Biology 198
PRINCIPLES OF BIOLOGY
Answers to Hardy-Weinberg practice questions
There are two types of percent concentration: percent by mass and percent by volume. PERCENT BY MASS. Percent by mass (m/m) is the mass of solute divided by the total mass of the solution, multiplied by 100%. Percent by mass = #'mass of solute'/'total mass of solution'# × 100% Example. What is the percent by mass of a solution that contains 26.5 g of glucose in 500 g of solution? Overall postsecondary enrollments decreased 0.5 percent or 83,803 students from spring 2019 (Table 1). The public sector enrollments (two- and four-year colleges combined), which enrolled nearly three-quarters of all postsecondary students, fell by 1.3 percent (163,964 students), compared to 1.9 percent (244,376 students) reported last year.
Updated: 21 August 2000
POPULATION GENETICS AND THEHARDY-WEINBERG LAW
ANSWERS TO SAMPLE QUESTIONS
Remember the basic formulas:
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
- PROBLEM #1.You have sampled a population in which you know that the percentage of the homozygousrecessive genotype (aa) is 36%. Using that 36%, calculate the following:
- The frequency of the 'aa' genotype. Answer: 36%, asgiven in the problem itself.
- The frequency of the 'a' allele. Answer: The frequency ofaa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36,then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequencyis 60%.
- The frequency of the 'A' allele. Answer:Since q = 0.6, andp + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
- The frequencies of the genotypes 'AA' and 'Aa.' Answer:The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So,using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 =0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
- The frequencies of the two possible phenotypes if 'A' is completely dominant over 'a.' Answers: Because 'A' is totally dominate over 'a', thedominant phenotype will show if either the homozygous 'AA' or heterozygous 'Aa' genotypesoccur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, thefrequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and therecessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and,in the first part of this question above, you have already shown that the recessive frequency is36%.
- PROBLEM #2.Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) havenormal blood cells that are easily infected with the malarialparasite. Thus, many of these individuals become very illfrom the parasite and many die. Individualshomozygous for the sickle-cell trait (ss) have red blood cells that readily collapse whendeoxygenated. Although malaria cannot grow in these red blood cells,individuals often die because of the genetic defect. However,individuals with the heterozygouscondition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these 'partially defective' red blood cells. Thus,heterozygotes tend to survive better than either of the homozygous conditions. If 9% of anAfrican population is born with a severe form of sickle-cell anemia (ss), what percentage of thepopulation will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cellgene? Answer: 9% =.09 = ss = q2. To find q,simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2(0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).
- PROBLEM #3.There are 100 students in a class. Ninety-six did well in the course whereas four blew it totallyand received a grade of F. Sorry. In the highly unlikely event that these traits are genetic ratherthan environmental, if these traits involve dominant and recessive alleles, and if the four (4%)represent the frequency of the homozygous recessive condition, please calculate the following:
- The frequency of the recessive allele. Answer: Since webelieve that the homozygous recessive for this gene(q2) represents 4% (i.e. = 0.04), the squareroot (q) is 0.2 (20%).
- The frequency of the dominant allele.
Answer: Since q = 0.2,and p + q = 1, then p = 0.8 (80%). - The frequency of heterozygous individuals. Answer: Thefrequency ofheterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that thefrequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).
- PROBLEM #4.Within a population of butterflies, the color brown (B) is dominant over thecolor white (b). And, 40% of all butterflies are white. Given this simple information, which issomething that is very likely to be on an exam, calculate the following:
- The percentage of butterflies in the population that are heterozygous.
- The frequency of homozygous dominant individuals. Answers: The first thing you'll need to do is obtain p and q. So,since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 =0.4. To determine q, which is the frequency of the recessive allele in the population, simply takethe square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So,q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, toanswer our questions. First, what is the percentage of butterflies in thepopulation that are heterozygous? Well, thatwould be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency ofhomozygous dominant individuals? That would be p2 or(0.37)2 = 0.14.
