Which nutrient is found in meat, fish and pulses? When digested it forms amino acids

B. Would all insertion or deletion mutations lead to a change in the amino acid

lyudmila [28]

Answer:

Would all insertion or deletion mutations lead to a change in the amino acid sequence? Yes because any new nucleotide that is inserted / deleted shifts the nucleotides, changing the letters of multiple codons.

6 0

3 years ago

State any five characteristics of bacteria

WINSTONCH [101]

Answer:

ningún núcleo, primera célula en la tierra, ADN lineal, tiene ribosomas, tiene una membrana.

Explanation:

3 0

3 years ago

Read 2 more answers

Which charterstic makes the nine amino acids essential

Andreas93 [3]

The essential amino acids cannot be produced in the human body and hence these amino acids have to be obtained through the diet. Hence these amino acids have this characteristic that these cannot be produced in the human body and have to be obtained through diet this makes them essential. Hence those nine amino acids are considered as the essential amino acids.

7 0

3 years ago

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

4 years ago

An additional gene, gene W, was also examined. A test cross was made between true-breeding EEWW flies and EEWW flies. The result

Debora [2.8K]

This question is incorrect but here is the correct question below;

An additional gene,gene W was also examined. a test cross was made between true breeding EEWW flies and eeww flies. The resulting F₁ generation was then crossed with eeww flies. 100 offspring in the F₂ generation were examined and it was discovered that the E and W genes were not linked.

Which is the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked?

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

b) Linked: 25% Eeww, 50% eeWw; not linked:parental genotypes EeWw and eeww.

c) Linked genotypes (EeWw and eeww) and recombinant genotype ( Eeww & eeWw) in the F₂ generation are nearly the same irrespective of their linkage.

d) Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw.

Answer:

a) Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

Explanation:

a test cross was made between true breeding EEWW flies and eeww flies

If EEWW self crossed, we have the following ( EW, EW, EW, EW)

Also, for eeww, we have ( ew, ew, ew, ew)

EW EW EW EW

ew EWew EWew EWew EWew

All offspring are (EWew)

The question goes further by saying "The resulting F₁ generation was then crossed with eeww flies".

And we are asked to find the correct genotype of the F₂ offspring if the genes were linked and if the genes were not linked

∴

To determine the offsprings of the linked genes we need to go by the definition and understand what linked genes are: Linked genes are genes that are physically close together on the same chromosomes. Effect of recombinantion on linked genes, results in gene swaps which occur in chromosomes that are homologous.

Having said that; If EWew × eeww

we have; EW & ew × ew & ew

EW ew

ew EeWw eeww

offspring that

are linked in ⇒ EeWw EeWw & eeww eeww

F₂ will be

$\frac{1}{2}$ = 50% of EeWw of the total 100 offspring in the F₂ cross

$\frac{1}{2}$ = 50% of eeww of the total 100 offspring in the F₂ cross

∴ Linked genes = 50% EeWw and 50% eeww.

For unlinked genes; If EWew × eeww

if rearrangement occurs in EWew and EWew self crossed, we have ( EW,Ew,eW,ew) as the traits needed for the unlinked gene F₂ crossing.

Also ewew will be (ew, ew, ew, ew).

EW Ew eW ew

ew EeWw Eeww eeWw eeww

We have the following results for the unlinked genes

$\frac{1}{4}$ = EeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = Eeww 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeWw 25% of the total 100 offspring in the F₂ cross

$\frac{1}{4}$ = eeww 25% of the total 100 offspring in the F₂ cross

∴ not linked: 25% EeWw, 25% Eeww, 25% eeWw and 25% eeww.

3 0

3 years ago