The cell responsible for secreting the matrix of bone is the__________a. osteoclast. b. chondroblast. c. chondrocyte. d. osteoblast.

Answer:

complementation test

Explanation:

The easiest examination to differentiate between the two possibilities is the complementation test. The test is simple to perform: two mutants cross and F1 is analyzed. If F1 expresses the wild-type phenotype, we conclude that each mutation is in one of two possible genes necessary for the wild-type phenotype. When it is shown that it is genetically shown that two (or more) genes control a phenotype, the genes are said to form a complementation group. Otherwise, if F1 does not express the wild type phenotype, but rather a mutant phenotype, we conclude that both mutations occur in the same gene.

These two results can be explained considering the importance of genes for phenotypic function. If two separate genes are involved, each mutant will have an injury to one gene while maintaining a wild-type copy of the second gene. When F1 occurs, it will express the mutant allele of gene A and the wild-type allele of gene B (each contributed by one of the mutant parents). F1 will also express the wild type allele for gene A and the mutant allele for gene B (contributed by the other mutant parent). Because F1 is expressing the two necessary wild-type alleles, the wild-type phenotype is observed.

Final answer:

The statement is false because fungi are not typically identified at the genus and species level by visualization of their nucleus under a transmission electron microscope, but rather by morphological characteristics and other methods.

Explanation:

The correct answer to the question is False. While fungi are eukaryotic organisms that do contain a nucleus, transmission electron microscopy of a nucleus is not typically used for identifying fungi down to the genus and species level. In practice, fungi are more commonly identified by their morphological characteristics, which can be observed under lower magnification microscopy, along with genetic sequencing and biochemical tests. For instance, the morphology of hyphae and the structure of mycelium, along with other characteristics like reproductive structures and spore types, are key attributes used in the identification of fungi. Transmission electron microscopy might be used for detailed cellular studies but not typically for standard identification at the genus and species level.

Answer:

D

Explanation:

Lamark believed in the inheritance of acquired characteristics. That is, that if an individual's body or organs changed in some way during the course of their life time (due to the environment or some other pressure), that these characteristics could be passed on to the next generation.

This was his theory of evolution. Therefore, if, slowly elephants started using their trunks to an advantage more, they would start to develop and grow. Lamark believed these acquired traits would be passed on to the next generation. We now know this not to be true, and random, selectively advantageous mutations in DNA are what cause evolution over time.

Answer:

The formation of favorable C-bond ionic or C-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or restore donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of H-bond in the H-bonding protein is offset by the energy required to break many interactions, with solvent going from the inter-molecular to the intramolecular state.

Answer:

The answer is:

The formation of favorable intermolecular ionic or H-bonding interactions in a folded protein replace interactions between solvent (water) and the ionic species (or H-bond donors and acceptors) in the unfolded state. The favorable ΔH obtained by formation of intramolecular bonds in the folded protein is offset by the energy required to break many interactions, with solvent going from the unfolded to the folded state.

Explanation:

Answer:

C. coccoid-shaped.

Explanation:

MreB protein serves as function as actin proteins do in the eukaryotic cells. MreB protein is involved in maintaining the cell shape. It is observed in many rod-shaped bacteria and archaea such as Bacillus subtilis. MreB polymerizes to form a spiral around the inside periphery of the cell and maintain the elongated shape of the cell. A bacterium that is not able to produce functional MreB protein cannot maintain the rod shape of its cell and is coccoid-shaped.

Answer:

Explanation:

a, The genetic counselor should  explain that color-blindness is a sex-linked disorder, that is its  expression is related to a particular sex-cells.

a) The percentage is 0% . This is because your husband is normal, so no sex-linked alleleis attached to his only X-chromosomes.However either of your X-chromosomes can bear the allele for color blind. Since a girl child sex-chromosomes combination is XX, and your husband X chromosomes is normal,hence your daughters can only be carries (XCX, XCX)like you  if they inherited any of your X -chromosomes color blind allele.But none will be colorblind.

b)Because you are carrier for this alllele, and your husband is normal;  chances are all your sons   be   colourblind  is 50% proportion(YXC/YXC).  This is because your husband Y chromosomes determines a male child, and you can  bear the defective(carrier allele)  for  colourblind  on either of your X-chromosmes to give a male child as XY. Thus either of the  male child  assuming  with Mendelian fashion of inheritance, will be colour blind in 50% proportion.

