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GENETIC CODE FILE 8: POINT MUTATIONS 1. In a mRNA sequence (wt) there is a triplet UUU. After a mutation, the triplet changes in UUA. What kind of mutation happened and which effects have the mutation on the protein encoded by the gene? To switch from UUU to UUA is necessary a mutation in the DNA From TTT to TTA AAA AAT This is a Transversion: it refers to the substitution of a purine for a pyrimidine or vice versa, in deoxyribonucleic acid (DNA). The aminoacid Phenylalanine encodes for UUU will be substitute with Leucine (UUA). This is a Missense mutation. The effect on the protein function is not predictable. 2. This is a sequence of wt mRNA 5’- AUG AGA CCC ACC…. What kind of effects has a mutation the substitute the fifth base from G to U? 5’– AUG AUA CCC ACC… In the second triplet the mutation changes the codon and the aminoacid encoded: Arg (Encoded by AGA) to Ile (encoded by AUA). What kind of effects has a deletion of the sixth base? 5’– AUG AUA CCC ACC… The deletion causes a frameshift of the codon code. All the aminoacids after will be different. Compare the two previous mutations: First case: Probably the protein will be active (Missense mutation) Second case: the protein is totally different from the original protein and probably will not be active. (FrameShift Mutation). 3. A nucleotide sequence below 5 5' 3' 10 15 20 25 30 35 ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC TAAGCTACCCTACCGTCACGGTTTCACCACTACCG 3' 5' a) Knowing that the transcription of this sequence is from left to right (5’->3’), write the resulting mRNA sequence: 5’ AUUCGAUGGGAUGGCAGUGCCAAAGUGGUGAUGGC b) Knowing that this mRNA sequence contains the translation start codon, identify the starting codon and indicate the amino acid sequence of the resulting peptide. Met- Gly- Trp-Gln-Cys- Gln- Ser-Gly- Asp- Gly 5' 3' ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC TAAGCTACCCTACCGTCACG GTTTCACCACTACCG 3' 5' C) Find which consequences will have on the amino acid sequence: - a transition of the T base pair in position 18; In the DNA sequence: switch from TA to CG; In the mRNA sequence there is a switch between U and C thus between the triplet UGC (Cys) to CGC (Arg) that causes a MIS-SENSE mutation. 5' 3' ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC TAAGCTACCCTACCGTCACG GTTTCACCACTACCG 3' 5' - a transversion of the CG base pair in position 20; Case 1: from CG to GC mRNA: switch between C to G From UGC (Cys) to UGG (Trp) MIS-SENSE MUTATION: protein with one different aminoacid Case 2: from CG to AT mRNA: switch between C to A From UGC (Cys) to UGA (STOP) NON SENSE MUTATION thus a truncated protein 5' 3' ATTCGATGGGATGGCAGTGCCAAAGTGGTGATGGC TAAGCTACCCTACCGTCACGGTTTCACCACTACCG 3' 5' - an insertion of a base pair after pair 11 (AT) +1 AUG GGA NUG GCA GUG CCA AAG UGG UGA UGG C Met-Gly-aaX-Ala-Val-Pro-Lys-Trp-Stop After the insertion all the aminoacids will be different: frame-shift. The frame-shift creates a stop codon. d) How can we abolish the insertion mutation in position 11? If in a position near the first insertion a deletion happens, we can restore the correct frame of aminoacids. For example in position 15 (intragenic suppressionsuppressor mutation) +1 AUG-GGA-NUG-GAG-UGC-CAA-AGU-GGU-GAU-GGC Met-Gly-aaX-Glu-Cys-Gln-Ser-Gly-Asp-Gly The second mutation restore the correct reading frame. Only the aminoacids located between the two mutations will be different. DNA with insertion and subsequent suppressor mutation: 5’ ATTCGATGGGANTGGAGTGCCAAAGTGGTGATGGC 3’ 3’ TAAGCTACCCTNACCTCACGGTTTCACCACTACCG 5’ This is a polypeptidic sequence of a protein. Wild type and mutant sequences are compared. What type of mutations did happen? WT : Met-Arg-Phe-Thr… Mutant 