# Punnett Square: Monohybird Crosses

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1 Punnett Squares A Punnett square is a mathematical device used by geneticists to show combinations of gametes and to predict offspring ratios. There are a few fundamental concepts of Punnett squares that must be understood for them to predict correctly. Each trait is controlled by two genes. Due to the binary nature, genes can be represented by letters. An uppercase letter denotes the wild type (usual, dominant) form of the gene. A lowercase letter denotes the mutant (unusual, recessive) form of the gene. Punnett Square: Monohybird Crosses In Monohybird crosses (Punnett square configurations) there are three distinct combinations: e.g. BB, Bb, and bb. If a dominant gene is paired with a recessive gene (Bb), the dominant's trait will always be expressed phenotypic. Black fur in guinea pigs (B) is dominant over white fur (b). If a heterozygous female is mated with a homozygous recessive male what could be the genotype and phenotype of the F 1 generation. Step 1: Determine the Parental Genotypes

2 To set up a Punnett square, draw a large square, and then divide it into 4 equal sections (also squares). It should look something like this: Now you need two parents to mate, ones with a known genotype. Place one of the parents on top, and one on the left. You should get a something similar to this: Finally, take each letter in each column and combine it with each letter from each row in the corresponding square. You should now have a picture close to this: The Genotypical ratios of this cross are: l Bb: l bb. From this it is possible to determine the phenotypically probability: Dihybrid Crosses 50% (0.5) of the guinea pigs will have black fur 50% (0.5) will have white fur. In given context, a Punnett square with two different sets of characteristics being analyzed, they may be on a autosomal chromosome or Sex linked, e.g. X chromosome, Y chromosome, H gene, and h gene. To make a dihybrid Punnett square, all that is necessary is the analyzing of two different sets of genes at the same time First let us look at autosomal traits. Short in a rose plant is dominant and tall is recessive. Also, broad leaves are dominate over narrow leaves. What is the F 1 generation if a rose plant

3 heterozygous for both traits were crossed with a plant that is heterozygous for the short trait and narrow-leafed traits. Step 1: Define the symbols: S - short s- tall B - broad leaves b - narrow leaves Step 2: Determine the 2 genotypes of the parents SsBb X Ssbb Step 3: Determine the gamete for each parent and draw the Punnett Square Step 4: Determine your 2 genotypes and phenotypes from your Punnett square F1's genotypes are: SSBb, 2SsBb, SSbb, 2Ssbb, ssbb, ssbb F1's phenotypes are: Sex Linked Traits 3 short and broad-leafed plants (SSBb, SsBb) 3 short and narrow leafed plants (SSbb,Ssbb) 1 tall and broad leafed plant (ssbb) 1 tall and narrow leafed plant (ssbb) Sex-linked traits are traits that are carried on the sex chromosomes, either X or Y. Hemophilia is a recessive, sex-linked disorder on the X-chromosome. Therefore, a male with a recessive allele on the X-chromosome will always express it phenotypically. In order for a female to express a recessive trait, it must be in a homozygous state. Mate a carrier female (a female with the mutant gene, but who is unafflicted because she has the normal gene on her other X-chromosome) with a normal male. Step 1: Define the symbols: X H - normal X h - hemophilia Y 0 - does not have the allele

4 Step 2: Determine the genotypes of the parents: X H X h X X H Y 0 Step 3: Draw your Punnett square. The Punnett square should look like this: Genotype X H X H X H X h X H Y 0 F1 Generation Phenotype Normal female Carrier female Normal male X h Y 0 Hemophiliac male Due to the male total dependence on their X chromosome, males have a much greater chance of getting hemophilia than females. For a female to get hemophilia, she has to have either a hemophiliac father, carrier mother or bad luck, or hemophiliac mother and hemophiliac father. Co-dominant Traits Co-dominance is an interesting case where there is no real dominant or recessive allele. Instead, both alleles are fully expressed in the heterozygous form (the homozygous forms act normally). A good example of co-dominance is a person's blood type. A person with blood type AB is showing the results of having both the I A and I B co-dominant genes. The AB blood type expresses the characteristics of both blood types A and B; therefore, the alleles for blood type must be co-dominant. However, the allele O is recessive to both A and B. Therefore, the genotypes and phenotype of the human blood type are:

5 What would the genotypes and the phenotypes of the F1 generation when a Type A blood male is mated with a Type B blood type? Their first born child is Type 0. Answer: Type A person's genotypes may be I A I A or I A i. Type B person's genotypes may be I B I B or I B i. But since they have a child that is Type 0, whose genotype is ii (recessive), both parents have to have had the recessive gene (i) to pass on to the offspring. Hence, the Type A genotype has to be I A i, and the Type B has to be I B i. I A i X I B i The Fl generation: Genetopyes: I A I B, I B i, I A i, ii Phenotypes: Type AB, Type B, Type A, and Type O

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