No. A girl child can only be  a carrier for color blindness  if the  inherited the color blind gene from her carrier mother  or the father is color-blind. None of this is present in this scenario.  The two parents are normal Therefore there is no chance of their daughter emerging a carrier.

A genetic counselor would tell the parents that (a) the percentage of having a daughter who is color-blind is 0% and  (b) the percentage of having a son who is color-blind is 50%.

When considering the genetic possibilities for a couple where the female is a carrier of the color-blindness gene and the male has normal vision, genetic counselors would provide the following insights:

The percentage of having a daughter who is color-blind is 0%. This is because for a female to be color-blind, she must inherit two color-blindness genes – one from each parent. In this scenario, the father can only pass on a normal vision gene.

The percentage of having a son who is color-blind is 50%. Since males have only one X chromosome, inherited from their mother, they will be color-blind if that X chromosome carries the color-blindness gene.

For a different couple, both with normal vision, their daughter can still be a carrier for color-blindness if the mother is a carrier or if the father's mother is a carrier. This is because the mother can pass a carrier X chromosome to the daughter, or the father can pass an X chromosome that he inherited from his color-blindness carrier mother.

Answer:

2.54 g

Explanation:

Equation for the reaction was followed by the dehydration of:

Cyclohexanol ----> Cyclohexene

Given that:

mass of cyclohexanol = 3.1 g

mass of cyclohexene = 2.2 g &

the standard molar mass of cyclohexanol is known to be = 100.16 g/mol

Also, the standard molar mass for cyclohexene = 82.14 g/mol

From what we have above, we can calculate for their respective numbers of moles;

By the formula which say;

number of moles = [tex]\frac{mass(g)}{molar mass}[/tex]

number of moles of cyclohexanol = [tex]\frac{3.1(g)}{100.16g/mol}[/tex]

= 0.03095 moles

From the equation for the dehydration reaction above;

one mole of cyclohexanol gives one mole of cyclohexene

∴ 0.03095 moles of cyclohexanol is said  to be equal to 0.03095 moles of cyclohexene

Thus, having gotten that; we can proceed to calculate for our theoretical yield of cyclohexene.

Theoretical yield of cyclohexene =  (number of moles × molar mass) of cyclohexene

= (0.03095 g × 82.14 g/mol)

= 2.542233 g of cyclohexene

≅ 2.54 g to two significant figures.

Following the balanced equation and converting g to mol and back, we find that the theoretical yield for the dehydration reaction of cyclohexanol to cyclohexene is 2.54 g.

Firstly, we need to calculate the molar masses of cyclohexanol and cyclohexene. The molar mass of cyclohexanol (C6H12O) is approximately 100.16 g/mol while the molar mass of cyclohexene (C6H10) is about 82.14 g/mol. We then convert the masses of cyclohexanol and cyclohexene into moles. This is done by dividing the given mass of each compound by its respective molar mass. So, for cy Convert this theoretical yield in moles back to grams by multiplying by the molar mass of cyclohexene. So, the theoretical yield is 0.0309 mol x 82.14 g/mol = 2.54 g.

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Answer:

a) 9%.  

b) 16.8%.

Explanation:

a).

We are provided with the information that Two linked genes, A and B, are separated by 18 cM (centiMorgan). i.e the recombinant frequency is 18%

Also , the man's genotype is AB/ab... This only result to one explanation, that The man will definitely produce 18% of recombinant gametes which entails

9% Ab & 9% aB

i.e 0.09 Ab & 0.09 aB

On the other-hand, The mother ab/ab have tendency to produce just one single type of gamete which is ab

∴

The probability that their first child will be Ab/ab will be

Pr ( Ab/ab) = (0.09) x (1)

= 0.09

= 9%.  

b).