1: Met-Ile-Phe-Thr… Mutant 2: Met-Ser-Ile-Tyr We compare mutant 1 with the wt: the second amino acid is different Arg could be encoded by CGU, CGC, CGA, CGG, AGA, AGG Ile could be encoded by AUU, AUC, AUA It is probable that the codon encoding Arg could be AGA, and a mutation occoured orginating AUA Phe could be encoded by UUU/UUC and Thr by ACU/ACC/ACA/ACG, The possible DNA sequence will be: Sequence of wt: AUG – AGA – UUPy – ACN Py=C o T Sequence of mutant 1: AUG – AUA – UUPy – ACN - WT : Met-Arg-Phe-Thr… Mutant 1: Met-Ile-Phe-Thr… Mutant 2: Met-Ser-Ile-Tyr We compare mutant 2 with wt: all the amicoacids after the Methionine are different. A frameshift mutation caused by an insertion. AUG – AGA – UUPy – ACN – Ser is encoded by UCN or AGU-AGC , We can hypotesize the sequence: AUG –AGN–AUU-PyACWith N=U/C The third codon AUU = Ile The fourth codon will be UAC = Tyr 5An Escherichia coli mutant auxotroph for tryptophan (Trp-) has an amino acid substitution in tryptophan-synthetase: Glycine at position 210 is replaced by an Arginine. On the basis of the genetic code, find which kind of mutation (on the DNA) you think has caused the amino acid replacement Gly Codons : GGU GGC GGA GGG Arg Codons We can hypothesize a single base substitution First base G could switch to C or A GGU->CGU; GGC->CGC; GGA->CGA; GGG->CGG; GGA->AGA; GGG->AGG : CGU CGC CGA CGG AGA AGG In the Escherichia coli metA gene a base substitution occurred. Because of this mutation, in the mRNA a UAA codon is present inside the gene. - Which consequence will this mutation have on protein synthesis? The triplet UAA is a stop codon, Thus the protein synthesis will be interrupted. The primary transcript of chicken ovalbumin RNA is composed by 7 introns (white) and 8 exons (black): If the Ovoalbumin DNA is isolated, denaturated and hybridized to its cytoplasmic mRNA, which kind of structure do we expect? Exon INTRON mRNA Structure with loops corresponding to introns. if a deletion of a base pair occurs in the middle of the second intron, which will be the likely effects on the resulting polypeptide? We don’t have any effect if the mutation is not in the splicing site. if a deletion of a base pair occurs in the middle of the first exon, which will be the likely effects on the resulting polypeptide? We have a frame-shift effect or a truncated protein. FILE 9 A man has the chromosome 21 translocated on the 14. Draw the karyotype (only the chromosomes involved in the mutation). What kind of gametes will be produced by this person? Which will be the consequences on the progeny, if this man has a child together with a normal woman? 14 21 14 21 14-21 Karyotype 21 Pairing of homologous chromosomes during meiosis 14-21 21 14 GAMETES 14 14-21 14-21 Parent with translocation Normal Parent Normal Gamete Gametes (First parent) Gametes (First parent) 14-21 21 14 14-21 14 21 Trisomy : Down’s Syndrome 2 chromosomes 14, DOWN 3 chromosomes 21 SYNDROME 21 Monosomy : not compatible with life 2 chromosomes 14, 1 chromosomes 21 NOT VITAL 14 Trisomy : not compatible with life 3 chromosomes 14, 2 chromosomes 21 NOT VITAL Monosomy 14 n:ot compatible with life 1 chromosome 14, 2 chromosomes 21 21 Zigote healthy carrier: normal phenotype 14 21 14-21 Normal Zigote: normal phenotype 2 chromosomes 14, 2 chromosomes 21 2 chromosomes 14, 2 chromosomes 21 NOT VITAL CARRIER TRASLOCATION NORMAL Which will be the consequences on the progeny, if this man has a child together with a normal woman? 