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emmasim [6.3K]
3 years ago
14

I need help with this

Biology
1 answer:
pickupchik [31]3 years ago
6 0
It would be A. That’s the point of a flu shot. U inject the virus to make your body know it’s bad and when It comes along again to fight it hard.
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Which are parts of a seed?
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Answer:

2

Explanation:

endosperm, embryo, seed coat

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Explain how plants and animals help each other obtain energy.
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Chloroplasts use this energy<span> to create sugar molecules that </span>help<span> the </span>plants<span> grow and reproduce. ... </span>Plants<span> use the carbon dioxide and water, and the cycle begins again. In order to </span>obtain energy<span>, </span>animals<span> do not always have to eat </span>plants<span>. They can also get </span>energy<span> from eating </span>other animals<span> that eat </span>plants<span>.</span>
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2 years ago
Read 2 more answers
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.  

Remark  

Sheryl's Avatar  

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

6 0
3 years ago
What is the primary goal of human communities?
Alenkinab [10]
B.To Understand group behavior through observation
3 0
3 years ago
Suppose that a dominant allele (P) codes for a polka-dot tail and a recessive allele (p) codes for a solid colored tail. If two
andrew11 [14]
The correct answer is B. The parents are heterozygous with polka-dot tails, so they both have Pp genotype. The combination of their genotypes (Pp x Pp) could produce 4 genotypes PP, Pp, Pp and pp. Therefore, there is 25% chance for a PP genotype, 25% for a pp genotype and 50% for a Pp genotype. The PP and Pp genotypes produce a polka-dot tail phenotype and the pp genotype produces a solid coloured tail phenotype. In conclusion, there is 75% chance for a polka-dot tail (25% for a PP + 50% for a Pp) and a 25% for a solid coloured tail phenotype (25% for a pp).
3 0
2 years ago
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