If the father produces 18% of recombinant gametes which entails

9% Ab & 9% aB , this typically implies that the number of the non-recombinant gametes will be;

100%-18% = 82%  ( non-recombinant gametes)

i.e genotype AB/ab = 82%

AB =41%; ab = 41%

AB = 0.41 ; ab = 0.41

Now, the probability that their first two children will both be ab/ab:

Using Multiplication Rule to calculate the probability that their first two children (ab/ab); we have:

 (0.41)(1) ×(0.41)(1)

= 0.1681

= 16.8%.

The probability of the first child being Ab/ab as 25% and the probability of the first two children being ab/ab as 6.25%.

Probability of first child being Ab/ab:

Probability of first two children both being ab/ab:

Answer:

b. Gradual evolutionary change explains why organisms are well-suited to their environments.

Explanation:

Darwin and Larmack are both evolutionists  who  believed that the  present organisms  or offspring are descendants of previous organisms. And they are able to survive till today  from previous generation because they have certain  features  that make them resistant to  selective pressure in their environment. Lamarck called  concluded that  changes in the physical features of the present organisms from using a certain body part  more than others  in order to survive is the major reason for their existence and resistant to selective pressure. He called his theory Law of use and disuse .

Darwin believed that,  the ability to survive  by theses organisms  was due to  certain factor or trait(gene) which the descendants must have inherited from  the previous generations,. And these trait(gene) made them to resist selection pressures,and therefore  have higher competition for survival  than other organisms which lack these traits. Thus he concluded that nature must have selected these organisms with certain traits(genes) than others, He tagged his findings  theory if evolution by natural selection.

However both scientists believed that present organisms   are products of Gradual evolutionary changes of million of years ago.

Answer: Option B.

Gradual evolutionary chnges explain why organisms are well suited for their environment.

Explanation:

Lamarck and Darwin are both naturalist that formulate theories. Lamarck theory was called theory of acquired characteristics and Darwin theory was called theory of evolution by natural selection. The two scientist agree that gradual evolutionary changes I.e change in trait and characteristics explain why organisms are well suited in their environment. Lamarck theory describe changes that happen to organisms in their life time and how these changes are passed to their offsprings and Darwin theory supported these that organisms arise from natural selection and inherited variations which help them survive and reproduce in their environment.

Answer:

0.67

Explanation:

In this population there are:

N: 196

The frequency of the A1 is the amount of times that allele appears divided by the total number of alleles in the population.

The frequency of the A1 allele can be thus calculated using the following formula:

[tex]freq(A1)=\frac{2*A1A1\ + A1A2}{2*N} \\\\freq(A1)=\frac{2*104\ + 55}{2*196} \\\\\\freq(A1)= 0.67[/tex]

Answer:

Survival depends on interaction of many factors, which can vary greatly. For each factor a given species has a range of conditions. If environmental conditions exceed upper or lower limit of tolerance, death can result.

Explanation:

This is the Shelford's law of Tolerance.

This law states that the success of distribution of an organism can be controlled by some factors like topographic, climate, and biological requirements of plants and animals, in which their levels exceed the higher or lower limits of tolerance of the organism.

This was a law proposed by V. E. Shelford in 1911.

Answer:

mass(g) = 0.55114g

Explanation:

Given that;

Molarity (M) 100mM = 0.1M

Volume (V) = 1mL

                  =  1.0 × 10⁻³ Litre

numbers of moles =  Molarity × Volume

                               = 0.1M × 1.0 × 10⁻³ L

                               = 0.001

                               = 1.0 × 10⁻³ mol

Now, to determine the mass of ATP (in solid form) that we need to weigh;

we have our numbers of moles to be =[tex]\frac{mass(g)}{molarmass(g)}[/tex]

where;

numbers of moles = 1.0 × 10⁻³ mol

molar mass = 551.14

mass(g) = numbers of moles × molar mass

mass(g) = 1.0 × 10⁻³ mol × 551.14

mass(g) = 0.55114g

∴ 0.55114g ATP(in solid form) is required to be weighed  in order to make 1 mL of stock solution with a concentration of 100 mM ATP.

