Deviation of Mendel’s second law “Dihybrid” Part 1 

The shape is controlled by genes that are located in two loci on different chromosomes. As result of interaction of these genes, the chickens can have four different phenotypes. Their comb shape can be Pea, Rose, Walnut and Simple. Wyandotte have Rose comb. Brahmas have Pea comb. Leghorns have Simple comb. When crossed the pure lines of Wyandotte and Brahmas, the all offspring in the first generation had the Walnut comb. That is, the offspring had an absolutely new phenotype not like their parents. From crosses of chickens with Walnut comb in the second generation was obtained the offspring with the phenotypic ratio: 9 Walnut: 3 Rose: 3 Pea: 1 Single. Thus we have dealing with the case, when each dominant gene has an independent phenotypic expression, combination of two these genes in genotype leads to the development of a new phenotypic trait, and the lack of dominant genes gives another phenotypic trait. Let's mark the genes with letters "P" and "R". If there are dominant allele "P" in genotype and dominant allele "R" is lack, then the chickens have the Pea comb. And if there are dominant allele "R" in genotype and dominant allele "P" is lack, then the chickens have the Rose comb. Chickens with presence of both dominant alleles "P" and "R" in the
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genotype would have the Walnut comb, and double recessive homozygotes, i.e. with genotype "pprr" would have the Simple comb.
P:  Rose comb

x Pea
comb
phenotype
 RRpp  rrPP genotype
F1:   Walnut comb

 Phenotype F1
  RrPr  Genotype F1
F2:
     R-P- R-pp rrP- rrpp

Write all of genotype that produced from mating between “Rose comb x Pea comb”:

 RP Rp rP rp RP     RP     RP     RP    
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2. Linkage and Crossing-Over
According to Mendel’s principle of independent assortment, a dihybrid cross with unlinked markers ought to produce a 9:3:3:1 ratio. If a significant deviation from this ratio occurs, it may be evidence that for linkage, that is, that the loci are located close to each other on the same chromosome pair. During meiosis, a pair of synapsed chromosomes is made up of four chromatids, called a tetrad. The phenomenon of a cross - over occurs when homologous chromatids in the tetrad(one from each of the two parents) exchange segments of varying length during prophase. The point of crossover is known as a chiasma (pl. chiasmata). A tetrad typically has at least one chiasma along its length. Generally, the longer the chromosome, the greater the number of chiasmata. There are two theories on the physical nature of the process. The classical theory proposes that cross-over and formation of the chiasma occur first, followed by breakage and reunion with the reciprocal homologues. According to this theory, chiasma formation need not be accompanied by chromosome breakage. Alternatively, according to the chiasma type theory, breakage occurs first, and the broken strands then reunite. Chiasmata are thus evidence, but not the causes, of a cross-overs. Recent molecular evidence favors the latter theory, although neither is a completely satisfactory explanation of all of the evidence.

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Compare between Cic & Trans

Linkage: T.H. Morgan performed a series of experiments with Drosophila (1910-1914). He showed that two genes were found in coupling phase (Cis) or repulsion phase (Trans), because these were present on the same chromosome pair. Such genes are called linked genes and the phenomenon was called linkage.  Coupling phase (Cis) Repulsion phase (Trans)


Gametes of Coupling (Cis) Gametes of Repulsion (Trans)
 

The genes on a given chromosome cannot assort independently and are called linked genes. The state of many genes being present on the same chromosome is called Linkage.
Also by looking at the gametes that are most abundant you will be able to determine if the original cross was a coupling or repulsion phase cross. For a coupling phase cross, the most prevalent gametes will be those with two
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dominant alleles or those with two recessive alleles. For repulsion phase crosses, gametes containing one dominant and one recessive allele will be most abundant. Understanding this fact will be important when you actually calculate a linkage distance estimate from your data.

Crossing Over: Recombination results from crossing over. The process of crossing over occurs by a mechanism called breakage and reunion theory and comprises four steps: synapsis, tetrad formation, crossing over and disjunction.

