Tt is what type of genotype

Biochem Biophys Res Commun. The polymorphism Gln-Arg of the high-density lipoprotein-bound enzyme is an independent cardiovascular risk factor in non-insulin dependent diabetic patients. Mutation C to T in the methylenetetrahydrofolate reductase gene aggravates hyperhomocysteinemia in hemodialysis patients. Kidney Int. Relation between folate status, a common mutation in methylenetetrahydrofolate reductase, and plasma homocysteine concentrations.

Oxford University Press is a department of the University of Oxford. It furthers the University's objective of excellence in research, scholarship, and education by publishing worldwide. Sign In or Create an Account. Sign In. Advanced Search. Search Menu. Article Navigation. Close mobile search navigation Article Navigation. Volume Article Contents Abstract. The TT genotype of the methylenetetrahydrofolate reductase C T gene polymorphism is associated with the extent of coronary atherosclerosis in patients at high risk for coronary artery disease.

Gardemann , A. Oxford Academic. Google Scholar. Wilhelm Hehrlein. Revision received:. Cite Cite A. Select Format Select format.

Permissions Icon Permissions. Abstract Background There are conflicting results on the relationship of N 5 ,N 10 -methylenetetrahydrofolate reductase C T gene variation in coronary artery disease and myocardial infarction. Methods and Results We analysed this gene variation in male Caucasians whose coronary anatomy was defined by coronary angiography. Conclusion The present study extends previous observations by the finding that carriers of the N 5 ,N 10 -methylenetetrahydrofolate reductase C T TT genotype with various coronary high risk profiles had clearly higher coronary heart disease scores than individuals with at least one C T C allele.

Homocysteine metabolism, coronary artery disease, myocardial infarction, coronary risk factors. Download all slides. View Metrics. Email alerts Article activity alert.

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The optimal strategy for multivessel coronary revascularization. Leadless vs. Looking for your next opportunity? Physician-Scientist Faculty Position. Infectious Disease Physician. Enzyme G The chromosomes common to cell lines making this protein are: 2 and 9 Cell line C has chromosome 2 but does not make the protein.

Therefore, the gene for Enzyme G must be on chromosome 9. Enzyme AD The chromosomes common to cell lines making this protein are: 5 and 14 Cell line E has chromosome 5 but does not make the protein. Therefore, the gene for Enzyme AD must be on chromosome Enzyme H The chromosomes common to cell lines making this protein are: 2 and 9 Cell line C has chromosome 2 but does not make the protein. Therefore, the gene for Enzyme H must be also be on chromosome 9. The cross is outlined below; the children are expected to be unaffected females and ocular albinism males in ratio.

The man in the cross has the dominant alleles for both loci, so his daughters will all be phenotypically normal. The sons' phenotypes, however, will depend on which X chromosome they inherit from the woman, and on whether she has the dominant alleles in cis or in trans. You just have to realize that because the ratio of phenotypes is very different in females vs.

Other than that, the procedure is the same as above--you use just the male progeny to follow the recombination that occurred in the female parent. You know that the genes are on the X chromosome; you know the parental genotypes.

This problem is easily solved. Human chromosomes present in cell lines that do not have the insulin sequence can be eliminated from our list of possible candidates. Therefore, any chromosome that is found in cell lines D, E, or F can simply be crossed out from the list of possibilities eliminated candidates shown below as colored-out boxes.

Cell line Human insulin sequence present? Of the remaining candidate chromosomes, the only one that is present in cell lines A, B, and C is chromosome Therefore, the insulin gene must be located on chromosome However, she has a colorblind, hemophilic son, so her other X chromosome must have both recessive alleles. III-2 has both disorders; he is X gh Y. III-3, being colorblind, must be homozygous recessive for the color vision locus. One chromosome is X gH the one she got from her father, II-2 ; the chromosome she got from her mother II-1 is either X gh if there was no recombination between the two X's in her mother or X gH if there was recombination.

Considering the phenotype of III-3, the only possible recombinant gamete is X gH ; the probability of that is 0. Week 2. The diploid form, having two sets of chromsomes, can undergo meiosis. The haploid form has only one set to begin with, so it cannot undergo a reductional division.

Furthermore, since we are looking for a son, the sperm will have to be a Y-chromosome bearing one. Therefore, the genotype of the sperm has to be gaY. Therefore, at anaphase I, the Y chromosome has to segregate with the homologs carrying the recessive alleles, as diagrammed:. Note: In the interests of simplicity, crossing over has been ignored here. Also, the relative sizes of chromosomes and locations of genes is fictitious.

