The offspring from an FF X ff cross will all have the genotype Ff.
This is because they receive one allele from each parent, and in this case they can only receive an F from the FF parent and an f from the ff parent.
Wiki User
∙ 11y agoWiki User
∙ 9y agoIf you breed Ff and Ff together you will receive some of every phenotype. You would receive 50 percent Ff, 25 percent FF, and 25 percent ff.
Emeleke Bezezuh
Alright, I suppose I will do your homework for you.. Here is your punnet square: F F F FF FF f Ff ff Therefore, 3/4, or 75%, offspring will have the phenotype of having freckles, and 1/4, or 25% will have the phenotype of no freckles. And 2/4, or 50%, of the offspring will have the genotype for homozygous for freckles, 1/4, or 25%, of the offspring will carry a heterozygous trait for freckles, and 1/4, or 25%, of the offspring will have the phenotype for homozygous no freckles.
There different genotypes and two different colors Black fur is dominant --> F White fur is recessive --> f The parents are bot Ff (heterozygotes, and because black fur is dominant they have a black fur). If they mate, you get parents: Ff x Ff Offspring: FF Ff Ff ff so 25% will be homozygous for Black fur 2x25=50% will be heterozygous, and have a Black fur and 25% wil be homozygous for White fur. Hence, of their offspring, 75% will have a black fur and 25% will have a white fur
if you are doing 7th grade life science, it would be B)all dominant because it has FF, FF, Ff, and Ff. All of these has dominant F, so it is all dominant.
Punnett Squares are useful because they allow you to see the chance of what genes will be handed down to the offspring. They are a quick and easy way to determine the chance of a offspring having a certain characteristic
If all of the children have freckles, that means that both parents had dominant genotypes. (Mother; FF and Father; FF). Or, one parent could have a hybrid genotype. (For example, Mother; Ff and Father; FF). Based on the outcome of a Punnett Square, either one parent must have a hybrid and the other dominant, or both must have dominant genotypes.
all dominant
Alright, I suppose I will do your homework for you.. Here is your punnet square: F F F FF FF f Ff ff Therefore, 3/4, or 75%, offspring will have the phenotype of having freckles, and 1/4, or 25% will have the phenotype of no freckles. And 2/4, or 50%, of the offspring will have the genotype for homozygous for freckles, 1/4, or 25%, of the offspring will carry a heterozygous trait for freckles, and 1/4, or 25%, of the offspring will have the phenotype for homozygous no freckles.
There different genotypes and two different colors Black fur is dominant --> F White fur is recessive --> f The parents are bot Ff (heterozygotes, and because black fur is dominant they have a black fur). If they mate, you get parents: Ff x Ff Offspring: FF Ff Ff ff so 25% will be homozygous for Black fur 2x25=50% will be heterozygous, and have a Black fur and 25% wil be homozygous for White fur. Hence, of their offspring, 75% will have a black fur and 25% will have a white fur
if you are doing 7th grade life science, it would be B)all dominant because it has FF, FF, Ff, and Ff. All of these has dominant F, so it is all dominant.
Punnett Squares are useful because they allow you to see the chance of what genes will be handed down to the offspring. They are a quick and easy way to determine the chance of a offspring having a certain characteristic
FF-FF-FF-FF-FF-FF
If all of the children have freckles, that means that both parents had dominant genotypes. (Mother; FF and Father; FF). Or, one parent could have a hybrid genotype. (For example, Mother; Ff and Father; FF). Based on the outcome of a Punnett Square, either one parent must have a hybrid and the other dominant, or both must have dominant genotypes.
Apparently... It doesn't spell check when you type it. Ff, ff, ff. :D
ffffff
That's called the "MAC address" or "Hardware address". It's a unique set of numbers assigned to a network device. Anything that access the internet whether phones, wireless cards, computers, etc, all have a MAC address. The address with all one's (FF-FF-FF-FF-FF-FF) is the broadcast address. All devices on the physical segment will see and respond to a message with that destination.
ff ff ff
No. Cystic fibrosis is an autosomal recessive disease. Each parent would have to be a carrier of a CF mutation and would be Cc.