation:
Prokaryotes lack membrane-bound organelles and therefore, do not have a nucleus. Transcription and translation occur in the cytoplasm. As soon as the required length of mRNA is being formed, ribosomes join it and start the process of translation. Prokaryotic genes lack introns. The primary transcript formed by transcription in prokaryotes does not undergo splicing.
On the other hand, the process of transcription occurs in the nucleus in eukaryotes while translation occurs in the cytoplasm. The primary transcript formed by transcription in eukaryotes undergoes modifications to remove introns and to add a poly-A tail and 5' cap. Post-transcriptional modifications and spatial separation of two processes in eukaryotes result in slower translation than prokaryotes.
To convert factor I (fibrinogen) into a fibrin clot
Answer:
Kidneys : helps in forming urine
Ureter : They are connected to kidney and by peristaltic action transports urine from kidneys to bladder .
Urinary bladder : It is an hollow sac that holds the urine .
Urethra : transports urine from the bladder to outside of the body .
Explanation:
The human excretory system consists of : Kidney ,ureter ,urinary bladder and urethra .The figure is shown below .
Now , as far as functions are concerned we have :
Kidneys : They have excretory units called nephrons that further consist of :
- Bowman's capsule
- PCT (Proximal convoluted tubule )
- DCT (Distal convoluted tubule)
- CT (collecting tubule )
In nephrons urine is formed and via ureter it is transported to a hollow sac like structure called urinary bladder .
This bladder has muscles that cant remain contracted up to longer period )and then voluntary and involuntary it is excreted through urethra .
The process of passing urine is called Micturition .
Answer:
some plants have thorns some animals have spikes
Explanation:
Answer:
1/4
Explanation:
Given that the F2 cross produces offspring according to the expected 9:3:3:1 ratio.
YyRr X YyRr:
Let us break the dihybrid cross into individual monohybrid crosses.
Yy X Yy :
Y y
Y YY Yy
y Yy yy
2/4 or 1/2 of the offspring are heterozygous for the trait.
Similarly, in Rr X Rr cross, 1/2 of the offspring are heterozygous for the trait.
Together, 1/2 * 1/2 = 1/4 of F2 plants are expected to be heterozygous for both traits.
