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A recessive allele on the X chromosome is responsible for red-green color blindness in humans. A woman with normal vision whose

Igoryamba

Answer:

In a cross of a homozygous prevailing (AA) individual and a homozygous recessive(AA) person,

a. on the off chance that they have a kid, what is the likelihood that the kid will be heterozygous?

If you cross somebody who is homozygous predominant with one who is homozygous latent, the entirety of the posterity will be heterozygous (AA). Accordingly, there is a 100% possibility of delivering a posterity that is heterozygous.

b. what is the likelihood that the youngster will be homozygous latent?

There is no way of creating a kid that is homozygous passive. The entirety of the posterity will be heterozygous. In this way, the likelihood is zero.

c. in the event that they have two kids, what is the likelihood of the first being homozygous prevailing and the second being homozygous passive?

Again, this can't be determined, on the grounds that there is zero chance of delivering a posterity that is homozygous prevailing or homozygous passive. The likelihood is zero. In any case, for no reason in particular, suppose that the punnet square uncovered that there was 1/4 possibility of creating a kid that is homozygous passive. To decide the likelihood that the subsequent youngster will likewise be homozygous passive, you complete this straightforward duplication:

1/4 x 1/4 = 1/16

In this manner, there would be a 1/16 possibility of the subsequent kid being homozygous passive. In the event that you needed to discover the likelihood of the third kid being homozygous passive, you would do the accompanying:

1/4 x 1/4 x 1/4 = 1/64

Thus, there would be a 1/64 possibility of the third kid being homozygous latent. Once more, I was simply giving in model, yet it doesn't have any significant bearing right now, the entirety of the posterity will be heterozygous.

Another question...A latent allele on the X chromosome is answerable for red-green visual weakness in people. A typical lady whose father is visually challenged weds a partially blind man. What is the likelihood that this couples child will be visually challenged?

XX = female

XY = male

Let C = typical vision (predominant)

Let c = red-green visual weakness (latent)

A typical lady what father's identity is' partially blind must be a transporter, or heterozygous. This implies she acquired the ordinary allele from her mom and the visually challenged allele from her dad.

Genotypes:

Ordinary lady - Xc

Partially blind man - Xc Y

On the off chance that these two mate, here are the accompanying prospects:

half of the female posterity will be bearers with ordinary vision (Xc)

half of the female posterity will be homozygous passive and partially blind (Xc)

half of the male posterity will have typical vision (XC Y)

half of the male posterity will be visually challenged (Xc Y)

In this manner, the likelihood that the couple's child will be partially blind is half, or 1/2.

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Sheryl addressed this Was this answer accommodating?

XX= lady, XY=man

Alleles:

XC=normal; Xc=colorblind

Typical Genotypes:

XC (typical lady)

Xc (typical lady, yet bearer)

Xc (visually challenged lady)

XC Y (typical man)

Xc Y (visually challenged man)

Lady has typical vision, yet her dad is visually challenged. In this way, she needed to get XC from her mother (must have one, she's ordinary) and Xc from her father (it was everything he could give her): Xc

Man is visually challenged: Xc Y

Xc Y

XC Y

Xc Y

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Although the foxp2 amino acid sequence is highly conserved, humans have two unique amino acids that chimpanzees don't have. thes

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Deoxyribose differs from glucose because it:

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Explain how mutations in the photoreceptor genes result in different vision defects.

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Answer:

The human eye is the prime organ of the body, which associates with the photons of light and allows one to see various things. The unique cells found in retina, which does activity of seeing are cones and rods cells. Rods help to see in dim light vision, while on the other hand, cone cells are unique in recognizing different colors.

These cells comprise photoreceptor proteins that help in trapping photons at particular wavelength. Mutation in the gene encrypting for these proteins results in permanent or temporary vision issues. The extremity of defects relies upon the degree to which mutation takes place.

The mutation in rod cells photoreceptor proteins leads to night blindness and retinitis pigmentosa. Retinitis pigmentosa refers to an inherited disorder that takes place because of early loss of rod cell, which destructs retina. On the other hand, night blindness does not mean complete blindness night, however, inadequate tendency to see in low light.

Identically, the mutation in the cone cell also results in vision issues, known as red color blindness and tritanopia. Tritanopia refers to a kind of color blindness, which originates because of insensitivity of blue receiving protein gene towards blue light. On the other hand, red color blindness refers to insensitivity of red receiving cone cells in captivating long-wavelength photons.

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