Let's complete the question by adding the missing piece of information

The mutation results in the breed's distinctive point markings (ears, mask, tail and legs) and lighter body color. Use this information to explain the pattern of the cat's fur pigmentation.

Answer:

The mutation of the TYR gene results in the enzyme tyrosinase to be heat susceptible. Tyrosinase takes part in the production of melanin to give darker fur in colder areas. The areas like the tail, legs, ears, and face do lack as much body heat and so will get darker.

Explanation:

A unique protein (enzyme), known as tyrosinase, is the major workhorse in the development of the melanin. A research team from the University of California, USA, led by L. A. Lyons, discovered that Siamese cats have tyrosinase that went through mutation due to the changes in the DNA helix and is temperature-sensitive as it's activity reduces with a rise in temperature. This explains why cat’s warm parts of the body are coated with white, melanin-lacking hair since Tyrosinase is deactivated in these regions and melanin is not developed – hair is white-colored. On the other hand, in cooler boundary the enzyme is active and the melanin is formed – hair has dark color.

Answer:

The cell division is divided into two main phase - interphase and M (mitosis) phase. The interphase is further divided into G1, S phase and G2 phase.

G1 (gap 1 phase): The G1 phase is marked by the complete growth of the cell. The cells accquire different proteins and factors that are necessary for the cells to undergo the process of DNA rep[lication of the synthesis phase.

G2 phase (gap 2 phase) : During the G2 phase the cells prepares itself to undergo the M phase and underwent through the different phases of the cell cycle. The nuclear envelope forms around DNA and centrosomes are fully developed.

G0 phase: The G0 phase is also known as the extended phase of the G1 cycle. The G0 phase occurs when the cells are completely ready but do not undergo G1 phase, this might occur when the tissue is under the generation process.

The cells will go through the G1, S, G2 and M phase of the cell cycle. This is not necessary that all cells undergo the G0 phase of the cell cycle.

Kangaroo rats in the Sonoran Desert would be most likely to develop new adaptations through natural selection if their environment undergoes significant changes that challenge their survival, leading to the evolution of advantageous traits that are passed on to offspring.

The kangaroo rats of the Sonoran desert would be most likely to develop new adaptations by natural selection under circumstances where there are significant changes in their environment that create new challenges or pressures for survival. The key to this process is variation within the kangaroo rat population and the selective pressures that favor certain traits over others. For example, if the desert climate were to become even drier or if a new predator were introduced, those individuals with traits that improve their chances of survival, like even more efficient water conservation or better evasion tactics, would be more likely to survive and reproduce. These advantageous traits would be passed on to the next generation, leading to a population that is better adapted to the new conditions. This process over many generations results in evolution and the development of new adaptations.

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Answer:

21 Mating

Explanation:

Ans A)The recessive lethal allele is tightly linked to, and thus co-segregates with, the recessive B allele, which is lethal in the homozygous state (bb)

Ans B)

1% of the gametes will be recombinant (Bl or bL)

0.5% of the gametes will be bL

0.5% of the gametes will be Bl

white animals will have the following genotype

genotype probability

bL/bL(0.005)(0.005) = 0.000025

bL/bl(0.005)(0.495) = 0.002475

bl/bL(0.495)(0.005) = 0.002475

Probability of white animal =0.004975

1/0.004975 = 201 therefore on average you'd have to look at 201 progeny before seeing a white animal so on average you'd have to do 21 mating

Answer:

The body tubes are long, semi-rigid tubes, usually made of _polyvinyl_ chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end (sticking out of mouth) has a standard _15_ mm adaptor.

