**PUNNETT SQUARE PROBLEMS (or how to do a cross) STEP 1**

Show transcribed image text How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCC? a. 4 b. 6 c. 8 d. 10 e. 12 A sexually reproducing animal has two unlinked genes, one for head shape (H) and one for tail length(T). Its genotype is HtTt. Which of the following genotypes is possible in a gamete from this organism? a. …... A faster way to do it is by (2^n) where 2 is the number of alleles and n is the number of genes, so we have 2^3 = 8. so we have 8 possible combinations. Assuming you are asking how many different types of gametes can an individual with the genotype AaBbCc make.

**How many gametes Aa Bb Cc science.answers.com**

As you calculate the genotype you take 2 and raise it to the number of heterozygous gene pairs, which in our case is 3; so it would read, 2^3=8 different gametes. If we took the phenotype you... 21/03/2010 · the 2^n rule is applicable for finding the number of distinct genotypes of sperm or egg. The reasoning is that by independent assortment, a true heterozygote AaBbCc will have: (2 possible genotypes for A) x (2 possible genotypes for B) x (2 possible genotypes for C). You should exclude homozygous genes from the n count, as these only result in one possible genotype. If you want to …

**How many genetically different types of gametes can you**

The number of gametes formed is decided by the number of heterozygous alleles present in the given genotype. 2^n is the formula used to find it out, where n=number of heterozygous alleles present in the genotype. Say for example, in the above genotype Aa & Bb are the 2 heterozygous alleles, so here n=2. Putting the values in the formula , we get 2^2=4. Hence 4 types of gametes are formed.... The number of individuals scored with respect to each biochemcal phenotype are given above the band. The allelic frequency of each band can be determined in a similar manner as in t he previous examples. p is the frequence of the 6.5 kb fragment in the population. Its value is

**how to determine the possible gametes for a genotyope**

There are two options for each allele so 2^n = gamete types, i.e. two alleles raised to the number of genes in the cross tetrahybrid has 2^4 = 16 gametes Next 2^2n = offspring, i.e.... Once all six recombinant gametes have been classified, the total number of crossovers between the three loci can be added up and crossover percentage determined between each pair of loci. This number will reflect gene distance but one more step is needed to complete the process.

## How To Find The Number Of Gametes Through Genotype

### Gamete Wikipedia

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## How To Find The Number Of Gametes Through Genotype

### Thus, instead of combination of your original 46 chromosomes, you can get countless recombinations of your alleles to form gametes. It would be nearly impossible to calculate the total possible number …

- A dihybrid that is AB/ab can produce four possible “output” gametes through meiosis. Output gametes that match one of the input gametes are nonrecombinant (AB or ab). Output gametes with new combinations of alleles are recombinant (Ab or aB).
- Review of Population Genetics Equations 1. Hardy-Weinberg Equation: p2 always a number between 0 and 1. Adding ws to the Hardy-Weinberg equation allows you to predict the effect of selection on gene and allele frequencies in the next generation. Take the Hardy-Weinberg equation and multiply each term (the frequency of each genotype) by the fitness of that genotype. Add those up and you …
- We expect two classes for the gametes 1.Parental class (genotype of parent cells) 2.Non-parental (recombinant) genotype DOF = classes (parental & recombinant) -1 For linkage analysis, the DOF will ALWAYS be 1. 13 Applying the chi square test to a linkage study Total 50 100 a B 11 22 A b 8 16 a b 14 28 A B 17 34 Genotype Experiment 1 Experiment 2 Class Observed/Expected Observed/Expected
- A faster way to do it is by (2^n) where 2 is the number of alleles and n is the number of genes, so we have 2^3 = 8. so we have 8 possible combinations. Assuming you are asking how many different types of gametes can an individual with the genotype AaBbCc make.

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