Explain why meiosis leads to significant genetic variation while mitosis does not
Questions and Answers
Assignment 1
Question One
If thymine makes up 21 percent of the bases in a certain DNA samples, what percentage of bases is guanine?
There are four bases in a DNA , that is Adenine, Thymine, Guanine and cytosine. Guanine pairs together with cytosine and Adenine pairs together with Thymine. Therefore, given the percentage of one base pair, the percentage of other three bases can be calculated using Chargaff’s rule. Chargaff’s rule states that the concentration of each base pair is always equal to its pair, in other words their ratio is 1:1.
In this case the concentration of thymine is 21%, therefore the concentration of Adenine is 21%. The percentage of both A and T is 21+21= 42%
The percentage of other bases will be= 100- 42=58/2= 29%
The percentage of the other three bases is
A=21%
G=29%
C=29%
Question Two
A virus has a DNA with amount of G that does not equal the amount of C, and the amount of A does not equal the amount of T. What can you conclude about the DNA in this virus?
The DNA in the virus is a single stranded DNA where Chargaff’s rule of molar ratio of A:T and G: C does not apply.
Question Three
Explain why meiosis leads to significant genetic variation while mitosis does not.
During meiosis chromosomes are paired very close together which enables crossing over of genes from one chromosome to another thereby resulting to genetic variation. On the other hand, mitosis involves the duplication of the exact genetic material that are copied to identical cells.
Question Four
The ability to taste the chemical phenylthiocarbamide is an autosomal dominant phenotype, and the inability to taste it is recessive. If a non-taster man with a taster father marries a taster woman who, in a previous marriage, had a non-taster son, what is the probability that their first child will be a non-taster boy? Explain your answer.
The chemical Phenylthiocarbamide tastes bitter to most people but tasteless to some people. In this case, inability to taste the chemical is referred to as autosomal recessive trait and individuals with TT or Tt genotypes have the taster phenotype and individuals with tt genotype have the non-taster phenotype. Therefore, the non-taster phenotype is referred to as recessive trait.
Considering the family pedigree, the man and the woman are heterozygous (Tt), thus p(taster)= ¾, p(nontaster)= ¼
Besides, p(girl) =1/2 and p(boy)=1/2
Therefore, p(taster) = 3/4. p(nontaster) = 1/4. Also, p(girl) = 1/2, p(boy) = 1/2
Thus, the probability that their first child will be a non-taster boy will be ¼* ½ = 1/8
Question Five
In pea plants there are two alleles for the pod color gene: the dominant allele (Y) causes green- colored pods, and the recessive allele (y) causes yellow-colored pods. You are presented with the results of a monohybrid cross in which approximately half of the progeny are yellow and half are green. Which of the following shows the most likely parental genotypes for this cross?
Answer is D: nYy x yy
Question Six
Hemophilia is caused by an X-linked mutation in humans. If a woman whose paternal uncle (father´s brother) was a hemophiliac marries a man whose brother is also a hemophiliac, what is the probability of their first child having hemophilia (note: assume no other hemophilia in the pedigree and no hidden carriers). Explain your answer and draw a pedigree with genotypes for all individuals.
Each son has a 25% probability of having hemophilia, and each daughter also has a 25% probability of having hemophilia. The order of birth does not affect the probability of having hemophilia. Since the normal woman’s father had hemophilia, she is a carrier. She inherited her father’s Xh chromosome and has the heterozygous genotype XXh.
X Xh
Xh XhX XhXh
Y XY XhY
Genotypic ratio is 1:1:1:1
XhX= 25%
Question Seven
In foxes, two alleles of a single gene, P and p, may result in lethality (never born) (PP), platinum coat (Pp), or silver coat (pp). What ratio is obtained when platinum foxes are interbred? Is the P allele behaving dominantly or recessively in causing lethality? In causing platinum coat color?
1) Pp x Pp = PP 25%, Pp 50%, pp 25%
– recessively for lethality, if it was dominant, then Pp’s would die too
– dominantly for platinum coat since a pp would be platinum if p were dominant
2) AB would make him not the father (since types A and B could carry the O allele)
– if he were type 0 and none of the other possible fathers were type 0, then he would be thefather
3) crossover by definition occurs between two different chromosomes – chunks of two chromosomes switch places onto the analogous location on the other chromosome.
Question Eight
A man’s grandfather has galactosemia, a rare autosomal recessive disease. The man married with a woman whose sister had galactosemia. The woman is now pregnant with their first child.
Draw the pedigree as described.
