Which of these would most likely happen if there is a sudden increase in the number of sunspots?
C. Earth's climate would become warmer
The correct answer is option C.
The sudden increase in sunspots will make the Earth more warmer because the more the sunspots, more energy is delivered to the atmosphere. This increases the global temperatures of the Earth.
Answer:
The correct answer is C.
Explanation:
Sunspots increase and decrease through an average cycle of eleven years.
There is a debate of how much this sunspot would affect the earth`s temperature, the earth`s climate is sensitive to very weak changes in the Sun`s energy. During maximum sunspot activity, there`s a very slight increase in the energy output of the sun. Ultraviolet radiation increases during high sunspot activity, and this can have a large effect on Earth`s atmosphere.
I hope it helps!
b. insertion mutations can only be silent mutations.
c. insertion mutations can affect many amino acids in the protein.
d. insertion mutations are chromosomal mutations?
The study of cells is called cell biology. There are two types of a cell on the basis of their number and these are unicellular and multicellular.
The correct answer to the question is option C that is insertion mutations can affect many amino acids in the protein.
The formation of protein is called translation. This addition of genes changes the sequence of RNA which alters the protein formation while in point mutation it changes only one amino acid.
Hence, the correct answer is option C.
For more information about the transcription, refer to the link:-
Answer:
the sun
Explanation:
The correct answer is the option (A) heparin.
DVT refers to a condition called the deep vein thrombosis when a blood clot occurs in the deep veins of the legs in the body. PE refers to the Pulmonary embolism when a blood clot is observed in the artery of the lungs.
The drugs used in the treatment of DVT and PE are called the anticogulants which interfere with the process of blood clot formation. Heparin and warfarin are the two older anticoagulants used in the prevention and treatment and prevention of DVT and PE. Heparin is given intravenously to treat DVT and PE whereas, warfarin is taken as a tablet or pill through the mouth. Heparin allows the body's clot lysis mechanism to break down the existing clots and it is also a natural anticoagulant produced by the basophils and the mast cells of the blood.
Thus, heparin is the old drug that is used subcutaneously to prevent DVT in high-risk patients and intravenously to treat DVT and PE.
DNA 2: TAC CCG ACG GGC GAT AGT TTT
a) What types of mutation have occurred in each of the DNA 1 and DNA 2 sequences?
b) Explain how these mutations affect the amino acid sequence produced and the overall effect on the final protein. Which mutation will cause the most disruption to the overall protein structure and why?
(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?
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:
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:
Parental) 6cm x 30cm
F1) 18 cm
Parental) 18 cm x 18 cm
F2) 6, 10, 14, 18, 22, 26, 30
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:
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
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
abc AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc
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.
B. Mutation
C. Independent assortment
D. Random fertilization
E. All of the above
Answer:
The correct answer is E. All of the above
Explanation:
Variation in offspring can be contributed by many different events that change the DNA sequence in gametic cells. These events are crossing over, mutation, independent assortment, and random fertilization.
During the gamete formation, the different genes are assorted independently from the other gametes which results in forming various possible combinations of genes in a gamete increasing the genetic variation.
Crossing over occurs during meiosis between the homologous chromosome that results in exchange of genetic material that brings genetic variation in gametes. During random fertilization, the two gametes fuse and genetic material of two different individual comes to make an offspring which forms a genetical different offspring.
Mutation in the gametic cell is transferred to the offspring during fertilization. Therefore all contributes to variation in offspring produced by sexual reproduction.