- PROBLEM #5.A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sidedindividuals. Assume that red is totally recessive. Please calculate the following:
- The allele frequencies of each allele. Answer: Well, beforeyou start, note that the allelic frequencies are p and q, and be sure to note that we don't have niceround numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessiveindividuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355.
- The expected genotype frequencies. Answer: Well, AA =p2 = (0.355)2 = 0.126; Aa = 2(p)(q) =2(0.355)(0.645) = 0.458; and finally aa =q2 = (0.645)2 = 0.416 (you already knew this frompart A above).
- The number of heterozygous individuals that you would predict to be in this population.
Answer: That would be 0.458 x 953 = about 436. - The expected phenotype frequencies. Answer: Well, the 'A'phenotype = 0.126 + 0.458 = 0.584 and the 'a' phenotype = 0.416 (youalready knew this from part A above).
- Conditions happen to be really good this year for breeding and next year there are 1,245young 'potential' Biology instructors. Assuming that all of the Hardy-Weinberg conditions aremet, how many of these would you expect to be red-sided and how many tan-sided?
Answer: Simply put, The 'A' phenotype =0.584 x 1,245 = 727 tan-sided and the 'a' phenotype = 0.416 x 1,245 = 518 red-sided ( or1,245 - 727 = 518).
- PROBLEM #6.A very large population of randomly-mating laboratory mice contains 35% white mice. Whitecoloring is caused by the double recessive genotype, 'aa'. Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice,which = 0.35 and represents the frequency of the aagenotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 - qthen 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate thefrequency of the remaining genotypes in the population (AA and Aa individuals). AA =p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa =q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they shouldequal 1.
- PROBLEM #7.After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter aplane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a desertedisland. No one finds you and you start a new population totally isolated from the rest of theworld. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele(c). Assuming that the frequency of this allele does not change as the population grows, whatwill be the incidence of cystic fibrosis on your island? Answer:There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc orp2 = (.05)2 = 0.0025 or 0.25% of the F1 populationwill be born with cysticfibrosis.
- PROBLEM #8.You sample 1,000 individuals from a large population for the MN blood group, which caneasily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). Theyare typed accordingly:
BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY M MM 490 0.49 MN MN 420 0.42 N NN 90 0.09 Using the data provide above, calculate the following:- The frequency of each allele in the population. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of theM allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3.
- Supposing the matings are random, the frequencies of the matings.
Answer: This is a little harder to figure out. Try setting up a 'Punnettsquare' type arrangement using the 3 genotypes and multiplying thenumbers in a manner something like this: MM (0.49) MN (0.42) NN (0.09) MM (0.49) 0.2401* 0.2058 0.0441 MN (0.42) 0.2058 0.1764* 0.0378 NN (0.09) 0.0441 0.0378 0.0081* Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e.the probabilities of matings occurring between these genotypes is TWICE that of the other three'unique' combinations. Thus, three of the possibilities must be doubled.MM x MM = 0.49 x 0.49 = 0.2401
MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
MN x MN = 0.42 x 0.42 = 0.1764
MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
NN x NN = 0.09 x 0.09 = 0.0081 - The probability of each genotype resulting from each potential cross.
Answer: You may wish to do a simple Punnett's square monohybridcross and, if you do, you'll come out with the following result: MM x MM = 1.0 MM
MM x MN = 0.5 MM 0.5 MN
MM x NN = 1.0 MN
MN x MN = 0.25 MM 0.5 MN 0.25 NN
MN x NN = 0.5 MN 0.5 NN
NN x NN = 1.0 NN
- PROBLEM #9.Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasianpopulation of the United States. Please calculate the following.
- The frequency of the recessive allele in the population.
Answer: We know from the above that q2 is 1/2,500 or0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: thefrequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%). - The frequency of the dominant allele in the population.
Answer: The frequency of the dominant (normal) allele in thepopulation (p) is simply 1 - 0.02 = 0.98 (or 98%). - The percentage of heterozygous individuals (carriers) in the population.
Answer: Since 2pq equals the frequency of heterozygotes or carriers,then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.
- The frequency of the recessive allele in the population.