Answer from teacher: The pedigree is wrong.
redraw it to get it right
b. What is the probability that this child will have galactosemia?
¼
c. If the first child has galactosemia, what is the probability that the second child will have it?
¼*1/4 =1/8
Complementaryquestions from teacher
Figure out the probability of both parents being heterozygotes. What is the probability that both parents are heterozygotes?
The man: grandfather = gg, what is then his father ?,
What is the probability that he got the allele g from his father?
If both man and woman are Gghow big is the risk of having a child who is gg¼ ?
Question Nine
Two strains of mice (A and B) are true-breeding (homozygous in all loci) for white coat color. When strain A mice are mated with strain B mice, the F1 offspring all have a brownish coat color. In the F2 offspring, 9/16 of the mice have a brownish coat color, 7/16 have a white coat color. Explain this with diagrams
From the following pedigree, the inbreeding coefficient is (1/2)3 (1 + FC) + (1/2)4 (1 + FB) = 3/8 because FB = FC = 1.
Questions from teacher:
You must explain step by step how to solve the problem.
Assignment 2
Question one
If a geneticist were to closely examine the make-up of a single autosomal chromosome from one of your cells, that chromosome would be found to be: (Explain your choice)
Answer is b: derived entirely from genes from just one of your grandparents, this is because it is only one copy of autosomal dominant inheritance allele on the X chromosome that is needed for an individual to be susceptible to an x-linked disorder.
Answer from teacher: Wrong answer:
?? This has nothing to do with the X chromosome. Remember that in all cells (except the germ cells) you have a chromosome from dad and a chromosome from mom!
Question Two
Color vision depends upon the dominant alleles of three genes – the R gene and the G gene are both on the X chromosome, while the B gene is autosomal. Recessive mutations in any of the three genes can cause color-blindness. Suppose a colorblind man marries a colorblind woman and all of their children have normal vision. What is the genotype of the woman? (Explain your choice)
The answer is b: rrGGbb— The woman with one mutant allele are heterozygous which means they are carriers not color. Besides, females have 2 X-chromosomes and both must carry the mutant allele for them to pass to their children.
Answer from teacher: Wrong answer:
No, remember it is enough to have recessive allele in one of the three genes to become color blind. Think about the differencebetween X-linked and autosomalinheritance.
Question Three
In peas, tall (T) is dominant to short (t) and red flowers (R) is dominant to white flowers (r). A tall plant with red flowers was testcrossed to a short pea plant with white flowers. Half the progeny are tall plants with red flowers and half are short plant with white flowers. Explain the results. Write the parental and F1 genotype.
The parental genotypes Tt Rr
The F1 genotype will be
Tt Rr
Tt TtTt TtRr
Rr RrTt RrRr
After testcrossing the F1 genotype will be TtTt, TtRr, RrTt and RrRr
Answer from teacher:
Punnet Square is WRONG!!!: Think about what’s in each germ cell, an allele from each gene. Here are the two genes T and R. Sometimes with the recessive allele t, sometimes with the dominant T, the same for R.
The condition explained is a dihybrid cross which involves two characters in a single plant where by the pea plant is either tall with white colored flower or short with white colored flowers.
WRONG ANSWER:
No, a test crossingalwaysmeanscrossingwithsomeonewho is homozygous recessive.
Question Four
In the pedigree presented below, the vertical lines stand for steroid sulfatase deficiency, and the horizontal lines stand for ornithine transcarbamylase deficiency. These are separate conditions caused by two different genes.
Is there any evidence in the pedigree that the genes determining the deficiencies are linked?Explain your answer.
No, when an affected daughter of non-founding parents has an affected father, we cannot determine whether the DOMINANT disease is autosomal or x-linked. The affected father can transmit either an autosomal dominant allele, or an X-linked dominant allele to his daughter.
Answer from teacher: This pedigree shows recessive X-linked inheritance for both diseases, can they be linked?
If the genes are linked, is there any evidence in the pedigree of crossingover between them? Yes, because in the bottom family, the two conditions are inherited separately. The parents are both carriers and therefore each have only one version of the gene that, if paired, leads to those conditions. If there were no crossing over and the genes were linked, then each child would have to have either conditions or neither.
WRONG ANSWER:
No bothparents in the lower part do not have the diseasealleles, they come from the same individual. Print genotypes in the pedigree to view.
Question Five
If Y/Y . O/O is crossed to y/y . o/o, and the F1 is testcrossed, what percentage of the testcross progeny will be Y/y . O/o if the two genes are
The cross is Y/Y. O/O x y/y. o/o , thus the F1 would be Y/y. O/o.
