A recessive allele on the X chromosome is responsible for red-green color blindness in humans. A woman with normal vision whose father is color blind marries a color-blind male. What is the probability that this couple's first son will be color blind?

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

meiosis is the first to occur after the nuclei of opposite of opposite mating types fuse