1/6+1/6+1/6: ZIGOTEs are not compatible with life 1/6 1/6 1/6 2In humans trisomy of chromosome 21 is responsible of the Down syndrome. - which gametes originated an affected person? - draw a scheme of the meiotic stages that can give rise to the mutated gamete and indicate the name of this process. - Chromosomes are distributed to gametes incorrectly - The gametes either are missing or have an extra chromosome 21 - It is caused by the NONDISJUNCTION of chromosome 21 during meiosis (homologous chromosomes or sister chromatids). 3. A man carries a heterozygous paracentric inversion. A B C D E F G H a b c d g f e h Draw a scheme of homologous chromosomal pairing during meiosis F E ABCD Draw only one cromatide for each chromosome. f e g G H h abcd Which gametes are produced (in particular which gametes are missing)? Explain why. Parental gametes A B C D E F G H a b c d g f e h F RECOMBINATION (CROSSING-OVER) E ABCD f e g G H h abcd A B C D RECOMBINANT GAMETES h E f g e F G d c b a H We will have PARENTAL GAMETES but we don’t have vital recombinant gametes produced by a crossing over happens in the inverted region Effect: a DICENTRIC CHROMOSOME (TWO CENTROMERES) and a ACENTRIC FRAGMENT CHROMOSOME (lost). Does the presence of the mutation change the fertility of this man? No, if inversion is not extended and is limited to a small part of chromosome. 4. Deletion of a small region on Y chromosome in humans can prevent the individual development as a male. How can you explain this result? The deletion is on a locus of Y chromosome where the SRY gene (Sex determining Region Y) is located. It encodes for the TDF, Testis Determining Factor. Missing of this gene prevent the development as a male, thus the individual develops as a FEMALE 6 - how can a triploid organism originate? Draw a scheme of meiosis process in a triploid cell with n = 3. From the cross between gamete n + gamete 2n. Chromosomes A, B and C gamete 2n= AABBCC gamete n = ABC individual 3n=AAABBBCCC In a triploid cell which gametes are produced? BB (1/2) CC (1/2) C (1/2) AABBCC (1/8) AABBC (1/8) B (1/2) CC (1/2) C (1/2) AABCC (1/8) AABC (1/8) BB (1/2) CC (1/2) C (1/2) ABBCC (1/8) ABBC (1/8) B (1/2) CC(1/2) C (1/2) ABCC (1/8) ABC (1/8) AA (1/2) A (1/2) 2/8 are gametes that could generate an individual 6/8 are not compatible with life 7Asiatic cotton and American wild cotton have both 26 chromosomes. The American cultivated cotton, that is derived from the previous species by alloploidy, has 52 chromosomes. Explain, with a scheme, how it originates. Alloploidy=A hybrid individual having two or more sets of chromosomes derived from two different species. We hypotize that both species 2n = 26 ASIATIC COTTON (A) = 13 chromosomes AMERICAN COTTON (B) = 13 chromosomes If in the hybrid we have 26 chromosomes from two different species the plant is vital but it is sterile. If in the hybrid a doubling of chromosomes occurs, we have an allopolyploid that is fertile because each chromosome has its homologous. 2 A + 2 B = 26 + 26 = 52 Chromosomes FILE 10 1. In Escherichia coli, the lac (lactose) operon, is made of the following genes and sites. Specify what is the function of the ones indicated below: - promoting site - operator site - repressor gene - structural genes. PROMOTER DNA region where RNA polymerase sits to start transcription. REPRESSOR gene that encodes for a protein that negatively regulates transcription. OPERATOR DNA locus where the repressor could bind to stop transcription. STRUCTURAL GENES Genes that are usefull for a cellular function; for example metabolism of lactose. 