Genetically speaking, there are roughly 8 well known 'Brown' mice. They all have different genes, which creates different shades. You should probably find out what shade of brown this 'Brown' mouse in question is.
There is only one type of Black genetic colouring in mice.
Generally speaking, if we were just using the common Agouti brown mouse (wild colour) & crossed it to Black, then the entire litter should've resulted in Agouti mice. However, Agouti being represented by a dominant A-, & the Black being represented by a recessive AA, crossing A- to AA would result in an entire litter of Aa = Agouti carring Black.
If you crossed one of those offspring back to the Black parent, you would get Aa & AA in the litter, resulting in the litter being 50% black, 50% Agouti.
However, if the 'Brown' mouse wasn't Agouti, & was actually another brown called Chocolate (I'm serious), then crossing Black (which in this case will be BB) to Chocolate (bb) would result in an entire litter of Blacks carrying Chocolate (Bb) . If you crossed 2 of the offspring together, you would get a mix of Blacks & Chocolates.
NB: Black is a dominant gene to all other genes but Agouti.
If two white sheep produce a black offspring, the parent's genotypes for color must be heterozygous, meaning both parents carry one allele for white (let's denote it as "W") and one allele for black (denote it as "B"). This results in the genotype "WB" for each parent, allowing for the possibility of producing a "BB" genotype (black) offspring when both parents contribute the black allele. Therefore, the parents are most likely "WB" (white) and "WB" (white).
No, black widows need to mate in order to produce fertilized eggs that will develop into viable offspring. Without mating, the eggs will not be fertilized and will not hatch into spiderlings.
The "B" allele codes for a black coat color, so offspring with a BB or Bb genotype will both have a black coat due to the dominant nature of the "B" allele. The presence of at least one "B" allele ensures that the offspring will express the black coat color phenotype.
The probable phenotype of the offspring would be black, as black color is dominant over chestnut color. The genotype of the offspring would be heterozygous for black (Bb), since one parent is homozygous black (BB) and the other is homozygous chestnut (bb).
Black stools can indicate the presence of blood in the stool, which could be a sign of a bleeding ulcer or other gastrointestinal issues. It's important to consult a healthcare provider if you have black stools as it may require further investigation.
Yes
No, rattlesnakes and rat snakes cannot interbreed.
Incomplete dominance occurs when the offspring's phenotype is a blend of the parents' traits, such as when a red flower and a white flower produce pink offspring. Codominance, on the other hand, results in both parental traits being expressed equally in the offspring, like when a black chicken and a white chicken produce offspring with both black and white feathers.
Yes, it is possible for a cross between two black guinea pigs to produce offspring that are not white. If both black guinea pigs carry recessive genes for coat color, such as the gene for white fur, there is a chance their offspring could inherit those genes and be white. However, if both parents are homozygous for the black coat color, all offspring will be black. Thus, the outcome depends on the specific genetic makeup of the parent guinea pigs.
If two white sheep produce a black offspring, the parent's genotypes for color must be heterozygous, meaning both parents carry one allele for white (let's denote it as "W") and one allele for black (denote it as "B"). This results in the genotype "WB" for each parent, allowing for the possibility of producing a "BB" genotype (black) offspring when both parents contribute the black allele. Therefore, the parents are most likely "WB" (white) and "WB" (white).
If you crossed a black hen/roo with a white hen/roo, the offspring would NOT be blue. If he was recessive white, the chicks would be black. If he was dominant white, the chicks would be white with some black "smudges" or "spots". The only way to produce 100% blue chicks is to cross a black roo over splash hens or a splash rooster over black hens.
No, not always. The offspring can be white, black, black and white or shades of both.
This is an example of incomplete dominance.
none black bears have no offspring in one year
No, black widows need to mate in order to produce fertilized eggs that will develop into viable offspring. Without mating, the eggs will not be fertilized and will not hatch into spiderlings.
the average offspring of a bear is 2. But the offspring depends on the species.
To answer this, first determine the genotypes of the parents. The female is homozygous dominant, which means it carries two copies of the black allele (which we will designate B), so its genotype we can represent as BB. The male is homozygous for the recessive trait (white, designated as b), so its genotype can be represented as bb. So the cross looks like this: BB X bb Next, we must determine the types of gametes each parent can produce. In this case it's easy, because both are homozygous. The female can only produce B gametes, while the male can only produce b gametes. Since the offspring carry one allele from each parent, all of the offspring can have only one genotype: Bb. Since black (B) is the dominant allele, and every offspring carries the dominant allele, all eight of the offspring from this cross will be black.