- PROBLEM #10.In a given population, only the 'A' and 'B' alleles are present in the ABO system;there are no individuals with type 'O' blood or with O alleles inthis particular population. If 200 people have type A blood,75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of thispopulation (i.e., what are p and q)? Answer: To calculate theallele frequencies for A and B, we need to remember that theindividuals with type A blood are homozygous AA, individuals with type AB blood areheterozygous AB, and individuals with type B blood are homozygous BB. Beamer 2 1 3 download free. The frequency of Aequals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number ofindividuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.
- PROBLEM #11.The ability to taste PTC is due to a single dominate allele 'T'. You sampled 215individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 couldnot. Calculate all of the potential frequencies. Answer: First,lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Takingthe square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. Youalready know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 =0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and0.495 and these should equal 1.0 or very close to it. This type ofproblem may be on the exam.
- PROBLEM #12. (You will not have thistype of problem on the exam)What allelic frequency will generate twice as many recessive homozygotes as heterozygotes?Answer: We need to solve for the following equation:q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or anotherway of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q,we can substitute 1 - q for p and, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 -4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, weget -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is0.8 =q. I cannot imagine you getting this type of problem in this generalbiology course although if you take algebra good luck.
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Kansas State University | Biology Division
Demonstration of the effect of a neutral density filter. Note that the photograph was exposed for the view through the filter, and thus the remainder of the scene is overexposed. If the exposure had instead been set for the unfiltered background, it would appear properly exposed while the view through the filter would be dark.
Set of ND filters.
In photography and optics, a neutral-density filter, or ND filter, is a filter that reduces or modifies the intensity of all wavelengths, or colors, of light equally, giving no changes in hue of color rendition. It can be a colorless (clear) or grey filter, and is denoted by Wratten number 96. The purpose of a standard photographic neutral-density filter is to reduce the amount of light entering the lens. Doing so allows the photographer to select combinations of aperture, exposure time and sensor sensitivity that would otherwise produce overexposed pictures. This is done to achieve effects such as a shallower depth of field or motion blur of a subject in a wider range of situations and atmospheric conditions.
For example, one might wish to photograph a waterfall at a slow shutter speed to create a deliberate motion-blur effect. The photographer might determine that to obtain the desired effect, a shutter speed of ten seconds was needed. On a very bright day, there might be so much light that even at minimal film speed and a minimal aperture, the ten-second shutter speed would let in too much light, and the photo would be overexposed. In this situation, applying an appropriate neutral-density filter is the equivalent of stopping down one or more additional stops, allowing the slower shutter speed and the desired motion-blur effect.
Mechanism[edit]
For an ND filter with optical densityd, the fraction of the optical power transmitted through the filter can be calculated as
where I is the intensity after the filter, and I0 is the incident intensity.[1]
Uses[edit]
Comparison of two pictures showing the result of using an ND filter at a landscape. The first one uses only a polarizer, and the second one a polarizer and a 1000× ND filter (ND3.0), which allowed the second shot to have a much longer exposure, smoothing any motion.
The use of an ND filter allows the photographer to use a larger aperture that is at or below the diffraction limit, which varies depending on the size of the sensory medium (film or digital) and for many cameras is between f/8 and f/11, with smaller sensory medium sizes needing larger-sized apertures, and larger ones able to use smaller apertures. ND filters can also be used to reduce the depth of field of an image (by allowing the use of a larger aperture) where otherwise not possible due to a maximal shutter speed limit.
Instead of reducing the aperture to limit light, the photographer can add a ND filter to limit light, and can then set the shutter speed according to the particular motion desired (blur of water movement, for example) and the aperture set as needed (small aperture for maximal sharpness or large aperture for narrow depth of field (subject in focus and background out of focus)). Using a digital camera, the photographer can see the image right away and choose the best ND filter to use for the scene being captured by first knowing the best aperture to use for maximal sharpness desired. The shutter speed would be selected by finding the desired blur from subject movement. The camera would be set up for these in manual mode, and then the overall exposure adjusted darker by adjusting either aperture or shutter speed, noting the number of stops needed to bring the exposure to that which is desired. That offset would then be the amount of stops needed in the ND filter to use for that scene.