So another way of stating the question is -- In which of these matings are all of the daughters heterozygous? Because daughter received X h from each parent, the father must be X h Y. The mother transmitted one hemophilia allele to the daughter and one normal allele to the son -- so she must be a carrier, X H X h. Affected children have unaffected parents, so the disease cannot be dominant assuming complete expressivity and penetrance.

Women and men are affected, so it cannot be sex-limited or Y-linked. If one assumes that the disease is fairly common, then it could be autosomal recessive. However, given that there are many more affected men than affected women, a more probable explanation is that it is X-linked recessive.

Alternatively, it could be sex-influenced -- dominant in males, but recessive in females. Again, affected children have unaffected parents, so the trait cannot be dominant. It could be autosomal recessive.

Furthermore, only men have been affected in theis pedigree, arguing against a simple autosomal recessive pattern. The fact that only men are affected in this pedigree suggests sex-linkage. But affected men have unaffected sons, so it is not Y-linked.

It could be X-linked recessive -- but the trait appears to be passed on from father to son in one instance IV-8 to V So it could be X-linked recessive only if IV-7 is a carrier. It could also be sex-limited phenotype expressed in men , but as with autosomal recessive, we'd have to assume that the disease is common.

We could as a matter of parsimony say that the most probable mode of inheritance is X-linked recessive or sex-influenced, but leave open the possibility that it is autosomal recessive or sex-limited if the disease proves to be common. The fact that phenotypes in the F1 are skewed with respect to sex immediately suggests that the trait must be sex-linked.

The trait is not passed father-to-son F1 males are normal , so it cannot be Y-linked. That leaves X-linked inheritance. The F1 males get their X chromosomes from the parental females. That there is only one phenotype amongst the F1 males tells us that the parental females must be homozygous for the normal allele. That means that the F1 females must all be heterozygotes getting a normal X from the mother and a squiggly-eye X from the father.

But these heteroozygous F1 females are all squiggly-eyed. Therefore, the squiggly-eye phenotype must be X-linked dominant. The F1 x F1 cross would give squiggly eye females, squiggly eye males, normal females, and normal males in ratio, as shown below:. The key here is in realizing that because these are independently assorting traits, we can look at each trait separately-- a The cross here is AABbDdee x AaBbddEe.

We are asked to calculate what fraction of the progeny will have the phenotype ABde. Because these are independently assorting traits, we can calculate the fraction of progeny that will have phenotype A, then the fraction that will have phenotype B, etc. Using the same logic as above For a dihybrid cross, we expect to see a ratio of phenotypes in the offspring--clearly not the case here.

Because nothing is mentioned about males vs. To sort out the puzzle, therefore, we could begin by looking at each phenotype separately and seeing if that helps. The observed progeny are creeper white , creeper yellow , normal white , and normal yellow chickens in ratio. Let's look at creeper vs. When we do that, we find that the ratio is 8 creeper : 4 normal , i. Where have we seen a heterozygote x heterozygote cross giving a ratio before?

That's right, if creeper is dominant over normal and creeper is lethal when homozygous, we'd get a ratio of creeper : normal in the progeny.

How about white vs. Here, the ratio is 9 white : 3 yellow, a simple ratio. Therefore, white must be dominant and yellow is recessive.

This one is a little tricky. A common mistake is to misinterpret the question to think that the first two children are boys -- when in fact, all we know is that at least two children in any order are boys. If the sex of the children is written in order of birth as B for boy or G girl , the possible 3-children families with at least 2 boys are:. We use the binomial distribution to solve this one. Substituting the probabilities of unaffected and affected children, we get:.

But an easier way is to find the probability of less than two affected children, then subtract that value from 1 Try it. For progeny, the expected numbers are:.

Therefore, the null hypothesis that the deviation from expected values is just due to chance cannot be rejected. What are the possibilities here? Possibility 1: the cross was homozygous purple x homozygous purple ; there should be no white-flower progeny. If the seed merchant picks just one seed at random and grows it up, and it makes white flowers -- she knows it must have been a heterzygote x heterozygote cross. However, if she picks one seed, and it makes a purple-flower plant -- can she then say that it must have been a homozygote x homozygote cross?

Suppose she picks two seeds? In other words, she needs to sample n seeds such that. In turn, we have to know the genotypes of their parents, and so on. III-4 is unaffected; the only way she can have an affected child is she is heterozygous Dd. What is the probability of that?

