Question:
Auricularia is the larval form of:
Auricularia is the larval form of:
Show Hint
Echinoderms exhibit indirect development with distinct larval forms.
Updated On: Mar 11, 2025
- Sea stars
- Sea cucumbers
- Sea urchins
- Sea lilies
Hide Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Understanding Echinoderm Larvae
- Auricularia larva is found in Holothuroidea (Sea Cucumbers).
- Other Echinoderm larval forms include:
- Bipinnaria in Sea stars.
- Pluteus in Sea urchins.
- Doliolaria in Sea lilies.
- Auricularia larva is found in Holothuroidea (Sea Cucumbers).
- Other Echinoderm larval forms include:
- Bipinnaria in Sea stars.
- Pluteus in Sea urchins.
- Doliolaria in Sea lilies.
Was this answer helpful?
0
0
Top Questions on Zoology
Assertion (A): Some adult gastropods (e.g., snails) are asymmetrical. Reason (R): Torsion takes place during the development of gastropods.
Match the following:
Study the following and pick up the correct combinations:
Match the following:
- Symmetry of sea anemone is:
View More Questions
Questions Asked in TS EAMCET exam
- If \(\alpha, \beta, \gamma\) are the roots of the equation \(2x^3 - 3x^2 + 5x - 7 = 0\), then \(\Sigma \alpha^2 \beta^2\) is:
- TS EAMCET - 2024
- Quadratic Equations
- x and y are two complex numbers such that \(|x| = |y| = 1\). If \( \text{Arg}(x) = 2\alpha \), \( \text{Arg}(y) = 3\beta \) and \( \alpha + \beta = \frac{\pi}{36} \), then \( x^6 y^4 + \frac{1}{x^6 y^4} \) is:
- TS EAMCET - 2024
- complex numbers
If \( \sqrt{5} - i\sqrt{15} = r(\cos\theta + i\sin\theta), -\pi < \theta < \pi, \) then
\[ r^2(\sec\theta + 3\csc^2\theta) = \]
- TS EAMCET - 2024
- complex numbers
The system of simultaneous linear equations :
\[ \begin{array}{rcl} x - 2y + 3z &=& 4 \\ 2x + 3y + z &=& 6 \\ 3x + y - 2z &=& 7 \end{array} \]
- TS EAMCET - 2024
- System of Linear Equations
Calculate the determinant of the matrix:
- TS EAMCET - 2024
- matrix transformation
View More Questions