What are the only two mammals that lay eggs?

Answers

Answer 1
Answer: The platypus and the echidna.
Answer 2
Answer: 1. Duck-billed platypus
2. spiny anteater

Related Questions

Which of the following are adaptations of plants for photosynthesis?A. thylakoid membranes stacked into granaB. broad, flat leavesC. stomata for gas exchangeD. xylem vessels for water transport to the leaves. E. All of these are adaptations for photosynthesis.
What are physical properties? Give some examples.
Study the graph below. At what depth does the thermocline begin? A line graph relating ocean depth to water temperature. At 0 meters, temperature is approximately 27°C. At 100 meters, temperature is approximately 24°C. At 200 meters, temperature is approximately 11°C. At 300 meters, temperature is approximately 8°C.
pepsin is found in the stomach. Salivary amylase is found in the mouth. And arginase is found in the liver. What does the graph indicate about the relative acidity of these three locations?
Which of the following statements correctly describes why a series of closely spaced action potentials causes a sustained contraction rather than a series of closely spaced twitches?a. Ca2+ ions are released quickly from troponin, keeping the Ca2+ concentration in the cytosol high between closely spaced action potentials.b. Release of Ca2+ from the sarcoplasmic reticulum through channels is slow compared to the uptake of Ca2+ into the SR via ATP-dependent pumps, resulting in Ca2+ slowly trickling into the sarcomeres between closely spaced action potentials.c. When a series of action potentials is closely spaced, there is not sufficient time for Ca2+ uptake into the sarcoplasmic reticulum between action potentials, and Ca2+ remains bound to troponin throughout the series.d. Fewer Ca2+ ions are released from the sarcoplasmic reticulum as a result of several closely spaced action potentials than as a result of a single action potential.

Multicellular organisms use cell division, mitosis, for growth and the maintenance and repair of cells and tissues. There are few cells in the body that do not undergo mitosis: most somatic cells divide regularly, some more than others. Single-celled organisms may use cell division as their method of reproduction. Regardless of the reason for mitosis, the process ensures genetic continuity. Consider the model of the cell cycle. Which detail(s) from the model best support the argument that cell division promotes genetic continuity?

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Final answer:

The model of the cell cycle provides details about how cell division promotes genetic continuity, including DNA duplication, accurate segregation of chromosomes, and cytokinesis.

Explanation:

The model of the cell cycle provides several details that support the argument that cell division promotes genetic continuity:

  1. During interphase, the cell grows and the nuclear DNA is duplicated. This ensures that each daughter cell receives an exact copy of the genetic material, maintaining genetic continuity.
  2. In the mitotic phase, the duplicated chromosomes are segregated and distributed into daughter nuclei. This process ensures that the genetic information is accurately passed on to the next generation of cells.
  3. Following mitosis, the cytoplasm is divided through cytokinesis, resulting in two genetically identical daughter cells. This further ensures that each cell receives a complete set of genetic information.

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Draw three water molecules and label any parts or interactions

Answers

Answer:

The roman numerals in blue are the labels I will write below.

Explanation:

i. lone pair of electrons on oxygen atom

ii. covalent bond between oxygen and hydrogen

iii. partial charges that are due to oxygen's high electronegativity. Oxgen will pull the pair of electrons in the covalent bond towards itself and that obtaining a partial negative charge, while hydrogen obtains a partial positive charge.

iv. hydrogen bonds between partially negative oxygen and partially positive hydrogen.

PS. EACH WATER MOLECULE CAN MAKE UP TO 4 HYDROGEN BONDS. ONE WITH EACH HYDROGEN AND TWO FOR EACH ELECTRON LONE PAIR ON OXYGEN. (refer to the middle water molecule in the diagram)

Final answer:

Draw three water molecules and label their parts and interactions, including the hydrogen bond.

Explanation:

  1. Draw three water molecules, each consisting of two hydrogen (H) atoms and one oxygen (O) atom.

  2. Label the parts of each water molecule:

    • The two hydrogen atoms as 'H'.
    • The oxygen atom as 'O'.
  3. Label the interactions between water molecules:

    • The hydrogen bond between the hydrogen atom of one water molecule and the oxygen atom of another water molecule.

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Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross. The progeny from this testcross are:S U Tu 2
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17__
Total 230
a. Determine the order of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. Draw the chromosomes of the parents used in the testcross.

