BIOLOGY COLLEGE

Helpp me . Is it A,B,C,D

Answers

Answer 1

Answer:

Answer:

The answer is C. The GPS makes a task easier

Answer 2

Answer:

Answer:

Its C

Explanation:

Related Questions

how do you find this day?what are the positive things that happened?what are those things that made your irritated or upset?
In humans alkaptonuria is a metabolic disorder in whichaffected persons produce black urine. Alkapotonuria results from anallele(a) that is recessive to the allele for normal metabolism(A). Sally has a normal metabolism, but her brother hasalcaptonuria. Sally's father has alcoptonuria, and her mother hasnormal metabolism. a) Give the genotype of Sally,her mother,father and herbrother. b) If Sally's parents have another child what is theprobabilty that this child will have alkaptonuria? c) If Sally marries a man with alkaptonuria, what is theprobability that their child will have alkaptonuria?
Rough er is connected to the_________ membrane and to ______ er
The plants in. are true breeding
Compare and contrast aspergillosis and mucormycosis.

) How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Answers

Answer:

$5 * 10^(10)$

Explanation:

The question is not complete. Remaining part of the question is as follows - Minimal growth medium for bacteria such as E. coli includes various salts with characteristic concentrations in the mM range and a carbon source. The carbon source is typically glucose and it is used at 0.5% (a concentration of 0.5 g/100 mL). For nitrogen, minimal medium contains ammonium chloride (NH4Cl) with a concentration of 0.1g/100 mL

How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Solution -

We will first find the mass concentration of 0.5 g/100 mL of solution.

$(0.5)/(100)$ gram per ml of glucose

The chemical formula of glucose is $C_6H_(12)O_6$

The molecular weight of glucose molecule is $180$ grams per mole

Now, we will find the number of moles of glucose in a 5 ml medium -

$((0.5)/(100) * 5)/(180) \n1.39 * 10^(-4)$ mole

The number of carbon atom in each glucose molecule is equal to six, thus, number of minimal carbon mole is equal to

$1.39 * 10^(-4) * 6\n= 8.34* 10^(-4)$ mole

Number of carbon atoms is equal to

$8.34* 10^(-4) * 6.023 * 10^(23)\n= 5 * 10^(20)\n$ Carbons

One bacteria has $10^(10)$ carbon molecule.Thus, $5$ ml medium will have $5 * 10^(10)$ bacteria

For the following biomolecular interactions, choose the main weak force interactions that mediates them. Interaction between hydrophobic (nonpolar) amino acid side chains van der Waals interaction
Repulsion between negatively charged backbone of DNA double helix van der Waals interaction
Interaction between water and carbonyl oxygen of guanine hydrogen bond

Answers

Answer:

Interaction between hydrophobic (nonpolar) amino acid side chains is due to hydrophobic interactions.

Repulsion between negatively charged backbone of DNA double helix is due to electrostatic repulsion forces.

Interaction between water and carbonyl oxygen of guanine is due hydrogen bonding

Explanation:

The hydrophobic effect refers to the tendency of non-polar substances forms aggregates in an aqueous solutions in order to avoid interactions with water molecules. This results in a favorable entropy increase.

For non-polar amino acids, hydrophobic interactions are important in stabilizing their structure wherein they are densely packed within the core of a protein while those with with polar groups are found in the exterior.

The phosphate backbone of DNA is negatively charged. Since like charges repel, therefore, electrostatic repulsion forces of the ions are responsible the repulsion of the backbone DNA double helix.

Hydrogen bonds are formed between electronegative oxygen atoms and hydrogen atoms of neighboring molecules due to the partial negative and positive charges on each of the atoms.

This hydrogen bonding is responsible for the interaction between water and carbonyl oxygen of guanine as the hydrogen atoms of the water molecules are attracted to the oxygen atoms of the guanine molecules due to the partial charges on each of the atoms.

What is the difference between Gene Pool, Genetic Drift and Gene Flow?

Answers

Answer:

Genetic drift is referred to the changes in allele frequency in a gene pool. Gene flow is the process of alleles moving from one population to another.

Explanation:

Which of the following statements about the functional consequences of single base-pair mutations are TRUE? Select the three correct answers. A) Mutations that occur in the intron sequences cannot affect the protein coding sequence of a gene. B) Insertions or deletions of single base-pairs within a protein coding sequence probably change the amino acid sequence encoded by a gene. C) Nucleotide substitutions within a coding sequences always change the amino acid sequence encoded by a gene. D) Missense mutations always change the amino acid sequence encoded by a gene. E) Nucleotide substitutions outside of the coding sequence can affect the expression of a gene.

Answers

Answer:

these statements are true

B) Insertions or deletions of single base-pairs within a protein coding sequence probably change the amino acid sequence

D) Missense mutations always change the amino acid sequence encoded by a gene

E) Nucleotide substitutions outside of the coding sequence can affect the expression of a gene.