2. What would be the result of a base substitution that inactivates the following genes: LacZ- Since the gene encodes for b-galattosidase enzyme, a mutation in this gene probably inactivates the function of the gene. LacIThis gene encodes for the repressor of lactose operon. The mutation will have different effects depending on the protein domain where it occours: a) If the mutation inactivates the protein (frame-shift, stop codon, missense), we have the absence of the repressor and constitutive transcription of the structural genes (recessive mutation LacI-) b) If the mutation alters the allosteric domain of the protein where the inducer binds, we have constitutive repression of the structural genes because the repressor is bound to its site and is not influenced by the presence of lactose. (dominant mutation, LacIs) 3. What would happen if a base deletion occurs in the operator region? OPERATOR is the DNA locus bound by the repressor to stop transcription. After the mutation in the operator, the repressor could be unable to recognize the locus. In fact the repressor (protein) is able to recognize the Operator (sequence). Thus, we have the constitutive expression of the genes. The mutant in operator constitutive lacOc (cis DOMINANT) Cis indicates that the mutation acts on the nearest genes. 4. Which of the following genotypes will be able to produce β-galactosidase and/or permease in the presence of lactose? bgal perm. Genotype 1 Lac I+ Lac P+ Lac O+ Lac Z+ Lac Y+ Lac A+ + ….. + …… Genotype 2 Lac I+ Lac P+ Lac O+ Lac Z+ Lac Y- Lac A+ + ….. …… Genotype 3 Lac I- Lac P+ Lac O+ Lac Z+ Lac Y+ Lac A+ + ..…. + …… Genotype 4 Lac I+ Lac P+ Lac Oc Lac Z+ Lac Y+ Lac A+ + …… + …… Genotypes 3 and 4 -> constitutive transcription of Lac operon (no repression) 4. If the lactose is not present, in which mutants the expression of genes change? bgal perm. Genotype 1 Lac I+ Lac P+ Lac O+ Lac Z+ Lac Y+ Lac A+ - -….. …… Genotype 2 Lac I+ Lac P+ Lac O+ Lac Z+ Lac Y- Lac A+ - ….. …… Genotype 3 Lac I- Lac P+ Lac O+ Lac Z+ Lac Y+ Lac A+ + ..…. + …… Genotype 4 Lac I+ Lac P+ Lac Oc Lac Z+ Lac Y+ Lac A+ + …… + …… Genotypes 1 and 2 will not express the genes (REPRESSION) Genotypes 3 and 4 will express constitutively the enzymes 5. What does it mean that the Oc mutation is dominant in cis? How can I demonstrate it? CIS dominant mutation: it expresses the dominant phenotype but it affects only the expression of genes on the same DNA molecule where the mutation occurs. LacOc, affects only neighbouring genes (for exemple a plasmid) To demonstrate it we construct an heterozygote (diploid) with: The mutation lacZ- (Bgal enzyme) located in cis to lacOc no production of bgalattosidase The gene lacZ+ in trans (on plasmid) with wt lacO lacOc lacZPhenotype in absence of induction: mutation is cis-dominant -> no b-gal activity lacO+ lacZ+ mutation is trans-dominant -> b-gal activity 6. What kind of phenotype will has a bacteria LacI+ O+ Z+ Y+ A+ carrying on a plasmid lacIs O+ Z+ Y+ A+ ? Which conditions can we use to analyse its phenotype? The bacteria will be: LacI+ O+ Z+ Y+ A+ / LacIs O+ Z+ Y+ A+ heterozygote for gene LacI+ /LacIS LacI+ encodes for the repressor that is able to bind lactose (induction and derepression) and is able to recognise sequence of Operator (repression) LacIS encodes for a repressor with a mutation that unable the protein to interact with lactose (constitutive repression). LacIS repressor is always bound on the Operator repressing the transcription of Operon Lac : the operon Lac genes are not transcribed and expressed In order to analyse the phenotype I’ll grow the mutant in a condition where I can detect the expression of enzyme…with the presence of lactose (induction). We expect that the bacteria with the plasmid does not express the structural gene it does not grow if the unique source of carbon is Lactose. (dominant mutation) The wt bacteria (control) express the structural genes and grow on Lactose. 