Likewise, III-5 has to be heterozygous Dd for their child to be affected. Answers to selections from Affected women have unaffected sons e. Likewise, affected men have unaffected daughters e. Thus, the mode of inheritance that best explains the observed pedigree is autosomal dominant. As described in lecture refer to the part on evidence for random segregation of homologs in meiosis , meiosis in the exceptional females XXY, homozygous for the X-linked white allele can give four kinds of gametes because the two X chromosomes can pair up during synapsis, or an X and a Y--in which case the lone X could segregate either with the other X or with the Y.

Some of these eggs can give rise to fertile red-eyed males and white-eyed females, the secondary exceptions. NOTE: The grid above shows only the kinds of progeny that can be formed, not the relative numbers.

Because synapsis of the two X chromosomes is more probable than synapsis of an X with a Y, the "Y is unpaired" outcome of meiosis I see the diagram above is more probable than the "X is unpaired" outcome.

Therefore, gamete types 1 and 2 are much more abundant than gamete types 3 and 4, and the progeny numbers are skewed accordingly. The term representing the probability of 3 albino and 2 normal children is 10a 3 b 2. Substituting the values of a and b , we get:. Week 1. Therefore, if there are F2 progeny, of them should be homozygous TT or tt -- i. The F1 consist of tall plants only, so the unknown must be homozygous TT ; the cross is shown. See below for why it can't be heterozygous Tt.

The only way a tall plant can yield short progeny after selfing i. Therefore, what the question is asking is: what fraction of the tall plants are heterozygous? Refer to the crosses shown in answer 2 for these questions. Note: You are not looking for tall plants that give only short progeny upon selfing is that even possible? If the cross is TT x Tt , the progeny are TT and Tt plants in equal proportions, so half of these progeny will yield short plants upon selfing.

In the two couples in generation I, we don't know which individual has free earlobes and which has attached, so I have chosen to display them as sex-unspecified but one member of each couple with attached earlobes. The child in generation III is again sex-unspecified, but has attached lobes and must therefore be homozygous recessive ff.

The parents in generation II must be heterozygous Ff. If the normal wing phenotype were dominant, the progeny would all show the normal phenotype. However, there are curly-wing flies in the progeny. Therefore, curly-wing C must be dominant over normal wing c. Furthermore, two phenotypes curly and normal are seen in the F1, and in ratio; therefore, the curly-wing parent must be a heterozygote.

Two T alleles, 2 t alleles. The simplest approach is a trial-and-error method: interpret each cross one at a time, and see if your interpretation is consistent with the interpretation of the previous crosses.

To begin with, it is clear that there are three phenotypes, so just for simplicity, I am going to assign them 3 allele designations R, B, W, for Red, Blue, and White and assume that they are alleles of the same determinant.

I may have to revise this initial hypothesis later on--e. Cross a -- Red 1 selfed -- yields a ratio of red and blue-flowered plants in the progeny. This looks like a typical heterozygous F1 cross, with R being dominant and B recessive. So I'm tentatively assigning Red 1 a genotype of RW. Cross c -- Blue selfed -- gives a ratio of blue:white; blue must be dominant over white and the genotype of the blue-flowered plant must be BW.

We are now in a position to predict the results of the remaining crosses, and seeing if our predictions are met. Cross e -- Red 1 x Blue -- should be RB x BW , which should give a ratio of red:blue draw Punnett squares if you're uncertain about this. Again, that's what we see. So our initial hypothesis appears to be sound as far as we can tell from the data provided.

We can predict the results of cross h :. A potential source of error in this problem is to simply add the number of recessive individuals from the two populations and to derive q from that -- i. However, doing so would ignore the contribution of recessive alleles from the heterozygotes in each population. If only black cats are left standing after the virus goes through, then only the recessive black allele will be left in the population; the frequency of the black allele in the next generation will be 1.

The d allele will be more frequent, as the forward mutation D to d occurs at a higher rate than the back mutation. If conditional alleles are not available, an alternative strategy is to cross heterozygotes and to ask if one quarter of the progeny show the phenotype: The logic here is that if mut and torso are mutations in the same gene for example , then Cross 1 is a monohybrid cross; one quarter of the progeny should be homozygous recessive, giving the mutant phenotype.

Any combination of genetic and environmental factors. F1 will be AaBb--which has 2 additive alleles, so the height will be 30 cm. Each parent has 4 additive alleles; since the F1 also have 4 additive alleles, the parents must be each be homozygous; the additive alleles of one parent are not present in the other. An 18 cm plant has 2 additive alleles; any genotype such as AAbbccdd or aaBBccdd would work. This one can be solved only if we make the assumption that everyubody gets to mate, and that all crosses produce equal numbers of progeny.