Explanation:

B) insertion or deletion of nucleotide in coding region of DNA can change the amino acid sequence encoded.

D) missense mutation means that the change in nucleotide sequence resulting in change in amino acid sequence encoded by gene.

E) A mutation may alter the promoter of a gene, thereby affecting the rate of transcription.

Final answer:

The correct statements regarding single base pair mutations are: insertions or deletions within a protein coding sequence likely change the encoded amino acid sequence; missense mutations always change the encoded amino acid sequence; nucleotide substitutions outside of the coding sequence can impact gene expression.

Explanation:

Your question focuses on single base-pair mutations and their functional consequences. After considering the provided statements, the three correct answers are: B) Insertions or deletions of single base-pairs within a protein coding sequence probably change the amino acid sequence encoded by a gene. D) Missense mutations always change the amino acid sequence encoded by a gene. E) Nucleotide substitutions outside of the coding sequence can affect the expression of a gene.

A missense mutation is a type of mutation where a single base pair is changed, leading to the substitution of a different amino acid in the resulting protein. This change can alter the function of the protein. On the other hand, insertions or deletions of single base-pairs can often lead to a frameshift mutation, which alters the reading frame and potentially changes every subsequent amino acid in the encoded protein, often resulting in a nonfunctional protein. Lastly, nucleotide substitutions outside of the coding sequence can also affect gene expression by changing areas involved in the regulation of gene transcription.

Learn more about Single Base-Pair Mutations here:

brainly.com/question/30875580

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A few organisms have evolved to use mostly asexual reproduction. what must be true about these?

Answers

I cannot find the statements, I will explain the asexual reproduction so you can answer the question by your own:

Asexual reproduction is as simple a cell multiplication as a photocopying machine reproduced in one or more copies an original document. In all cases, we obtain a clone of cells all having the same hereditary information.

The great advantage of asexual reproduction is its speed.
It produces a new organism that is genetically identical to the parent you do not have to look for a partner
Energy can be deployed to produce potentially large numbers of offspringOffspring are usually adapted to their environment by the success of their parent
Helps to quickly and effectively colonize a parent-friendly habitat, thanks to a large number of descendants he can breed in a short timeThe offspring are often already multicellular and more viable.

The inconveniences are:
Does not cause a genetic diversity of organisms of the same species.
The species does not adapt at all or very slowly depending on the changing circumstances.
There is only one parent to care for the offspring.
The parent disappears sometimes because his body is no more (binary fission, fragmentation).
An asexual species may be suddenly destroyed by a disaster that affects all genetically identical organisms of this species.

Tail length in mammals is a heritable trait. A pig with a 6 cm tail was mated to a pig with a 30 cm tail. All F1 offspring had a tail of length 18 cm. An F2 generation was made by crossing F1 individuals. This resulted in many piglets whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). The majority of pigs had 18 cm tails, 1/64 had 6 cm tails, and 1/64 had 30 cm tails.These results are consistent with what genetic model?

(a) Two genes, each with two alleles that show dominance
(b) Two genes, each with two alleles that act additively
(c) Three genes, each with two alleles that show dominance
(d) Three genes, each with two alleles that act additively

An F2 piglet with an 18 cm tail is heterozygous at each of the tail-length loci, and is mated to an F2 piglet with a 6 cm tail. Write down the predicted offspring genotypes and calculate the predicted tail lengths. What is the expected frequency of piglets with a 14 cm tail length?

Answers

Answer:

The answer is (d) Three genes, each with two alleles that act additively.

Explanation:

1/ The genotype of P is AABBCC and aabbcc, for example. Then, F1 should be AaBbCc.

The genotype of individuals with n heterozygous pair is 2^n. Thus, in this case, the number of genotype in F2 should be 2^3 * 2^3 = 8 * 8 = 64. We can get this conclusion by analyzing the number of 6 cm tails in F2: 1/64 with genotype aabbcc, and the number of 30 cm tails: 1/64 with genotype AABBCC. These two genotypes is as same as the ancestor in this experiment.

The genotype of an F2 piglet with an 18 cm tail is AaBbCc. If these genes show dominance, the tail lenght of F2 will be 30 cm. And there are 7 possible phenotypes. Thus, we can conclude that the genes act additively.

2/ 18 cm tail F2: AaBbCc, and 6 cm tail: aabbcc.

The offspring genotypes are:

18 cm AaBbCc, 14 cm AaBbcc, 14 cm AabbCc, 10 cm Aabbcc
14 cm aaBbCc, 10 cm aaBbcc, 10 cm aabbCc, 6 cm aabbcc

The frequency of piglets with 14 cm tail length should be 3/8 = 37.5%.

Answer and Explanation:

1) These results are consistent with option (d)Three genes, each with two alleles that act additively .