7. Two bacteria have a Trp- phenotype, then? The Trp- bacteria are unable to synthesize Tryptophan How can I verify whether the two mutations are in the same gene? I complement them: I produce bacteria carrying both mutations one on a plasmid, the other on the chromosome. To analyse the phenotype I plate them on a medium without Tryptophan CASE 1: If bacteria grow, the two mutations complement each other, because they affect two different genes. CASE 2: If bacteria don’t grow, the two mutations do not complement each other, because they affect the same gene. POPULATION GENETICS The study of the allele frequencies in a population The Hardy Weinberg equilibrium “Allele and genotype frequencies in a population tend to remain constant in the absence of disturbing influences” -non-random mating -mutations -selection -limited population size -random genetic drift Conditions for HW equilibrium are never met in nature. The equations p+q=1 p=frequency of allele A q=frequency of allele a p2 + 2pq + q2 =1 p2=frequency of AA q2=frequency of aa 2pq=frequency of Aa An Example Assume a population in which 36% of the population are homozygous for a certain recessive allele a. Assume the population is at equilibrium. What is the frequency of the recessive allele in this population? 36%aa q2=0.36 q=f(a)= 0.6 What is the frequency of the dominant allele in this population? q =0.6 p = f(A) =1-0.6=0.4 What percentage of the population are homozygous for the dominant allele A? p2=f(AA)=0.4 x 0.4=0.16 16% What percentage of the population are heterozygous for the trait? 2pq=f(Aa)=2 x 0.6 x 0.4 =0.48 48% Why do we have to start the problem with the percentage or numeber of the homozygous recessive in the population? Because only the recessive phenotype indicates only the homozygous recessive genotype…the dominant phenotype indicates heterozygous and homozygous 1 - Determine the gene and genotypic frequencies of the following populations: I n° of individuals with genotype: AA Aa aa 100 400 500 II 300 200 500 III 700 200 100 Genotypic frequency of AA 0,1 Aa aa A a 0,4 0,5 0,3 0,7 In this case we know the number of heterozygous We think about the number of alleles p =f (A) = (2 x 100) + 400 = 0,3 2 x 1000 q = f(a)=1 – 0,3 = 0,7 Gene Frequency of 1 - Determine the gene and genotypic frequencies of the following populations: Genotypic frequency of AA Aa I n° of individuals with genotype: AA Aa aa 100 400 500 II 300 200 500 0,3 0,2 III 700 200 100 p = (2 x 300) + 200 = 0,4 2 x 1000 q = 1 – 0,4 = 0,6 aa 0,5 Gene Frequency of A a 0,4 0,6 1 - Determine the gene and genotypic frequencies of the following populations: I n° of individuals with genotype: AA Aa aa 100 400 500 II 300 200 500 III 700 200 100 p = (2 x 700) + 200 = 0,8 2 x 1000 q = 1 – 0,8 = 0,2 Genotypic frequency of AA 0,7 Aa 0,2 aa 0,1 Gene Frequency of A 0,8 (A(AA)+A(AA))+A(Aa)/(total alleles) a 0,2 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it. Population 1 p = 0,3 q = 0,7 f(AA) = p2 = 0,09 f(Aa) = 2pq = 0,42 f(aa) = q2 = 0,49 Genotipi Xo Xa (Xo – Xa)2 : Xa AA 100 90 1,11 Aa 400 420 0,95 aa 500 490 0,20 c2 = 2,26 Degree of Freedom = 2(p,q)-1= 1 P = 0,10-0,25 Population is in equilibrium 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it. Population II p = 0,4 q = 0,6 f(AA) = p2 = 0,16 f(Aa) = 2pq = 0,48 f(aa) = q2 = 0,36 Genotipi Xo Xa AA 300 160 Aa 200 480 aa 500 360 Expected and observed values are too different. The population is not in equilibrium (Xo – Xa)2 : Xa 2 - Are the populations of exercise 1 at equilibrium? Where necessary, use statistical methods to verify it. Population III p = 0,8 q = 0,2 f(AA) = p2 = 0,64 f(Aa) = 2pq = 0,32 f(aa) = q2 = 0,04 Genotipi Xo Xa (Xo – Xa)2 : Xa AA 700 640 5,6 Aa 200 320 45 aa 100 40 90 c2 = 140,6 GL = 2– 1 = 1 P <0,01 The population is not at equilibrium 3 - The frequencies of LM and LN alleles in a group of 200 black Americans were 0,8 e 0,2, respectively. Calculate the expected frequencies