SHOW WITH PUNNETSQUARE !!!
unlinked –
If the genes are unlinked, all four progeny classes from the testcross (including y/y ; o/o) would equal 25 %
SHOW WITH PUNNET SQUARE, SHOW CALCULATIONS!!!
completely linked (no crossing-over at all)
With completely linked genes, the F1 would produce only Y O and y o gametes. Therefore, there would be a 50% chance of having y o/y o progeny from a testcross of this F1.
25 map units apart? Explain your answer.
If the two genes are linked and 25 map units apart, 25% of the testcross progeny should be recombinants. This is because the F1 is Y O/y o, y o is one of the parental classes and it should equal 1/2 of the total parental.
Question Six
A rice breeder obtained a triple heterozygote carrying the three recessive alleles for albino flowers (al), brown awns (b), and fuzzy leaves (fu), all paired with their normal wild-type alleles. This triple heterozygote was test-crossed. The progeny phenotypes were:
Are any of the genes linked? If so, draw a map labeled with map distances. (Don´t bother with a correction for multiple crossovers)
In this case I have to find the recombinants…for each
170+150+5+3 =328 recombinants between al and fu
42+38 + 5+3 = 88 recombinants between fu and b
The recombinants are the ones that are not parental…so for example between al and fu, u have to select the progeny numbers that dont have al/fu+ and al+/fu (the parentals) … so you get 5+3+42+38=88
The triple heterozygote was originally made by crossing two pure lines. What were their genotypes?
fu Al 440 b fu al 422 b FU AL 18 B FU AL 442 B FU al 424 B fu al 26 B fu AL 20 b FU al 24 Total 1816
Question SevenThe father of the mermaid Ariel came from the sunken city of Atlantis; Ariel’s mother came from Earth (Gothenburg). People from Atlantis have blue blood (determined by allele B), webbed fingers (determined by allele F), and gills (determined by allele G). All these alleles are dominant to normal Earth alleles. The three loci are autosomal, and they are linked as shown in the linkage map below: B F G 15 m.u. 18 m.u. If Ariel marries an Earth man and there is no genetic interference, what proportion of their children will have blue blood and gills but Earth fingers (no webbed fingers). Assume that Ariel’s dad, mother and husband are homozygous. Explain your answer.
Map distance = % recombination = (# in SCO phenotypes + # in DCO phenotypes x 100) (total # progeny)
(ba) Map distance = 112 + 102 + 18 + 15 x 100 = 24.7% = 24.7 m.u. 1000 (ac) Map distance = 66 + 59 + 18 + 15 x 100 = 15.8% = 15.8 m.u. 1000 (bc) Map distance = 24.7 m.u. + 15.8 m.u. = 40.5 m.u
There is some probability that a cross-over will certainly occur to form between Allele a and loci b which is equal to the map distance between a and b . Besides, another independent probability that a cross-over will occur between b and loci c which in this case will be equal to the map distance between b and c. The probability of a double cross-over is the product of these two independent probabilities.
Assignment 4
Question one
Match each with the correct definition
Transcription factor- J. Binds DNA to regulate gene expression
SNP- A. A DNA substitution present at a frequency greater than 1%
Ribosome- D. Has the anticodon UAC
Alternative splicing- H. Process by which multiple unique proteins are produced from a single gene
Primary transcript- B. Will ultimately be exported out of the nucleus
tRNAMet- . Contains a “P” site and an “A” site
Proteasome- F. Its substrates are ubiquitinated proteins
Stop codon- I. Serves to terminate translation
Degenerate- G. Describes the triplet code
Spliceosome- E. Is composed of proteins and small RNAs
Question two
Why is it essential that DNA be spooled around histones in the nuclei of eukaryotic cells?
DNA is wrapped around a group of proteins known as histones to form nucleosome in the nuclei of eukaryotic cells.
This is essential as it helps to compact the DNA strands and impact chromatin regulation. Chromatin in this case refers to a combination of DNA and protein thereby making up the contents of cell nucleus.
Question three
If thymine makes up 14% of the bases in a specific DNA molecule, what are the percentages of the other three bases (A, G and C) in this molecule?
There are four bases in a DNA , that is Adenine, Thymine, Guanine and cytosine. Guanine pairs together with cytosine and Adenine pairs together with Thymine. Therefore, given the percentage of one base pair, the percentage of other three bases can be calculated using Chargaff’s rule. Chargaff’s rule states that the concentration of each base pair is always equal to its pair, in other words their ratio is 1:1.
