Three questions to answer you can do however much you want but PLEASE ANSWER
1 answer:
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Answer and Explanation:
the total length of the gene is 6,000 base pairs (or) 6.0 kb and it has five exons, four introns and a 1,000 base pair length flanking region for the transcribed region.
Hence, each exon has (1500/5) = 300 by
each intron has (3500/4) = 875 by and
flanking segment (5' and 3') length for the transcribed region would be (1000/2) = 500 bp.
The gene segment is as shown in the <em>1st diagram attached below </em>
In eukaryotes, transcription takes place in the nucleus which results in the synthesis of hnRNA (heteronuclear ribonucleic acid). This hnRNA (pre-RNA) has both introns and exons, however when it is synthesized or while it is being synthesized the Spliceosomal complex removes introns from hnRNA and becomes converted into mRNA (messenger RNA) in the nucleus itself.
(<em>second diagram explains further)</em>
Now the synthesized mRNA reaches the cytoplasm. So, if mRNA collected from the nucleus and cytoplasm will have the same length of about 2,500 base pairs only. Therefore, after the commencement of northern blot analysis both mRNA will yield the same kind of result and have unique length.
Answer: The probability that individual 1 is heterozygous is 1.
The probability that individual 3 is heterozygous is 1.
The probability that individual 4 is affected is 0.
Explanation: Since the father was said to be affected, the genotype of the father will be homozygous recessive. Using (A) to represent the dominant allele and (a) to represent the recessive allele, we can say that the father's genotype is aa.
Since the genotype of the mother was not given, we assume that she has wild-type genotype which is homozygous dominant allele because no contrary information was given. We can then say that her genotype is AA.
A cross between aa and AA will produce offspring with genotype Aa only. Therefore, we can conclude that 1 - 4 have heterozygous alleles.
Check the attached image for better illustration.