Explanation:

An endotracheal tube (ET) is a flexible plastic tube that is inserted into the trachea through the mouth to help a patient breathe. Then it is connected to a ventilator for oxygen supply to patients's lungs. This process of insertion of tube is refereed as endotracheal intubation.

The body tubes in question are usually part of endotracheal tubes, made of polyvinyl chloride or a similar plastic. The proximal end of these tubes that sticks out of the mouth has a standard 15 mm adaptor.

The body tubes referred to in the question are typically part of medical devices known as endotracheal (ET) tubes, used to secure a patient’s airway during operations or critical care. These tubes are long, semi-rigid tubes, typically made of polyvinyl chloride or some other type of plastic. A typical ET tube has nine basic parts. The proximal end, which sticks out of the patient's mouth, has a standard 15 mm adaptor. This universal size allows it to attach to a variety of equipment like ventilators or bag valve masks.

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Answer:

(a) 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)

(b) 100%, bitter, yellow spotted.

(c) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.

(d) 1:1:1:1 bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots.

Explanation:

(a) The parental cross is BBSS x bbss.

We know that the F1 will be all BbSs (see attached punnet square). The only possible gametes the parental generation can pass on to the F1 are BS and bs, respectively.

The F1 intercross is therefore BbSs x BbSs. the alleles that each can pass on are BS, Bs, bS, or bs. The punnett square attached shows the phenotype ratio is 9:3:3:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. (see colours showing the genotype on the punnett square)

(b) An F1 plant is BbSs. If it is backcrossed with a bitter, yellow spotted parent (BBSS), the cross is BbSs x BBSS (see attached punnett square). The resulting genotypes all give a bitter, yellow spotted phenotypes, so the phenotype of the offspring is 100% bitter, yellow spotted

(c) An F1 plant is BbSs, if it is backcrossed with a sweet, non spotted parent (bbss), the cross is BbSs x bbss. (see attached punnett square). The resulting phenotypes are 1:1:1:1 of bitter fruit, yellow spots: bitter fruit no spots: sweet fruit yellow spots: sweet fruit no spots. I.e. 25% of each genotype

(d) A test cross is a cross between an individual with a known genotype and another plant. The plants showing the recessive trait have a known genotype (they must be homozygous). So in this case, the F1 plant would be crossed with the homozygous recessive plants (BbSs x bbss). Therefore, the resulting phenotypes would be as in option c, 1:1:1:1

Answer:

Cephalization does not confer an advantage to free-floating or sea organisms. Many aquatic species display radial symmetry like the jelly fish for instance.

Aquatic animals are not cephalized because they must be able to find food, take caution and defend themselves from harm from all directions.

Answer:D. the DNA is usually condensed into a chromosome

Explanation:

The bacterial genome is found loose in the cytoplasm but with a form. This unique feature is called the chromosomal DNA and it is not contained within a nucleus in the bacterial cell.

Answer: Option D.

DNA is usually condensed into a chromosome.

Explanation:

Bacterial genome refers to the complete set of DNA or complete set of genetic material of bacteria. Bacteria chromosomes are circular DNA. The chromosomes are packed by histone proteins into a condensed structure called chromatin. The condensed DNA is supercooled.

Answer:

(A)  0.0625

(B)  25%

Explanation:

A

Given that ;

T is tall

t is short

R is red

r is tan.

If a short, tan male goat mates with a tall, red female goat of an unknown genotype;

Let's take it one after the other;

A short tan male goat = ttrr

The female goat is an unknown genotype, we will have to determine that but we were told that it is tall and red

∴ For tall it is T and red is R;

Definitely, the female genotype should be T_ R_;

the probability genotype of the female for each allele is either:

-   TT or Tt

-   RR or Rr

However,  from above;

the probability that the female will be Tt = [tex]\frac{1}{2}[/tex]

the probability that the female will be Rr = [tex]\frac{1}{2}[/tex]

the probability that the recessive allele 't' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]

the probability that the recessive allele 'r' will be transferred to the offspring = [tex]\frac{1}{2}[/tex]

∴ the probability that they would produce short, tan offspring (ttrr) will be;

= [tex](\frac{1}{2})^{4}[/tex]

= [tex](\frac{1}{16})[/tex]

= 0.0625

(B)

One of a pea plants is homozygous recessive for both genes, resulting in the cross Yy Rr × yy rr. Calculate the percentage of green and round offspring.