In this case thymine the concentration of thymine is 14%, therefore the concentration of Adenine is 14%. The percentage of both A and T is 14+14=28
The percentage of other bases will be= 100- 28=72/2= 36%
The percentage of the other three bases is
A=14%
G=36%
C=36%
Question Four
A scientist extracts and purifies different molecules needed for DNA replication. To the solution of the newly purified molecules she also adds DNA. Replication occurs, however, the new DNA molecules are defect. Half of the newly synthesized DNA molecules consist of one normal DNA strand paired with a number of DNA-fragments (about 200 nucleotides long). What molecule did the scientist forget to add to the reaction? Briefly explain your answer.
The scientists forgot to add the primary enzyme that is involved in DNA replication known 0as DNA polymerase. DNA polymerase is essential in joining nucleotides to synthesize the new complementary strand of the DNA. Besides, the enzyme proofreads each new DNA strand to make sure that there are no errors in DNA replication process.
Question five
What are the three key properties of the hereditary material?
-Hereditary materials is made up of four bases which includes A, G, C and T.
-Hereditary material is coiled up into chromosomes in the nucleus of a cell.
-It’s sequencing is unique in humans except in rare cases of identical twins.
Question SIX
If you have an artificial messenger RNA with the sequence AUGUUUUUUUUUU…, it will produce the following polypeptide in a cell-free, protein-synthesizing system: Met-Phe-Phe-Phe-… In your search for new antibiotics, you find one agent which blocks protein synthesis. When you try it with your artificial mRNA in a cell-free system, the product is Met-Phe. What step in the protein synthesis does this agent affect? Why?
Elongation step, this is because the elongation process of assembling the amino acids into a polypeptide cannot take place and it is prematurely terminated to form a small peptide Met-Phe instead of a polypeptide Met-Phe-Phe-Phe………… Phe.
Question Seven
Most proteins have more leucine than histidine residues, but more histidine than tryptophan residues. Correlate the number of codons for these amino acids with this information.
Leucine – 6 instances (CUU, CUA, CUC, CUG, UUA, UUG)
Histidine – 2 instances (CAU, CAC)
Tryptophan- 1 instance (UGG)
Question eight
The normal sequence of a protein is given here, along with several mutant versions of it. For each mutant, explain what type of mutation occurred in the coding DNA-sequence of the gene: Normal: Met-Gly-Glu-Thr-Lys-Val-Val-…-Pro
a) Mutant 1: Met-Gly–Deletion mutation- The mutant has deleted the section of the DNA to Met-Gly
b) Mutant 2: Met-Gly-Glu-Asp- Insertion mutation- The mutant has inserted extra base pair (Asp) to the DNA sequence.
c) Mutant 3: Met-Gly-Arg-Leu-Lys –Frameshift mutation- The mutant has led to insertion of amino acid (Arg and Leu) and deletions of amino acid (Thr) which leads to the altering of the DNA sequence.
d) Mutant 4: Met-Arg-Glu-Thr-Lys-Val-Val-…-Pro- Substitution mutation- The mutant has exchanged one amino acid (Gly) for another (Arg).
Question Nine
Many Noble prizes in category of physiology/medicine (also in chemistry) have been awarded to Genetics-related research. Choose one Genetics-related Noble prize and write a page about the prize: background to the prize, a bit about the Nobel Laureate, his/her research topic, why the prize went to this specific topic (what is important about it), when the prize was awarded, and any other information that you find interesting. Your explanation and understanding of the research part is important. (one page max)
One of the Genetics-related Noble prizes is the Nobel Prize in physiology and medicine for 2007 which was awarded by the Nobel Assembly at Karolinska Institute jointly to Mario R. Capecchi, Martin J. Evans and Oliver Smithies. The prize was as a result of their contribution to discoveries of principles for introducing specific gene modification in mice using embryonic stem cells.
Oliver Smithies and Mario Capecchi had the vision that homologous recombination could be utilized to modify genes in mammalian cells which made them to work closely towards this goal. Capecchi demonstrated that homologous recombination could take place between DNA and the chromosomes in mammalian cells. He showed that defective genes could be repaired by homologous recombination with another DNA.
It was these attempts that led Smithies to find out that endogenous genes could be targeted irrespective of their activity. This suggested that all genes may be accessible to modification by homologous recombination. The cell types that were initially studied by Smithies and Capecchi could not be utilized to make gene targeted animals as this needed another kind of a cell which could give rise to germ cells. This made Martin Evans to work with mouse embyonic carcinoma cells which could give rise to any cell. In his attempts he found out those chromosomally normal cells cultures could be established directly from mouse embryos. Besides, he could show that embryonic stem cells could contribute to the germ line.
