Answer:
Explanation:
To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------40 % Parental (Not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k = 30%
<u>ANSWER:</u>
The correct option is B- Qualitative.
<u>EXPLANATION:</u>
When the growth of corn has to be recorded, the quality of the corn grown in different soils needs to be recorded. Qualitative methods are best used to classify quality of anything.
Qualitative data basically is non-numerical data which in the case of corns can be the length of the corn plant, the sweetness of the corn kernel or the hardness or softness of the corn. This data cannot be recorded using numerical values and so other methods of data recording cannot be used.
Answer:
Valine-Leucine-Proline-Lysine-Histidine
Explanation:
The central dogma of biology is the process by which DNA is used to synthesize RNA and subsequently amino acid sequence (PROTEIN). The processes of transcription and translation is used in gene expression. Transcription is the process whereby the information encoded in a DNA molecule is used to synthesize a mRNA molecule. Transcription is catalyzed by RNA polymerase enzyme, which uses complementary base pairing rule i.e Adenine(A)-Thymine(T), Guanine(G)-Cytosine(C) pairing.
N.B: Thymine is replaced by Uracil in the mRNA
For the above DNA sequence: CAC GAC GGA TTC GTA, the mRNA sequence will be: GUG CUG CCU AAG CAU
Translation is the second process of gene expression which involves the synthesis of an amino acid sequence from an mRNA molecule. The mRNA is read in a group of three nucleotides called CODON. Each codon specifies an amino acid (see attached image for genetic code)
Based on the attached genetic code, an mRNA sequence: GUG CUG CCU AAG CAU will encode an amino acid sequence: Valine(Val) - Leucine (Leu) -Proline (Pro) -Lysine (Lys) - Histidine (His).
GUG specifies Valine amino acid
CUG specifies Leucine amino acid
CCU specifies Proline amino acid
AAG specifies Lysine amino acid
CAU specifies Histidine amino acid
Answer:Naturally occurring,inorganic substance.
Explanation:
The teacher would be well advised to help students encode information about the cell using both visual and verbal forms. This is because, multiple forms of encoding increases the probability of easy retrieval. Encoding in this case involves converting the names of the cell organelles to forms that can easily be remembered.
