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Your Genome. J1. p. chr1. m. J1. chr2. chr22. (and sex chromosomes). J1. 3 ’. 5 ’. 5 ’. 3 ’. 12K base pair. Transcription. ?. Splicing. Translation. Protein is 200 Amino Acids. J1. 3 ’. 5 ’. 5 ’. 3 ’. 12K base pair. Transcription. J1. 3 ’. 5 ’. 5 ’. 3 ’.
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Your Genome J1 p chr1 m J1 chr2 . . . chr22 (and sex chromosomes)
J1 3’ 5’ 5’ 3’ 12K base pair Transcription ? Splicing Translation Protein is 200 Amino Acids
J1 3’ 5’ 5’ 3’ 12K base pair Transcription
J1 3’ 5’ 5’ 3’ 12K base pair Transcription Pre-mRNA (12K bp)
J1 3’ 5’ 5’ 3’ 12K base pair Transcription Pre-mRNA Slicing mRNA AAAAAAAAAAAAAAAAA (~1000 bp)
J1 3’ 5’ 5’ 3’ 12K base pair Transcription Pre-mRNA Splicing mRNA AAAAAAAAAAAAAAAAA (1000 bp) Exons “exit” the nucleus UTR UTR AAAAAAAAAAAAAAAAA 600 bp AUG stop
J1 3’ 5’ 5’ 3’ Transcription Pre-mRNA Splicing mRNA AAAAAAAAAAAAAAAAA Exons “exit” the nucleus UTR UTR (1000 bp) AAAAAAAAAAAAAAAAA AUG stop Translation
We did a protein purification found the following: • J1 of 200 amino acids • J1 of 150 amino acids • J1 of 130 amino acids How is this possible? Pre-mRNA Alternative Splicing mRNA 200 AAs AAAAAAAAAAAAAAAAA or 150 AAs AAAAAAAAAAAAAAAAA or 130 AAs AAAAAAAAAAAAAAAAA
Product rule The probability of two or more independent eventsall occurring (event #1 and event #2 and…) = the product of the individual probabilities Example: what is the probability of throwing dice two times and getting a five each time? = 1/36 1/6 X 1/6
Sum rule The probability of any one of 2 or more mutually exclusive outcomes (outcome #1 or outcome #2 or…) = sum of the individual probabilities Example: what is the probability of throwing dice a single time and getting a six or a four? or = 1/3 1/6 + 1/6
Aa Aa Two events necessary: II-3 must be Aaand they must have aa child A a A a Product rule a = no pigment Example: Albinism… What is the probability that III-1 will be affected? P(II-3 is Aa) = 2/3 P(Aa x Aa giving aa) = 1/4 P(III-1 is aa) = 2/3 x 1/4 = 1/6
Sum rule The same pedigree… What is the probability that III-1 will be homozygous? Need III-1 = AA or aa. Possibilities: Aa father has aa child: 2/3 x 1/4 = 1/6 Aa father has AA child: 2/3 x 1/4 = 1/6 AA father has AA child: 1/3 x 1/2 = 1/6 Therefore, probability of homozygous child = 1/6 + 1/6 + 1/6 = 1/2