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BIMM 110 Final Exam questions and correct answers

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BIMM 110 Final Exam questions and correct answers

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BIMM 110 Final Exam questions and
correct answers

Rods - correct answer ✔✔ - photoreceptors that cannot distinguish between different
wavelengths (colors) of light

- more sensitive to low light, used for night vision

- focused on the outer areas of the retina



Cones - correct answer ✔✔ - photoreceptors that can see colors

- concentrated on the center of the retina, on the fovea

- 3 different types of cones that differ in the opsin they make



What are the 3 types of cones? - correct answer ✔✔ - L-cones (pro): sense long wavelength
(red); 560 nm

- M cones (deut): sense medium wavelength (green); 530 nm

- S-cones (tri): sense short wavelength (blue); 420 nm



What is the difference between a trichromat vs. a dichromat? - correct answer ✔✔ - trichromat:
has 3 photoreceptors (normal vision)

- dichromat: has 2 photoreceptors (two color vision, amorph)



Different types of dichromacies - correct answer ✔✔ - Protanopia: L-cone absent (cannot see
red)

- Deuteranopia: M-cone absent (cannot see green)

- Tritanopia: S-cone absent (cannot see blue)

,Anomalous trichomacies and the different types - correct answer ✔✔ - neomorph, absorption
patterns are different than they should be

- ex.) having R' and R instead of G and R

- Protanomaly: L cone defect, red-weak

- Deuteranomaly: M cone defect, green-weak

- Tritanomaly: S cone defect, blue-weak



Rod monochromacy (Achromatopsia) - correct answer ✔✔ - no cones at all; no color vision

- can only see shades of grey



Phototransduction in the light - correct answer ✔✔ 1) Light bleaches rhodopsin, causes a
conformational change in protein, and activates it

2) Activated rhodopsin activates transducin/protein

3) Transduction activates phosphodiesterase enzyme that degrades cGMP bringing the
concentration down (cGMP --> GMP)

4) Decrease in cGMP/ligand causes cyclic nucleotide gates to close (hyper-polarizing the cell
because Na+ and Ca2+ cannot enter the cell)

5) Causes less neurotransmitter to be released onto bipolar cell



Phototransduction in the dark - correct answer ✔✔ 1. Rhodopsin is not activated by light

2. Transducin G-protein is not activated and does not cause the conversion of cGMP to GMP

3. cGMP can continue to bind to the ion channel

4. Ion channel remains open and cations can enter the cell and depolarize it



Incidence of color vision deficiencies - correct answer ✔✔ For anomalous trichromacies and
dichromacies:

- protanomaly and deuteranomaly incidence is greater in males than in females

- tritanomaly affects males and females in roughly the same proportion

, Why is protanomaly/protanopia and deuteranomaly/deuteranopia more present in males than
females? - correct answer ✔✔ - the red and green opsin genes are present on the X
chromosome

- because males only have one X chromosome, a faulty opsin gene will result in that type of
colorblindness whereas females have an extra X chromosome that can compensate for the
mutation



Old World primates color vision - correct answer ✔✔ - the blue pigment allele is present on an
autosome

- the X chromosomes has both the red and green pigment allele

- as such, both male and female Old World primates have trichromatic vision



New World primates color vision - correct answer ✔✔ - the blue pigment allele is present on an
autosome

- the X chromosome only carries one pigment allele

- male primates will be dichromatic because they only have 2 pigment alleles

- females can either be dichromatic or trichromatic depending on if the X chromosomes carry
different pigment alleles



How did primate trichomacy evolve? - correct answer ✔✔ - an error in the recombination
process due to misalignment of chromosomes and unequal crossing over led to a duplication
mutation --> 2 pigment alleles on the same X chromosome --> trichromatic vision

- trichromatic vision was maintained because it conferred a selective advantage on male and
female primates.



What is haploinsufficiency? - correct answer ✔✔ - condition where having only one copy of a
wild-type gene does not produce a wild-type phenotype or is not sufficient enough for normal
function

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