出生前基因型

1、 Prenatal genotyping using RFLP The differences in RFLP patterns are used to diagnose genetic disorders prenatally and can also determine the genotype of parents and other family members who may be heterozygous carriers. For prenatal diagnosis fetal cells are obtained by amniocentesis The tools need

2、ed to perform these DNA diagnoses are first we need a restriction enzyme that will cut DNA in a manner that is informative for the disease in question, and second we need a nucleic acid probe that will detect that region 1kb 4kb sample A: probe xxxxxxxxxx Sample B : xxxxxxxxxxx The two DNA samples a

3、re cut with the appropriate enzyme and are hybridized with the correct probe. The probe hybridized to the 1 and 4 kb fragments in sample A and the 5 kb fragment in sample B. This is an example where a restriction site has been gained in sample A (or lost in sample B) by mutation. Sample A Sample B S

4、ample A Sample B - - 5.0 kb _ 5.0 kb _ 4.0 kb _ 4.0 kb _ 1.0 kb _ 1.0 kb _ - - Genotype Determination using RFLPs and a Gene Probe A DNA probe that hybridizes to the 5 end of the human beta globin gene (shown in blue on the diagram below) was used to identify RFLP pieces from members of a family in

5、which sickle-cell hemoglobin (HbS) was segregating. The normal HB allele (HbA) is cut at three places by a particular restriction enzyme (positions shown with red arrows). The HbS mutation destroys the internal restriction site so the HbS gene is cut in only two places. Thus, the probe hybridizes to

6、 a 1.15 kb DNA fragment from HbA DNA and hybridizes to a 1.35 kb fragment from HbS DNA. Linked DNA markers All of the DNA (RFLP or PCR) markers that we have discussed so far have been targeted to a specific gene. For many important traits, the actual gene of interest is not known. Therefore, probing

7、 for the presence of the normal or mutated allele is not possible. Instead the probe recognizes a sequence that is close to the actual gene of interest. Both the phenotype and the RFLP marker is followed in the generations. Recombination frequencies can be estimated and the distance between the mark

8、er and the gene of interest can be determined Marker 1 Marker 2 1 cM 5 cM 1 cM 5 cM Gene Marker 1 Gene Marker Gene Marker 1 Gene Marker 2 2 | _!_ ! ! | _!_ ! ! Theres only 1% chance that the linkage is broken, with marker 1 . DNA sequencing- Is the determination of the precise sequence of nucleotide

9、s in a sample of DNA. The most popular method for doing this is called the dideoxy method (chain termination method) DNA is synthesized from four deoxynucleotide triphosphates (dNTP) .Each new nucleotide is added to the 3 OH group of the last nucleotide added in the presence of DNA polymerase. The D

10、NA to be sequenced is prepared as a single strand. This template DNA is supplied with a primer having complementary sequence. DNA sample and the primer is separated into 4 tubes and a mixture of all 4 (deoxy) nucleotides which are radiolabelled are added in ample quantities to all 4 tubes plus DNA p

11、olymerase I. One dideoxynucleotide is then added in a very small quantity to each tube. Because all four normal nucleotides are present, chain elongation proceeds normally until, by chance, DNA polymerase inserts a dideoxy nucleotide At the end of the incubation period, the fragments are separated b

12、y length from longest to shortest. The resolution is so good that a difference of one nucleotide is enough to separate that strand from the next shorter and next longer strand. The fragments from each reaction tube are separated in four adjacent lanes by gel electrophoresis (Fig 16-13) 1) You recove

13、r a cloned DNA segment of interest, and determine that the insert is 1300 bp in length. To characterize this cloned segment you isolate the insert and decide to construct a restriction map. Digesting with enzymes EcoRI and BamHI and following electrophoresis, you determine the number and size of the

14、 fragments produced by EcoRI and BamHI alone and in combination as follows. Enzyme Res. Fragment sizes EcoRI 350bp, 950 BamHI 200bp, 1100bp EcoRI+ BamHI 150bp, 200bp 950bp 350 EcoRI 950 200 150 BamHI 2)An inbred strain of plants has a mean height of 24 cm. A second strain of the same species from a

15、different country also has a mean height of 24 cm. The F1 plants from a cross between these two strains are also 24 cm high. However, the F2 generation shows a wide range of heights; the majority are like the P1 and F1 plants, but approximately 4 of 1000 are only 12 cm high, and 4 of 1000 are 36 cm

16、high. (i)What mode of inheritance is occurring here? (ii)How many gene pairs are involved? (iii)How much does each gene contribute to the plant height? (iv)Indicate one possible set of genotypes of the P1 and F1 plants that could explain their heights. (v)Indicate one possible set of genotypes to ac

17、count for F2 plants that are 18 cm or 33 cm high. (i)What mode of inheritance is occurring here? Polygenic inheritance where a continuous trait is involved and alleles contribute additively to the phenotype is demonstrated (ii)How many gene pairs are involved? 4/1000 = 1/250 = 1/4n ; n=4 (iii)How mu

18、ch does each gene contribute to the plant height? 36-12 = 24/8 = 3 (iv)Indicate one possible set of genotypes of the P1 and F1 plants that could explain their heights. If the shortest is aabbccdd, is the base height then 24-12 = 12 ; 4 additive alleles should be in both P1 and F1. AABBccdd x aabbCCD

19、D AaBbCcDd (v)Indicate one possible set of genotypes to account for F2 plants that are 18 cm or 33 cm high. For plant with 18 ; 18-12 = 6 only two additive alleles AaBbccdd, etc For plant with 33 ; 33 12 = 21 only 7 additive alleles AABBCCDd etc 3) Trihybrid eggplants with hairy leaves, purple flowe

20、rs, and thorny stalks are testcrossed with plants having hairless leaves, white flowers, and smooth stalks. The progeny are: Phenotype Number hairy, purple, thorny 10 hairy, purple, smooth 132 hairy, white, thorny 81 hairy, white, smooth 998 hairless, purple, thorny 1020 hairless, purple, smooth 101

21、 hairless, white, thorny 150 hairless, white, smooth 8 Derive a genetic map of the three loci. What is the coefficient of coincidence? hairy, purple, thorny 10 hairy, purple, smooth 132 hairy, white, thorny 81 hairy, white, smooth 998 + w s w + s w s + hairless, purple, thorny 1020 h + + + h + + + h

22、 hairless, purple, smooth 101 incor corr incor hairless, white, thorny 150 hairless, white, smooth 8 total 2500 Bet. H and w = 150 + 132 + 10 + 8/2500 = 0.12 = 12mu Bet H and s = 81 + 101 + 10 + 8 /2500 = 0.08 = 8mu Observed dco = 18/2500 = 0.007 Exp dco = 0.12 x 0.08 = 0.009 c = 0.007/0.009 = 0.77

23、Squiggly-eyed male flies were crossed with normal females, and the resulting F1 progeny were: 122 normal males 131 squiggly-eyed females What is the mode of inheritance for squiggly eyes? What phenotype ratio do you expect for F2 (if you cross the F1 females with the F1 males)? The mode of inheritance for squiggly eyes are sex linked dominant Squiggly-eyed male flies were crossed with normal females 122 normal males 131 squiggly-eyed females ss x SY Ss sY F1 Ss x sY Ss ss