Neutral-density filters are often used to achieve motion-blur effects with slow shutter speeds
Examples of this use include:
- Blurring water motion (e.g. waterfalls, rivers, oceans).
- Reducing depth of field in very bright light (e.g. daylight).
- When using a flash on a camera with a focal-plane shutter, exposure time is limited to the maximal speed (often 1/250th of a second, at best), at which the entire film or sensor is exposed to light at one instant. Without an ND filter, this can result in the need to use f/8 or higher.
- Using a wider aperture to stay below the diffraction limit.
- Reduce the visibility of moving objects.
- Add motion blur to subjects.
- Extended time exposures.
![Note Note](https://anith.com/wp-content/uploads/2019/11/Handy-Note_7-1098x703.jpg)
Neutral-density filters are used to control exposure with photographic catadioptric lenses, since the use of a traditional iris diaphragm increases the ratio of the central obstruction found in those systems, leading to poor performance.
ND filters find applications in several high-precision laser experiments because the power of a laser cannot be adjusted without changing other properties of the laser light (e.g. collimation of the beam). Moreover, most lasers have a minimal power setting at which they can be operated. To achieve the desired light attenuation, one or more neutral-density filters can be placed in the path of the beam.
Large telescopes can cause the Moon and planets to become too bright and lose contrast. A neutral-density filter can increase the contrast and cut down the brightness, making the Moon easier to view.
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Varieties[edit]
A graduated ND filter is similar, except that the intensity varies across the surface of the filter. This is useful when one region of the image is bright and the rest is not, as in a picture of a sunset.
The transition area, or edge, is available in different variations (soft, hard, attenuator). The most common is a soft edge and provides a smooth transition from the ND side and the clear side. Hard-edge filters have a sharp transition from ND to clear, and the attenuator edge changes gradually over most of the filter, so the transition is less noticeable.
Another type of ND filter configuration is the ND-filter wheel. It consists of two perforated glass disks that have progressively denser coating applied around the perforation on the face of each disk. When the two disks are counter-rotated in front of each other, they gradually and evenly go from 100% transmission to 0% transmission. These are used on catadioptric telescopes mentioned above and in any system that is required to work at 100% of its aperture (usually because the system is required to work at its maximal angular resolution).
In practice, ND filters are not perfect, as they do not reduce the intensity of all wavelengths equally. This can sometimes create color casts in recorded images, particularly with inexpensive filters. More significantly, most ND filters are only specified over the visible region of the spectrum and do not proportionally block all wavelengths of ultraviolet or infrared radiation. This can be dangerous if using ND filters to view sources (such as the Sun or white-hot metal or glass), which emit intense invisible radiation, since the eye may be damaged even though the source does not look bright when viewed through the filter. Special filters must be used if such sources are to be safely viewed.
An inexpensive, homemade alternative to professional ND filters can be made from a piece of welder's glass. Depending on the rating of the welder's glass, this can have the effect of a 10-stop filter.
Specialist neutral-density filters[edit]
Mailtags 5 1 8 download free. The two most common are the variable ND filters and the extreme ND filters such as the Lee Big Stopper.
![Handy note 1 0 9 percent 3 Handy note 1 0 9 percent 3](https://images.firstpost.com/wp-content/uploads/2018/08/Samsung-Galaxy-Note-9-1280-720-15.jpg)
Variable neutral-density filter[edit]
The main disadvantage of neutral-density filters is that different situations might require a range of different filters. This can become an expensive proposition, especially if using screw filters with different lens filter sizes, which would require carrying a set for each diameter of lens carried (although inexpensive step-up rings can eliminate this requirement). To counter this problem, some manufacturers have created variable ND filters. These can work by placing two polarizing filters together, at least one of which can rotate. The rear polarizing filter cuts out light in one plane. As the front element is rotated, it cuts out an increasing amount of the remaining light, the closer the front filters comes to being perpendicular to the rear filter. By using this technique, the amount of light reaching the sensor can be varied with almost infinite control.