for individuals with M, N and MN blood group. If the population is at equilibrium: Group M = LM LM = p2 = 0,8 x 0,8 = 0,64 x 200 = 128 Group MN = LM LN = 2pq = 2 x 0,8 x 0,2 = 0,32 x 200 = 64 Group N = LN LN = q2 = 0,2 x 0,2 = 0,04 x 200 = 8 Population I f(aa) = 810/1000 = 0,81 Since f(aa) = q2 f(a) = √0,81 = 0,9 f(A) 1 – 0,9 = 0,1 Population II f(aa) = 360/1000 = 0,36 f(a) = √0,36 = 0,6 f(A) 1 – 0,6 = 0,4 Population III f(aa) = 490/1000 = 0,49 f(a) = √0,49 = 0,7 f(A) 1 – 0,7 = 0,3 The individuals with phenotype A can be AA or Aa. We are sure that individuals with phenotype a are: aa We start from this information (aa) to figure out f(A) 5 - What is the expected frequency for dominant homozygous and heterozygous genotypes in a population at equilibrium in which the homozygous recessive genotype frequency is 0,09? f(aa) = 0,09 f(a) = √0,09 = 0,3 f(A) = 1 – 0,3 = 0,7 f(AA) = 0,7 x 0,7 = 0,49 f(Aa) = 2 x 0,7 x 0,3 = 0,42 6 - if in a population at equilibrium the frequency of Rh- phenotype is 0,0025, which is the expected frequency of heterozygous individuals? f(Rh-) = 0,0025 = f(dd) f(d) = √0,0025 = 0,05 f(D) = 1 – 0,05 = 0,95 f(Dd) = 2 x 0,95 x 0,05 = 0,095 7 In Drosophila melanogaster the w recessive sex-linked allele, is responsible of the white colour of the eyes. In a population the frequency of this allele is 0,3. Which frequencies of male and female with white-eyes is expected if the population is at equilibrium? Male white eyes = w Y Female white eyes = w w Frequency: Males white eyes = f(w) = 0,3 Females white eyes = f(ww) = q2 = 0,3 x 0,3 = 0,09 If alleles are X-linked, females may be homozygous or heterozygous, but males carry only a single allele for each X-linked locus. For X-linked alleles in females, the H-W frequencies are the same as those for autosomal loci. In males, however, the frequencies of the genotypes will be p and q, the same as the frequencies of the alleles in the population. 8 - Daltonism is due to a sex-linked recessive allele. In a population the daltonic male frequency is 0,1; which is the one of daltonic females? Daltonic MALE = d Y (XY) emyzygote freq dY = freq allele d Daltonic FEMALE = d d (XX) Frequency Daltonic Males = 0,1 Frequency allele d in MALES = 0,1 Daltonic FEMALES = f(d d) = 0,1 x 0,1 = 0,01 Healty carriers= 2*0.1*0.9=0.18 Fitness involves the ability of organisms— or, more rarely, populations or species— to survive and reproduce in the environment in which they find themselves . The consequence of this survival and reproduction is that organisms contribute genes to the next generation. Fitness= Its lifetime reproductive success Fitness absolute 200/100 = 2 Fitness relative (W) 2/2 = 1 400/200 = 2 100/100 = 1 2/2 = 1 The selection is against the phenotype homozygous recessive 1/2= 0,5 10 - A population is composed of: AA 250, Aa 500, aa 250 individuals, when a selection factor acts against the homozygote recessive and its fitness decreases to 0. The variation of genic frequency (selection effect) could be figured out taking into consideration the contribution of the gametes of different genotypes in the next generation. Consider the previous population, try to figure out the variation of genetic frequency of a after a generation of selection. 