Their discoveries was significant as it led to the creation of an immensely powerful technology known as gene targeting in mice which is now being applied to almost all fields of biomedicine from basic research to the development of new medicines.
Assignment 5 Alternative to PopGenLab
Question One
What are the forces that can change the frequency of an allele in a population?
The frequency of an allele in a population can be altered or changes by different forces which include genetic drift, mutation, natural selection and migration.
Question two
Evaluate and comment on the following statement: “Inbreeding increases the frequency of recessive alleles in a population”.
Inbreeding increases the frequency of recessive alleles in a population by decreasing the genetic variation and increasing homozygosity in the genomes of their offsprings. This therefore means that the chances of recessive alleles to pair are significantly higher.
Question three
Calculate the frequencies of the AA, Aa, and aa genotypes after one generation if the initial population consists of 0.2 AA, 0.6 Aa, and 0.2 aa genotypes. What genotypes will occur after a second generation?
Frequency of A= p= 0.2 +0.6/2= (0.2 +0.3) = 0.5
Then in the second generation, the genotype frequency that will occur will be,
Frequency of AA= (0.5)2 = (0.5)2 = 0.25 OR 25%
Frequency of Aa = 2pq =2pq= 2 (0.5)(0.5)= 0.5 or 50%
Frequency of aa= q2 = (0.5)2 = 0.25 or 25%
Question four
If the recessive trait albinism (a) is present in 1/40,000 individuals in a population at equilibrium, calculate the frequency of
(a) The recessive mutant allele (heterozygotes in the population
In assuming the population is in H-W equilibrium, then the frequency of individuals with the albino genotype is the Frequency of (aa) =Q2
Frequency of the recessive mutant allele will be 1/40000=0.000025 Square root= 0.005
Therefore the frequency allele of albinism is 0.005
(C)The frequency of the normal allele
The frequency of the normal allele is equal to 1-q = 1 -0.005= 0.995
(d) mating between heterozygotes
. We’d then predict that the frequency of Hopi who are homozygous normal (genotype AA) is p2 , which is 0.873. In other words, 87.3% of the population, or an estimated 5238 people, should be homozygous normal. The frequency of carriers we’d predict to be 2pq, which is 0.123. So 12.3%, or 737 people, should be carriers of albinism, if the population is in H-W.
Wrong answer
Answer from teacher:
Hereyoushouldcalculate the risk of twoheterozygotesAamating. (Aa x Aa) Aa = 2pq
Question five
Tay-Sachs disease is caused by mutations in a gene on chromosome 15 that encodes a lysosomal enzyme. Tay-Sachs is inherited as an autosomal recessive condition. Among Jews of central European ancestry, about 1 in 3600 children is born with the disease. What proportion of the individuals in this population are carriers?
There are 1 in 2 chance for one to be the carrier of TaySach disorder. Thus the proportion of carriers of individuals in the given population will be
3600/2 X 1= 1800
The proportion of the carriers in the population given will be 1800
Answer from teacher: Wrong answer:
No: keep in mind that q = √ (1/3600)
What is then 2pq ???
Question six
In a population, the D è d mutation rate is 4 x 10–6 . If p = 0.8 today, what will p be after 50,000 generations?
0.65
Answer from teacher: Correct answer!!
But you must show how you did the calculations
Question seven
Eugenics is the term employed for the selective breeding of humans to bring about improvements in populations. As a eugenic measure, it has been suggested that individuals suffering from serious genetic disorders should be prevented (sometimes by force of law) from reproducing (by sterilization, if necessary) in order to reduce the frequency of the disorder in future generations. Suppose that such a recessive trait is present in the population at the frequency of 1 in 40,000, and that affected individuals do not reproduce. In 10 generations, or about 250 years, what would be the frequency of the condition? Are the eugenic measures affective in this case
1:40000
In 10 generations
40000*10=400,000
The frequency of the condition will be 1 in 400,000.
The eugenic measures will be effective in this case as the frequency of the condition to the population will be reduced to 1 in 400,000 after 250 years.
ANSWER FROM TEACHER: NO: This is wrong. I
t was q = √(1/40 000) at the start !!!
Howmanyafter 10 generations? Use the followingequation: q_g=q_0/(1+〖gq〗_0 )