For Color, Let;

Y = Yellow

y = Green

For shape, Let;

R = Round

r = wrinkled

If a cross occur between  Yy Rr × yy rr.   (i.e yellowround and green wrinkled)

the percentage of green and round offspring will be;

If Yy Rr self crossed; we have the following traits (YR, Yr, yR, yr)

yy rr will result to (yr, yr)

                   YR                    Yr                    yR                    yr

yr                 YyRr                 Yyrr                 yyRr                 Yyrr

yr                 YyRr                 Yyrr                 yyRr                 Yyrr

                   Yellow              Yellow            Green              Green

                   Round               Wrinkled        Round              Wrinkled

Therefore, the percentage of Green and Round offsprings from above is:

= [tex]\frac{2}{8}*100%[/tex]%

=[tex]\frac{1}{4}*100[/tex]%

= 25%

Answer:

The statements in options A,B,C,D and E are all correct.

A. Invertebrates lack a back bone

B. Ecdysozoans typically posses a flexible exoskeleton that protects these animals from harmful external factors like water loss.

C. Deuterstomia, Lophotrochozoa and Ecdysozoa are classes of Bilaterians.

D. Animal A is a bilateral because bilateral embryos are triploblastic possessing three germ layers, are bilaterally symmetrical and lack a back bone.

E. All animal from studies are found to have a common ancestry.

Answer: Phosphodiester bond

Explanation:

The backbone of DNA consists of deoxyribose nucleotides linked by phosphodiester bridges.

The 3'-hydroxyl of the adjacent sugar of one deoxyribonucleotide is joined to the 5'-hydroxyl of the adjacent sugar by an internucleotide linkage called a phosphodiester bond.

Thus, phosphodiester bond is the answer

Answer:

Phosphodiester bond

Explanation:

Nucleotides are linked by phosphodiester bonds that are formed between the 5' phosphate group of one nucleotide and the 3' hydroxyl group of another nucleotide in the DNA.

The tree with the most red trait marks best shows which branches are evolving.

In phylogenetic trees, the gain or loss of a trait is represented by a red trait mark, which shows evolution. The tree that best shows which branches are evolving is the one with the most red trait marks. This indicates that those branches have undergone the most changes in traits over time.

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Answer:

a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra

c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.

Explanation:

Mitochondria contain their own DNA known as mtDNA, a strand which is inherited maternally. mtDNA can be used to find maternal ancestry and cannot be passed down by a paternal parent.

This information provides evidence given that the remains of the teenage girl had mtDNA that was identical to the mother, Tsarina Alexandra. It also gave evidence as to why Anna Anderson's claim of being Anastasia were false as the mtDNA of her and Tsarina Alexandra were not similar.

As to why the other answers cannot support the hypothesis, mtDNA is a maternally passed-down strand making the father, Tsar Nicholas Romanov, from having identical mtDNA to his children. On the other hand, if the teenage girl were to not have matching mtDNA to that of Tsuarina's mtDNA, this would disprove the theory of the girl being related to her.

While the teenage boy also had matching mtDNA with the girl, this only proves that the two were related to one another but not necessarily related to Tsarina unless the mtDNA of all three were proven identical.

The statements supporting the hypothesis concerning the killing of Princess Anastasia with her family in 1918 are stated below:

a. The mtDNA of Anna Anderson bore no resemblance to that of Tsarina Alexandra.  

c. Tsarina Alexandria's mtDNA was found to be a match to that of the teenage girl.  

e. The mtDNA of the teenage girl and boy matched each other.

Thus, the 2007 discovery unequivocally confirmed the hypothesis that Princess Anastasia was killed alongside her family in 1918 during the Russian revolution.