Quantitative heritability: Refers to the transmission of a phenotypic trait in which expression depends on the additive effect of a series of genes.

Polygenic heritability occurs when a trait is due to the action of more than one gene that can also have more than two alleles. This can cause many different combinations that are the reason for genotypic graduation.

Quantitative traits are those that can be measure, such as longitude, weight, eggs laid per female, among others. These characters do not group individuals by precise and clear categories. Instead, they group individuals in many different categories that depend on how the genes were intercrossed and distributed during meiosis. The result depends on how each allele of each gene contributed to the final phenotype and genotype.

In the exposed example, there are 7 different phenotypes (6, 10, 14, 18, 22, 26, 30) and the majority of F2 individuals have 18-lengthed tails. This might be the heterozygotic phenotype for all the genes.The rest of the phenotypes are possible combinations of genes and their alleles. There are six possible combinations (apart from the 18 lenght form). This leads to 3 genes and two alleles in each gene.

So, until now we have:

A pig with a 6 cm tail was mated to a pig with a 30 cm tail. All F1 offspring had a tail of length 18 cm. This is:

Parental) 6cm x 30cm

F1) 18 cm

An F2 generation was made by crossing F1 individuals. This resulted in many piglets whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). This is:

Parental) 18 cm x 18 cm

F2) 6, 10, 14, 18, 22, 26, 30

The majority of pigs had 18 cm tails, this means that 18 cm phenotype is a heterozygote for each gene

1/64 had 6 cm tails, and 1/64 had 30 cm tails, this means that these two phenotypes are the extreme traits, that is the recessive homozygote and the dominant homozygote.

There are 7 phenotypes, one of them is the recessive form, the other is the dominant form and the majority is the heterozygotic form for every intervening gene. There are three genes with two alleles each:

Gene 1: allele A and a

Gene 2: allele B and b

Gene 3: Allele C and c

Phenotypes:

aabbcc: homozygotic recessive form

AABBCC: homozygotic dominant form

AaBbCc: heterozygotic form for every intervening gene

If the recessive form is 6cm length, and each dominant allele contributes in 4 cm to each phenotype, we get:

6 cm length = aabbcc (1/64)

10cm length = Aabbcc (A contributes 4cm)

14cm length = AaBbcc (A and B contribute 4 cm each)

18cm length = AaBbCc (The majority) (A, B and C contribute 4cm each)

22cm length = AABbCc

26cm length = AABBCc

30cm length = AABBCC (1/64)

2) An F2 piglet with an 18 cm tail is heterozygous at each of the tail-length loci, and is mated to an F2 piglet with a 6 cm tail.

Parental) AaBbCc x aabbcc

Gametes) ABC ABc AbC Abc aBC aBc abC abc

abc abc abc abc abc abc abc abc

Punnet square)

ABC ABc AbC Abc aBC aBc abC abc

abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc

F3 genotype and phenotype)

8/64 AaBbCc = 18 cm

8/64 AaBbcc = 14cm

8/64 AabbCc = 14cm

8/64 Aabbcc = 10cm

8/64 aaBbCc = 14cm

8/64 aaBbcc = 10cm

8/64 aabbCc = 10cm

8/64 aabbcc = 6cm

Each dominant allele contributes 4cm to the recessive homozygote form for each phenotype.

Other Questions

What makes ribosomal RNA useful as a molecular clock? (2 points) A large portion of the DNA ring is not vital to structure or function, allowing it to accumulate neutral mutations. Its rate of mutation increases over time as organisms continue to evolve and differentiate from each other. A slow mutation rate makes it useful for determining evolutionary relationships between ancient species. It is only found in select organisms, making it easier to compare relationships between species that have it.

Erythrocytes might the best of example of form equals function because they lack many organelles. What organelles are absent in erythrocytes?

In mice, the autosomal locus coding for the β-globin chain of hemoglobin is 1 m.u. from the albino locus. Assume for the moment that the same is true in humans. The disease sickle-cell anemia is the result of homozygosity for a particular mutation in the β-globin gene.a. A son is born to an albino man and a woman with sickle-cell anemia. What kinds of gametes will the son form, and in what proportions? (2 points)b. A daughter is born to a normal man and a woman who has both albinism and sickle-cell anemia. What kinds of gametes will the daughter form, and in what proportions? (2 points)c. If the son in part a grows up and marries the daughter in part b, what is the probability that a child of theirs will be an albino with sickle-cell anemia? (1 point)

As a ribosome translocates along an mRNA molecule by one codon, which of the following occurs? a. The polypeptide enters the E site. b. The tRNA that was in the A site departs from the ribosome via a tunnel. c. The tRNA that was in the P site moves into the A site. d. The tRNA that was in the A site moves to the E site and is released. e. The tRNA that was in the A site moves into the P site.