Answers

Answer and Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

  • S U Tu 2
  • s u Tu 70
  • S u Tu 21
  • s u tu 4
  • S U tu 82
  • s U tu 21
  • s U Tu 13
  • S u tu 17

The total number, N, of individuals is 230.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

  • s u TU (70 individuals)
  • S U tu (82 individuals)

Double recombinant)

  • S U Tu (2 individuals)
  • s u tu (4 individuals)

Comparing them we will realize that between

s u TU (parental)

s u tu (double recombinant)

and

S U tu (Parental)

S U TU (double recombinant)

They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- S ---- TU -----U ----

In a scheme it would be like:

Chromosome 1:

---s---TU---u--- (Parental chromatid)

---s---tu---u--- (Double Recombinant chromatid)

Chromosome 2

---S---tu---U--- (Parental chromatid)

---S---TU---U--- (Double Recombinant chromatid)

Now we will call Region I to the area between S and TU and Region II to the area between TU and U.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between S and TU genes, and P2 to the recombination frequency between TU and U.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

  • P1 = (R + DR) / N

        P1 = (21+17+4+2)/230

        P1 = 44/230

        P1 = 0.191

  • P2= = (R + DR) / N

        P2 = (21+13+4+2)/230

        P1 = 40/230

        P1 = 0.174

Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:

P1 + P2= Pt

0.191 + 0.174 = Pt

0.365=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.191 x 100 = 19.1 MU

GD2= P2 x 100 = 0.174 x 100 = 17.4 MU

GD3=Pt x 100 = 0.365 x 100 = 36.5 MU

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

  • observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
  • expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((2 + 4)/230)/0.174x0.191

CC=(6/230)/0.0332

CC=0.7857

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.7857

I = 0.2143            

Three of the most prominent definitions of species are the biological species concept, the phylogenetic species concept, and the morphological species (morphospecies) concept. Match each characteristic with the species concept(s) to which it applies. Species:
1. biological
2. morphological
3. phylogenetic
4. morphological and phylogenetic
5. biological, morphological and phylogenetic
Characteristics:
1) accommodates asexual reproduction
2) not applicable for extinct species
3) relies on similarities of structures
4) used by scientists in classification
5) species acceptance criteria can be subjective
6) based on evolutionary history

Answers

Answer:

Biological species concept defines species as individual in a population that can interbreed and produce viable offsprings. Therefore Biological species concept is not applicable for extinct species

Morphological means in relation to structure therefore, this concept relies on similarity of structure to classify species.

Phylogenetic refers to members having the same ancestor. It classifies species according to common ancestry. Hence it will be based on evolutionary history.

Morphological and phylogenetic- Will accommodate asexual reproduction and species acceptance criteria can be subjective.  

Biological, morphological and phylogenetic-Together they are used by scientists in classification

Final answer:

In the context of biological species, it does not support asexual reproduction and is not for extinct species. Morphological species, rely on structure similarities, are used for classification, and have subjective species acceptance. Phylogenetic species are based on evolutionary history, supports asexual reproduction, and has subjective species acceptance criteria.

Explanation:

The three significant concepts of species include the Biological species concept, Phylogenetic species concept, and Morphological species concept.

1) Biological species - This concept does not accommodate asexual reproduction (2) and is not applicable to extinct species (2).

2) Morphological species (or morphospecies) - This concept relies on similarities of structures (3), is used by scientists in classification (4) and the species acceptance criteria can be subjective (5).

3) Phylogenetic species - This concept is based on evolutionary history (6), can accommodate asexual reproduction (1) and the species acceptance criteria can be subjective (5).

The morphological and phylogenetic concepts are used by scientists in classification (4) and their species acceptance criteria can be subjective (5).

All three concepts - biological, morphological, and phylogenetic - are all considered in the broader understanding and definition of species.

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Help me please , easy 10 points ... just look at the picture & answer the question.

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Copper is metallic !!

A rolling ball has an initial velocity of 5 meters per second. 30 seconds later, its velocity is 2 meters per second. What is the acceleration of the ball?

Answers

Acceleration is defined as the velocity of an object over a period of time.
Mathematically, 
Acceleration, A = Velocity [V] /Time [T]
From the question given above, we are told that 
T = 30 seconds
Initial velocity, V1 = 5
Final velocity, V2 = 2
Change in velocity, V = V1 - V2 = 5 - 2 =3 
A = V / T = 3 / 30 = 0.1
Therefore A = 0.1 m/s^2.