Assuming that the allele frequencies are the same in men vs. Because these are already Hardy-Weinberg frequencies, there will be no change in allele frequencies in the next generation. The more heterozygous population is genetically more heterogeneous, and will therefore show higher heritability than the more homozygous population.

The population in the more uniform environment will show higher heritability. Use the frequencies of the various genotype classes-- 0. The promoter is defective, so there can be no transcription of the lac operon. The operator is mutated, so lac repressor cannot bind -- transcription of lac Z, Y, and A will be constitutively high regardless of whether lactose is present or absent. Transcription of lacZ and lacY will still be under normal inducible control; the lacA product alone will not be made.

Depending on the nature of the missense mutation, the lacY product lac permease may be functional or non-functional. A stop codon near the start of the lacY coding region would likely act as a polar mutation the ribosome would never get to the start codon of lacA , so the cell would produce neither lac permease not lac transacetylase.

Without CAP, no activation of lac gene transcription can occur regardless of whether glucose is present or absent, or whether lactose is present or absent. Since phosphoenol pyruvate PEP is one of the glycolysis products that inhibits cAMP production and thereby blocks CAP activation , the failure to produce PEP will result in reduced inhibition of CAP activation by glucose; lactose will induce lac operon transcription even in the presence of glucose. Constitutively high beta-galactosidase.

Constitutively low. An example of a polar mutation -- the mutation must result in premature termination of translation such that a truncated, non-functional protein B is made, and translation of gene C coding sequence does not occur. A key point to note here is that various types of mutations can occur in any gene. This is a zygotic gene; failure to produce hunchback protein results in los of anterior segments.

Because the two mutant strains showed complementation the F1 were able to see , the mutations must have been in separate genes ; the simplest explanation is that there are two genes involved. The F2 ratio indicates that we are dealing with a dihybrid ratio the fractions go in sixteenths. As with any dihybrid cross, one quarter of the progeny will be true breeding.

As with any independently assorting pair of genes,we can look at the the ratios for the two genes independently. Here, with respect to gene E , one quarter of the progeny show the homozygous recessive phenotype--therefore both parents must be heterozygous Ee for gene E. D and E will rescue because if either one is provided, E3 function will no longer be needed ; C will accumulate because there is no E3 to convert C to D.

Conversion of B to D cannot proceed, so D and F will each rescue this mutation. Purple flowers because of complementation-- the F1 will be heterozygous for each gene. Neither intermediate pyrimidine or thiazole rescues more than one mutation. An alternative pathway-- -- is also possible, but the cln-sic- double mutant phenotype argues against it.

The patches probably arose by mitotic recombination. Because the twin spots and lone spots occurred in ratio, the centromere- rd distance and the rd-b distance must be in the approximate ratio of Lone spots of rd phenotype could arise either from mitotic nondisjunction or by mitotic recombination with double crossover, one crossover between the centromere and rd and one crossover between rd and b.

The map is: If a recombinant sector has phenotype a alone, then the crossover must have occurred between a and all the other genes; if a sector has phenotypes a and b , then b must be between a and all the other genes, etc. Here, the mutant allele must be promoting cell proliferation by failing to block Protein A.

The result is unexpected in that we are seeing only the parental non-crossover products--we should see some recombinants. The fact that chromosomal material gets stretched between the spindle poles and breaks suggests that a dicentric chromosome must be formed -- indicating that the inversion must be a paracentric inversion: Note that the gene order shown is arbitrary; the question does not give information on the correct gene order or even which parent had the inversion.

There are two crossovers in the inversion loop, so the products will all be viable -- no reduction in fertility. The crossovers are both outside the inversion loop, so again, there will be no reduction in fertility.

Here, there is only one crossover within the inversion loop, the other crossover occurring outside the loop. One way of distinguishing between the hypotheses is to obtain DNA from affected individuals and do a Southern blot experiment on the DNA, cutting it with EcoRI and probing the blot with the 2.

The same 2. The Southern blot approach only detects fragment sizes, not chromosomal locations. There is just one way to get XYY progeny from normal XX and XY parents if nondisjunction happens only in one parent -- male nondisjunction in meiosis II to produce YY sperm marked with an arrow in the diagram.

The calico pattern is the result of X-inactivation. It should worse for females than for males. Castillo, M.

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