During prophase of meiosis I chromosomes physically exchange their segments. At zygotene homologous chromosomes are brought together and they are synapsed (side to side pairing). Each homologue has two chromatids at this stage. At pachytene chromatids exchange segments, resulting crossover types.  Finally, for two genes are right next to each other on the chromosome crossing over will be a very rare event.  Two types of gametes are possible when following genes on the same chromosomes. If crossing over does not occur, the products are ParentalGametes. If crossing over occurs, the products are Recombinant Gametes. The allelic composition of parental and recombinant gametes depends upon whether the original cross involved genes in coupling or
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repulsion phase. The figure above depicts the gamete composition for linked genes from coupling and repulsion crosses.
It is usually a simple matter to determine which of the gametes recombinant. These are the gametes that are found in the lowest frequency. This is the direct result of the reduced recombination that occurs between two genes that are located close to each other on the same chromosome. Also by looking at the gametes that are most abundant you will be able to determine if the original cross was a coupling or repulsion phase cross. For a coupling phase cross, the most prevalent gametes will be those with two dominant alleles or those with two recessive alleles. For repulsion phase crosses, gametes containing one dominant and one recessive allele will be most abundant. Understanding this fact will be important when you actually calculate a linkage distance estimate from your data.
1. test cross for members of the first generation with the father “recessive”  The important question is how many recombinant chromosomes will be produced. If the genes are far apart on the chromosome a cross over will occur every time that pairing occurs and an equal number of parental and recombinant chromosomes will be produced. Test cross data will then generate a 1:1:1:1 ratio. But as two genes are closer and closer on the chromosome, fewer cross over
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events will occur between them and thus fewer recombinant chromosomes will be derived. We then see a deviation from the expected 1:1:1:1 ratio.
1. Calculation of Recombinants gametes of “Test Cross”: Recombinants frequency ratio= (Number of recombinant progeny / total number of progeny) x 100
2. Calculating the degree of correlation of “The Second Generation data”:

Linkage degree = (2 x (A number of structures parental individuals)-1
 total number of progeny F2



References:
http://www.slideshare.net/zohaibkhan404/epistatic-gene-interaction-by-biotechnology-ciit-abbottabad https://tpsbiology11student2.wikispaces.com/Unit+3+-+Genetic+Processes







 


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1. Choose the right answer: a. In chickens the genotype rr ppproduces single combs; R_ P_, walnut comb;rr P_,pea comb; and,R_ pp, rose comb. Ifrr PP(pea) are crossed to RR pp(rose). Whatwould be the progeny’s comb type?  Rose  Single  Walnut  Pea  More than one of the above is true. 2. From the following, determine the genotype and phenotype of the offspring for each cross: •  (a) RR PP X rr pp   •  (b) Rr pp X Rr pp   •  (c) Rr Pp X rr pp  •  (d) Rr pp X rr Pp  •  (e) Rr Pp X Rr Pp   •  (f) rr pp X rr Pp  3. A female animal with genotypeA/a.B/b is crossed with a double-recessive male a/a.b/b. Their progeny include 442 A/a.B/b, 458 a/a.b/b, 46 A/a.b/b, and 54 a/a.B/b. Explain these proportions and draw the chromosomes of the dihybrid parent showing the positions of the genes and alleles.Determine the recombinent gametes and explain you data with chi square?



Exercise 10

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4. In Drosophila, black body (b) and purple eyes (pr) are recessive mutations located on chromosome II. A true-breeding wild type female is mated with a true-breeding black body, purple eye male. All the offspring are wild type. One of these offspring is then test-crossed with a black body, purple eye fly. The results of this test cross indicate that there was a recombination rate of 0.06.
(a) Which four gametes were produced by the wild type fly used in the test cross?
(b) What proportions of the four gametes would you expect from that parent? Keep in mind that they will not be in a 1:1:1:1 ratio because the genes are

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