count numbers of cells in brightfield image that have 3 proteins

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I have brightfield images of cells and floruoscent images as well (these fluoroscent channels shows the proteins the cells have in the.There are three channels-red, blue and green hence 3 proteins). . How can I count number of cells that have these proteins. The other issue I am facing is if you see the brghtfield image, the cells are very oddly shaped, have different levels of grey in them and some cross over the other, when I try to threshold them to binarize, some of them appear broken and I cannot use imfill directly( If I could use imfill, I would just intersect them and find overlapping area). Is there any solution to finding the number of cells that have the three proteins given these concerns?

Risposte (1)

Image Analyst
Image Analyst il 2 Dic 2020
Modificato: Image Analyst il 2 Dic 2020
Make your RGB image from the 3 channels
rgbImage = cat(3, redChannel, greenChannel, blueChannel);
then bring up the Color Thresholder on the Apps tab of the tool ribbon and threshold for purple/pink regions. Then tell it to Export the code.
To fill the blobs, try
mask = bwconvhull(mask, 'objects');
  2 Commenti
AAS
AAS il 2 Dic 2020
Thank you so much. I tried the conv hull for the brightfield image. This is what I get- Is there anything I can do about the circled regions?
IMAGE A: CONV HULL OF BRIGHT FIELD IMAGE:
IMAGE B: SIMPLE IMBINARIZE OF BRIGHT FIELD IMAGE:
IMAGE C: COLOR THRESHOLDED IMAGE FROM RGB IMAGE
IMAGE D: RGB IMAGE BEFORE THRESHOLDING
Basically If I can get Image A to work, I can probably do a Image A +Image C and count number of objects. Do you have any other suggestions?
Image Analyst
Image Analyst il 2 Dic 2020
Why does it matter? You're never going to be 100% accurate. Why don't you just measure the area fraction and be done with it? You're never going to get every single pixel correct with such a grainy image. I think the area fraction would correlate well with whatever you're doing, like how many bacteria get killed with your antiseptic or whatever.

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