1 1 250/1000 = 0,25 1 x 0,25 = 0,25 0 500/1000 = 0,5 250/1000 = 0,25 1 x 0,5 = 0,5 0,25 x 0 = 0 1000x2 total of alleles 2x250 aa=250a+250a 2(0.25+0.5) total of freq [2(0) + 0,5] /2(0,25 + 0,5 + 0) = 0,33 2x0 aa=0a+0a [2(250) + 500]/2(1000) = 0,5 q1 – q0 = 0,33 – 0,5 = -0,17 As expected the frequency of a allele decreases 11 – In a population there is a selection against the homozygous recessive with fitness = 0 but the genotypic frequencies are: AA 0,64; Aa 0,32; aa 0,04, Try to figure out the variation of genetic frequency of a after a generation of selection 1 1 0 0,64 0,32 0,04 0,64 x 1 = 0,64 0,32 x 1 = 0,32 0,04 x 0 = 0 [0,32 + 2(0,04)]/2(0,64 + 0,32 + 0,04) = 0,20 [0,32 + 2(0)]/2(0,64 + 0,32 + 0) = 0,16 q1 – q0 = 0,16 – 0,20 = -0,04 Previous population Dq = -0,17 (> of q0 (0,5)) This population q0 = 0,2 With the same selection coefficient, the effect of selection against homozygous recessive change on varying of genic frequency 12 – In a population the gene frequency of A is 0,3 and the frequency of a is 0,7, and there is a selection against the heterozygous (s =coefficient of selection=1-w= 0,4). Try to figure out the variation of genetic frequency of a after a generation of selection genotypes fitness AA 1 Aa aa 1 – 0,4 = 0,6 Genotypic frequency p2 = 0,32 = 0,09 Gametic contribution 1 x 0,09 = 0,09 2pq = 0,42 1 q2 = 0,72 = 0,49 0,6 x 0,42 = 0,252 1 x 0,49 = 0,49 Frequency of q0 [0,42 + 2(0.49)]/2(0.09+0,42+0,49)=0,7 Frequency of q1 [0,252 + 2(0,49)]/2(0,09 + 0,252 + 0,49) = 0,74 Dq q1 – q0 = 0,74 – 0,7 = 0,04 - What is the final result of this selection? The recessive allele will be fixed and the dominant allele will be delete from the population 13 – The population is composed of: 640 AA, 320 Aa and 40 aa, The fitness of homozygous dominant is 0,8 and the fitness of homozygous recessive is 0,1. Try to figure out the variation of genetic frequency of after a generation of selection genotypes fitness Genotypic frequency Gametic contribution AA Aa 0,8 1 640/1000=0,64 320/1000=0,32 0,64 x 0,8=0,51 0,32 x 1 = 0,32 aa 0,1 40/1000=0,04 0,04 x 0,1=0,004 Frequency of q0 [320+2(40)]/2(1000) = 0,2 Frequency of q1 [0,32+2(0,004)]/2(0,51+0,32+0,004) = 0,19 Dq 0,19 – 0,20 = -0,01 Considering people that become part of the family through marriage don’t have the mutant allele, calculate the probability that an affected (sick) child is born from the indicated cross ? Daltonism is a recessive x-linked disease. A girl has a daltonic father and a normal mother but the mum’s father is daltonic. Which is the probability that the girl is daltonic? Draw the tree Genes A and B are concatenated at 20 mu. Which gametes are produced by an individual: AaBb? If this individual is cross with a homozygous recessive, which phenotypic classes and which frequencies will we obtain? For each of the following crosses: - determine the genotype of the individuals used for the cross; - determine if there is concatenation of the analyzed genes ; - if genes are concatenated, determine their map distance; - draw a schematic picture of the map of crossed individuals. Parental Phenotypes AB x ab Progeny Phenotypes: AB 40 Ab 63 aB 58 Ab 37 Gene A, B and C are located in this order on the same chromosome. The distances are A-B 20 mapunits and B-C 10 mapunits. If an individual triple heterozygous is crossed with a recessive homozygous, which phenotypical classes are expected in the progeny and with which frequencies? If the interference is 1? Analysing a sample of individuals we have these phenotypic results MM 60 MN 92 NN 20 Determine allelic frequencies of M and N Determine if the population is at equilibrium (use statistical analysis) 7 Gene A, B and C are located in this order on the same chromosome. The distances are M-N 2 map units and N-O 8 map units (genes are in cis). If an individual triple heterozygous is crossed with a recessive homozygous, which phenotypical classes are expected in the progeny and with which frequencies? Draw the map M M 2 N n 8 O o x m m Parental MNO mno Recombinants1 Mno mNO Recombinants2 Mno mnO DR MnO mNo Freq DR= 0.02 x 0.08 =0.0016/2=0.0008 Freq Ric 1=0.02-0.0016=0.0184/2= 0.0092 Freq Ric 2 =0.08 – 0.0016= 0.0784/2 =0.0392 Parental = 1- 0.0016-0.0184-0.0784=0.9016/2= 0.450 n n o o