Answer:
strength
Explanation:
Skeletal system of the human body has 206 bones. Human skeleton is divided into appendicular skeleton and axial skeleton.
The skeletal system provide support to the other parts of body, chemicals like calcium are stored in the skeletal system and red bone marrow is the site of hematopoiesis. The skeletal system doesnot provide strength because muscles are mainly involved in providing strength to the body.
Thus, the correct answer is option (3).
Mendel was the first to explain the transmission of phenotypic characters and the independent assortment of the genes.
In the given data, F2 progeny of phenotypic traits are shown.
In the above observations, the gray body long wings and ebony body vestigial wing are parental combinations.
Also, the gray body and vestigial wings, and ebony body long wings are the recombinants.
The given ratio in the progeny indicates that gray bodies and long wings are expressed.
The genes for the two traits are independently assorted, which means that genes are unlinked present on the samechromosome.
Now,
For the F1 progeny:
The cross between Gl and hybrid will result in a 50% chance of flies having the gray body and vestigial wings.
b) In the above given F2 progeny, the cross between true gray body and long wings with true ebony body and vestigial wings, will result in the independent assortment of the genes.
Given:
For the F1 progeny, all the offspring will have genotype GgLl (Gray body and long wings but in heterozygous condition)
The above cross can be shown in the Punnett square, which is given in the attachment below.
To know more about Mendelian ratio, refer to the following link:
Answer:
Check the explanation
Explanation:
Consider the below table of F2 data of phenotypic flies: we can reconstruct the table as seen in the second attached image below:
From the above observations it is clear the Gray body long wings and Ebony body vestigial wings are the parental combinations and that the Gray body vestigial wings and Ebony body long wings are the recombinants.
Also the ratio indicates that Gray body is dominant over ebony body and that long wings is dominant over vestigial wings.
here the genes for the two characters have shown independent assortment which means that the genes are unlinked if located on the same chromosome or are located on different chromosomes.
Now F1 hybrid= GgLl (G for Grey and L for Long)
Cross between F1 hybrid and true breeding Gray vestigial (GGll)
GgLl x GG ll
Gametes-----------> GL Gl gL gl Gl
GL Gl gL gl
Gl GGLl GGll GgLl Ggll
(Gray long) (Gray vestigial) (gray Long) (Gray vestigial)
Therefore the probability of getting the flies with gray body vestigial wings= 2/4= 50%
b) The reason why the students obtained the above given F2 results involving a cross between true breeding Gray body and long wings with true breeding Ebony body and vestigial wings is that the genes for the two charcters asssort independently in F2 generation and that the genes are not linked as:
Parents------------------> GGLL x ggll
Gametes -----------------> GL gl
F1---------------------> GgLl (Gray long but in heterozous condition)
Now GgLl x GgLl
Gametes GL Gl gL gl GL Gl gL gl
Here gametes assorted independently and hence in F2 generation we got the above results (U can show the results in the form of punnett square.
4x-2y=12
plug in 0 to find y or x intercept: 4x-2(0)=12 4(0)-2y=12
4x=12 2y=12
x=2 y=-4
(2,0) (0,-4)
so the y-intercept is -4 and x-intercept is 2
Answer:
The number of hydrogen bonds involved will be 39
Explanation:
There are 4 types of bases that exist in a DNA, which are adenine, thymine, guanine and cytosine. Double stranded DNA molecules have these bases attaching specifically to one another; adenine only binds to thymine and vice versa while guanine only binds to cytosine and vice versa. These two bases bonding together are referred to as base pairs and the type of bond here is the hydrogen bond. There are double bonds between the adenine and thymine base pair while there are triple bonds between the guanine and cytosine base pairs.
Thus, when there are 13 cytosine bases in a DNA molecule, the number of hydrogen bonds present in the resulting base pairs will be 13 × 3 (because cytosine binds with a triple bond to guanine).
13 × 3 = 39
The number of hydrogen bonds involved will be 39
So, in a DNA molecule of 50 base pairs that contains 15 cytosine (C) bases, there would be a total of 115 hydrogen bonds involved in base pairing.
In DNA, base pairing occurs between complementary nitrogenous bases. Adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). Each base pair is connected by hydrogen bonds.
In a DNA molecule of 50 base pairs with 15 cytosine (C) bases, you can determine the number of hydrogen bonds involved in base pairing as follows:
1. Each adenine (A) pairs with thymine (T) and forms two hydrogen bonds.
2. Each cytosine (C) pairs with guanine (G) and forms three hydrogen bonds.
So, for the 15 cytosine (C) bases, you would have 15 pairs of C-G base pairs, and for the remaining 35 bases, you would have 35 pairs of A-T base pairs.
Total hydrogen bonds involved in base pairing:
(15 pairs of C-G base pairs * 3 hydrogen bonds per pair) + (35 pairs of A-T base pairs * 2 hydrogen bonds per pair)
= (15 * 3) + (35 * 2)
= 45 + 70
= 115 hydrogen bonds.
So, in a DNA molecule of 50 base pairs that contains 15 cytosine (C) bases, there would be a total of 115 hydrogen bonds involved in base pairing.
#SPJ3
A. thin brown stalks
B. the sperm
Answer:
The sporophyte stalk, called the seta, bears the sporangium (spore capsule) on its tip. One sporangium may produce up to a million spores.
Answer:
The correct answer is 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation.
Explanation:
ATP and NADPH molecules are synthesized during light reaction of the photosynthesis which is utilized during the reactions of the dark phases. Dark reaction or Calvin cycle (C3 cycle) is the cyclic pathway of producing glucose triose phosphates (3C) from carbon dioxide and water. This reaction proceeds into 3 phases: carboxylation, reduction and regeneration.
First ATP and NADPH are utilized during the reduction step in the reduction of 3-phosphoglycerate to glucose-3-phosphate by the transfer of phosphate group from 6 ATP to 3-phosphoglycerate and 6 NADPH reduction as it donates an electron. Regeneration step also uses 3 ATP in conversion of G3P to RUBP molecules.
A total of 9 ATP and 6 NADPH are utilized in producing 3C G3P molecule, So, to produce 6C sugar molecule 18 ATP AND 12 NADPH are used.
During chemiosmosis synthesis of ATP, 4 protons produce 1 molecule of ATP which could be able to generate 3 molecules of ATP for each pair of NADPH. so, noncyclic photophosphorylation or Z-scheme will be able to produce ATP in the absence of cyclic photophosphorylation.
Thus, 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation are the correct answer.
Answer:
cells have different shapes because they do different things
Explanation:
each cell type has has its own role to play in helping our bodies
to work properly