# Problem 1 The haploid human genome is ~3x10 9 base pairs. If we assume the final estimate of human genes will be 100,000 (It is currently ~20687 protein.

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Problem 1 The haploid human genome is ~3x10 9 base pairs. If we assume the final estimate of human genes will be 100,000 (It is currently ~20687 protein coding for Homo sapians and — 17,387 for C. elegans. Of those genes the function for ~5898 are unknown. Note: humans have between 10-100 trillion cells and C. elegans have 959 cells.) and that the average human gene coding sequence is 2kb, what can be concluded about the average spacing between genes measured from the central point of one gene to the central point of the next. What can be concluded about the proportion of the genome that codes for protein? What occupies the noncoding regions? The E. coli DNA polymerase III makes a mistake ~1 in 10 8 nucleotide. After DNA replication approximately how many cells out of 100,000 cells would E. coli DNA polymerase III make a nucleotide mistake in the gene coding sequence of the haploid human genome. The genome of the bacterium nicenumberas is 1000 kb in size, sequencing has shown it has 1000 genes and it has an average gene size of 1kb. After DNA replication approximately how many cells out of 100,000 would E. coli DNA polymerase III make a mistake in a gene of the nicenumberas bacterium genome.

Answer Problem 1 The average spacing between central points of adjacent genes must be 3,000,000kb/100,000genes which equal 30 kb/gene Since the protein-coding segment is only 2kb, it is ~2kb/30kb or 6% of the total sequence The remaining sequence must be composed of regulatory regions adjacent to genes, as well as repetitive DNA and nonrepetive spacer DNA between adjacent protein-coding regions and introns within the protein-coding regions Since only 6.667% of the genome is protein-coding, only 2x10 8 nucleotides code for protein (0.0667x3x10 9 or 2000*100,000). Because E. coli DNA polymerase makes a mistake approximately every 1x10 8,~2 genes would have a mistake in each cell (so of the 100,000 cell approximately every cell would have roughly two genes with a mistake in their sequence). For nicenumberas 100% of genome is genes (1000kb/1000gene=1000bp/gene so a 1000 bp gene/1000bp/gene*100=100%) In 100,000 approximately 1000 cells would have an error in one of its genes. (100,000 cellx1,000,000 bpx1/1x10 8 errors/bp)

Problem 2 Draw a pedigree from the words of the song I’m own grandpaw and indicate the degrees of separation from the proband Many, Many years ago when I was twenty-three I was married to a widow who was pretty as could be. This widow had a grown up daughter who had hair of red My father fell in love with her and soon, they too, were wed Chorus This made my dad my son-in-law, and changed my very life. For my daughter was my mother cause she was my father’s wife. To complicate the matter even though it brought me joy, I soon became the father of a bouncing baby boy. Chorus My little baby then became a brother-in-law to Dad, And so became my uncle, though it made me very sad. For if he was my uncle then that also made him brother Of the widow’s grown-up daughter who, of course, was my stepmother. Chorus Father’s wife then had a son who kept them on the run. And he became my grandchild for he was my daughter’s son. My wife is now my mother’s mother and it makes be blue Because, although she is my wife, she’s my grandmother too. Chorus Oh, if my wife is my grandmother, then I am her grandchild And every time I think of it it nearly drives me wild For now I have become the strangest case you ever saw; As husband of my grandmother I am my own grandpaw. Chorus I’m my own grandpaw It sounds funny, I know, But it really is so; I’m my own grandpaw

Answer to problem 2 1o1o 1o1o Proband I was married to a widow who was pretty as could be This widow had a grown up daughter who had hair of red My father fell in love with her and soon, they too, were wed 2o2o 1o1o I soon became the father of a bouncing baby boy Father’s wife then had a son who kept them on the run Degrees of separation based on number of different meiosis from proband

Problem 3 1 2 The accompanying pedigree was obtained for rare kidney disease. From the pedigree deduce what the mode of inheritance is, state your reasons. If individuals 1 and 2 marry, what is the probability that their first child will have the kidney disease.

Answer Problem 3 1 2 From the pedigree deduce what the mode of inheritance is, state your reasons Likely to be autosomal recessive. Affected individuals inherited from unaffected parents. It affects both males and females inherited the disease what is the probability that their first child will have the kidney disease.

Answer Problem 3 1 2 If individuals 1 and 2 marry, what is the probability that their first child will have the kidney disease. Both parents must be heterozygous for them to have a ¼ chance of passing the disease For individual 2, the parent is heterozygous because her father in homozygous For individual 1, has ½ chance because her father is heterozygous So the chance that any offspring will be affected is 1*1/2*1/4=1/8

Problem 4 Considering the accompanying pedigree of a rare autosomal recessive disease, PKU. A. List genotypes of many of the family members as possible B. If individuals A and B marry, what is the probability that their first child will have PKU C. If their first child is normal, what is the probability that their second child will have PKU? D. If their first child has the disease, what is the probability that their second child will be affected? A B I II III IV

Answer Problem 4 A. List genotypes of many of the family members as possible A B Because the disease PKU is a rare autosomal recessive disorder the affected males must be homozygous. Because the disease is rare it is a good assumption that the individuals marring into the family are homozygous for the wildtype p/p P/P P/p P/P I II III IV For II because the mothers are homozygous for the wildtype and the fathers are homozygous for the disease gene they must be heterozygous P/p P P pp The remaining family member could be homozygous for the wildtype or heterozygous P/?

Answer Problem 4 B. If individuals A and B marry, what is the probability that their first child will have PKU A B I II III IV We know that for the grandmother of individual A is heterozygous and the grandfather is homozygous. Therefore the chance that individual A’s father is heterozygous is ½. We know that individuals mother is homozygous. If individual A father (1/2 chance) has the mutant allele he has ½ chance of passing the allele on. Therefore individual a has ½*1/2=1/4 of being a carrier of the mutant allele. P/P P/p A P/P P/pP/P P P Pp 1/2 ¼ P/p 1/2 P/p

Answer Problem 4 B. If individuals A and B marry, what is the probability that their first child will have PKU A B I II III IV We know the grandmother and grandfather of individual B are heterozygous. Therefore the chance that individual B’s mother is heterozygous is 2/3. We know that individual B’s father is homozygous. If individual B mother (2/3 chance) has the mutant allele she has ½ chance of passing the allele on. Therefore individual B has 2/3*1/2=1/3 of being a carrier of the mutant allele. P/p A P/P p/pP/p P p Pp 2/3 X So if individual A is heterozygous ¼ and individual B is heterozygous 1/3. The chance of having a homozygous child is ¼*1/3*1/4=1/48 ¼ P/p 1/3 P/p P/pP/P p/pP/p P p Pp 1/48 ¼ chance to be homozygous 2/3 P/p

Answer Problem 4 C. If their first child is normal, what is the probability that their second child will have PKU? If the first child is normal we do not gain any additional information, so the probability of having an affected child is still 1/48 D. If their first child has the disease, what is the probability that their second child will be affected? If the first child has the disease we know that both parents must be heterozygous, therefore the chance for the second child is affected is ¼. A B I II III IV

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