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Final answer:

The chromosomes in a somatic cell of any organism are not all morphologically alike. During mitosis, chromosomes divide and sister chromatids separate, resulting in two nuclei with the same number of chromosomes as the parent cell.

Explanation:

a. False. The chromosomes in a somatic cell of any organism are not all morphologically alike. Somatic cells contain pairs of homologous chromosomes, which are matched pairs containing the same genes in identical locations along their lengths. This means that each homologous chromosome may have different variants called traits caused by differences in the DNA sequence for a gene. So, the chromosomes in a somatic cell can show morphological variation.

b. True. During mitosis, the chromosomes divide and the resulting sister chromatids separate at anaphase, ending up in two nuclei, each of which has the same number of chromosomes as the parental cell. This ensures that each daughter cell receives an identical set of chromosomes to the parent cell.

Answer:

They have contributed same allele for straight hair.

Explanation:

There are two conditions homozygous and heterozygous. In homozygous conditions both the allele of a particular gene is same in an offspring which means the offspring got two same alleles from its parent for a particular gene and if an offspring is heterozygous for a gene that suggests that the parent has contributed two different alleles for that gene in the offspring.

So If Micah is homozygous for straight hair then it can be concluded that his parents have contributed two same alleles for straight hair genes.

If Micah is homozygous for the straight hair gene, both of his parents must have contributed an allele for straight hair. They could be homozygous or heterozygous for this trait.

If Micah is homozygous for the gene that determines straight hair, it means he has two of the same type of allele for that gene - in this case, the allele for straight hair. This indicates that both of his parents must have contributed an allele for straight hair. The simplest scenario is that both of his parents are also homozygous for the straight hair allele. However, it's also possible that one or both parents are heterozygous, possessing one straight hair allele and one potentially recessive curly hair allele. In this scenario, they would have each passed the straight hair allele to Micah.

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Answer:

Here is the full question:

Each of two parents has the genotype blond divided by red​, which consists of the pair of alleles that determine hair color​, and each parent contributes one of those alleles to a child. Assume that if the child has at least one blond ​allele, that color will dominate and the​ child's hair color will be blond.

a. List the different possible outcomes. Assume that these outcomes are equally likely.

b. What is the probability that a child of these parents will have the red divided by red ​genotype?

c. What is the probability that the child will have blond hair color​?

Explanation:

a.) Remember that the pair of allele determines the hair color. So, each of the two parents have the genotype red/blond. Therefore, the possible outcomes are blond/blond, blond/red, red/blond, and red/red.

b.) Since there are four outcomes, We calculate this as P(red/red) = (1/1)/(1/4) = 0.25

c.) There are three outcomes of the child having a blond hair which are blond/blond, blond/red, and red/blond. The probability is therefore P(outcome of blond/total outcome) = 3/4 =0.75

Answer:

O ribosomes bound to rough endoplasmic reticulum, vesicles, Golgi apparatus, vesicles

Explanation:

→Ribosomes  bounded to R.ER are responsible for the mediation of the process of translation and elongation of the amino acids chains during protein synthesis for gebe expression. These R.ER bounded ribosomes are concerned for the synthesis of  proteins for excretion out of the cells.

→As the  protein chains  were  eleongated  on the ribosomes units they are pushed into the   lumen of R.ER into the  intracistern space.Therefore  R.ER is concerned with synthesis and packaging of protein synthesised  by the attached  ribosomes,

→These proteins are  transported in vesicles,( which move through the cytoskeleton of the Cystosol) which budded off from the R.ER membranes.These vessicles transport the synthesized proteins to the Golgi complex.

→The Golgi apparatus receive the  protein from the (R.ER), process it,  and sorted it out  into its vesicles(Golgi vesicles)  for onward  secretion out of the cells through excytosis from the plasma membranes. Example  of this sorting and processing by the Golgi apparatus is the formation of  Glycoprotein by addition of sugars molecules  to proteins .