The advantage of this approach is reduced bulk and expenses, but one drawback is a loss of image quality caused by both using two elements together and by combining two polarizing filters.
Extreme ND filters[edit]
To create ethereal looking landscapes and seascapes with extremely blurred water or other motion, the use of multiple stacked ND filters might be required. This had, as in the case of variable NDs, the effect of reducing image quality. To counter this, some manufacturers have produced high-quality extreme ND filters. Typically these are rated at a 10-stop reduction, allowing very slow shutter speeds even in relatively bright conditions.
ND filter ratings[edit]
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In photography, ND filters are quantified by their optical density or equivalently their f-stop reduction. In microscopy, the transmittance value is sometimes used. In astronomy, the fractional transmittance is sometimes used (eclipses).
Notation | Lens area opening, as fraction of the complete lens | f-stop reduction | Fractional transmittance | ||||
---|---|---|---|---|---|---|---|
Optical density | ND1number | ND.number | NDnumber | ||||
0.0 | 1 | 0 | 100% | 1 | |||
0.3 | ND 101 | ND 0.3 | ND2 | 1/2 | 1 | 50% | 0.5 |
0.6 | ND 102 | ND 0.6 | ND4 | 1/4 | 2 | 25% | 0.25 |
0.9 | ND 103 | ND 0.9 | ND8 | 1/8 | 3 | 12.5% | 0.125 |
1.2 | ND 104 | ND 1.2 | ND16 | 1/16 | 4 | 6.25% | 0.0625 |
1.5 | ND 105 | ND 1.5 | ND32 | 1/32 | 5 | 3.125% | 0.03125 |
1.8 | ND 106 | ND 1.8 | ND64 | 1/64 | 6 | 1.563% | 0.015625 |
2.0 | ND 2.0 | ND100 | 1/100 | 62⁄3 | 1% | 0.01 | |
2.1 | ND 107 | ND 2.1 | ND128 | 1/128 | 7 | 0.781% | 0.0078125 |
2.4 | ND 108 | ND 2.4 | ND256 | 1/256 | 8 | 0.391% | 0.00390625 |
2.6 | ND400 | 1/400 | 82⁄3 | 0.25% | 0.0025 | ||
2.7 | ND 109 | ND 2.7 | ND512 | 1/512 | 9 | 0.195% | 0.001953125 |
3.0 | ND 110 | ND 3.0 | ND1024 (also called ND1000) | 1/1024 | 10 | 0.1% | 0.001 |
3.3 | ND 111 | ND 3.3 | ND2048 | 1/2048 | 11 | 0.049% | 0.00048828125 |
3.6 | ND 112 | ND 3.6 | ND4096 | 1/4096 | 12 | 0.024% | 0.000244140625 |
3.8 | ND 3.8 | ND6310 | 1/6310 | 122⁄3 | 0.016% | 0.000158489319246 | |
3.9 | ND 113 | ND 3.9 | ND8192 | 1/8192 | 13 | 0.012% | 0.0001220703125 |
4.0 | ND 4.0 | ND10000 | 1/10000 | 131⁄3 | 0.01% | 0.0001 | |
5.0 | ND 5.0 | ND100000 | 1/100000 | 162⁄3 | 0.001% | 0.00001 |
- Note: Hoya, B+W, Cokin use code ND2 or ND2x, etc.; Lee, Tiffen use code 0.3ND, etc.; Leica uses code 1×, 4×, 8×, etc.[2]
- Note: ND 3.8 is the correct value for solar CCD exposure without risk of electronic damage.[citation needed]
- Note: ND 5.0 is the minimum for direct eye solar observation without damage of retina. A further check must be performed for the particular filter used, checking on the spectrogram that also UV and IR are mitigated with the same value.
See also[edit]
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References[edit]
- ^Hanke, Rudolph (1979). Filter-Faszination (in German). Monheim/Bayern. p. 70. ISBN3-88324-991-2.
- ^'CAMERA LENS FILTERS'. Retrieved June 12, 2014.
External links[edit]
Wikimedia Commons has media related to ND filters. |
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